1002 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

Open questions

OQ.4501. Let Hn = 1 + 1/2 + · · ·+ 1/n be the nth harmonic number.

(a) Calculate∞∑n=1

(−1)n(Hnn

)2.

(b) Let k ≥ 0 be an integer. Calculate∞∑n=1

(−1)nHnn · Hn+k

n+k .

Ovidiu Furdui

OQ.4502. Calculate the series∞∑n=1

1n

(1

n+1 − 1n+2 + 1

n+3 · · ·)2

and

∞∑n=1

(−1)n

n

(1

n+1 − 1n+2 + 1

n+3 · · ·)2

.

Ovidiu Furdui

OQ.4503. Let n ≥ 2 be an integer.Calculate In =

∫∞0

(arctanx

x

)ndx.

Remark. One can verify that I2 = π ln 2 and I3 =3π ln 2

2 − π3

16 , so theopen problem would be to calculate, if possible, in terms of well-knownconstants the value of In when n ≥ 4.

Ovidiu Furdui

OQ.4504. Calculate limn→∞

(2√2+ 4√4+···+ 2n√2n

1+ 3√3+···+ 2n−1√2n−1

)n.

Ovidiu Furdui

OQ.4505. Let k ≥ 3 be an integer. Calculate∞∑n=1

(ζ(k)− 1− 1

2k− · · ·− 1

nk

)2, where ζ denotes the Riemann zeta function.

Remark. The problem is motivated by problem 3.22 in [1] which states

that∞∑n=1

(ζ(2)− 1− 1

22− · · ·− 1

n2

)2= 3ζ(3)− 5

2ζ(4).

REFERENCE

[1] Ovidiu Furdui, Limits, Series and Fractional Part Integrals, Problems inMathematical Analysis, Springer, London, 2013

Ovidiu Furdui

Open questions 1003

OQ.4506. Calculate the series∞∑n=1

(−1)n(ζ(2)− 1− 1

22− · · ·− 1

n2

)2and

∞∑n=1

(−1)nn(ζ(2)− 1− 1

22− · · ·− 1

n2

)2, where ζ denotes the Riemann zeta

function.

Ovidiu Furdui

OQ.4507. Let k ≥ 4 be an integer and let i = j. Calculate∞∑

n1,n2,...,nk=1

ninj

(n1+n2+···+nk)!and

∞∑n1,n2,...,nk=1

n2i

(n1+n2+···+nk)!.

Remark. The cases when k = 2, 3 are recorded as problems 3.113 and 3.117in [1]. For another related open problem the reader is referred to [1, Problem3.137, p. 163]F.

REFERENCE

[1] Ovidiu Furdui, Limits, Series and Fractional Part Integrals , Problems inMathematical Analysis, Springer, London, 2013

Ovidiu Furdui

OQ.4508. Let k ≥ 3 be an integer. Calculate∫ 10 ln2

(k√1 + x− k

√1− x

)dx

and∫ 10 ln2

(k√1 + x+ k

√1− x

)dx.

Remark. We mention that, when k = 2 both integrals can be expressed asrational linear combination of 1, ln 2, ln2 2, π and Catalan’s constant.

Ovidiu Furdui

OQ.4509. Calculate∫ 10

∫ 10

∫ 10

∫ 10

{x+yz+t

}dxdydzdt and

∫ 10

∫ 10

∫ 10

∫ 10

{xyzt

}dxdydzdt, where {a} = a− ⌊a⌋ denotes the fractional part

of a.

Ovidiu Furdui

OQ.4510. Find all primes pk (k = 1, 2, ..., 2n+ 1) such that2n+1∑k=1

p2nk is the

product of n consecutive integers.

Mihaly Bencze

1004 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

OQ.4511. Find all positive integers n and r for whichr∑

k=1

((n− k)! + (n+ k)!) is a perfect square.

Mihaly Bencze

OQ.4512. Solve in N the equationn∏

k=1

(m+ k) + 1 = pr.

Mihaly Bencze

OQ.4513. Denote d (n) the number of divisors of n. Compute∑1≤i<j≤n

1d(i+n2)d(j+n2)

.

Mihaly Bencze

OQ.4514. Let a1, a2, ..., an be the side of a convex polygon. Prove thatn∑

k=1

ak ≤ ∑cyclic

n−1

√a1a2...an

a1+a2+...+an−1−an.

Mihaly Bencze

OQ.4515. If xk > 0 (k = 1, 2, ..., n), then

(∑

cyclic

(x1x2...xn−1)n−1

)n∑

k=1

1

xn2−2n+2k

≥(

n∑k=1

x2k

)(n∑

k=1

1xk

).

Mihaly Bencze

OQ.4516. Determine all prime p such that for all positive integer m there

are integers ak ∈ N (k = 1, 2, ..., n) such thatn∑

k=1

anr

k ≡ m (mod p) where

r ∈ N.

Mihaly Bencze

OQ.4517. Denote Fk the kth Fibonacci number. Determine all m,n ∈ N forwhich Fpm − Fpn is divisible by p where p ≥ 3 is a prime.

Mihaly Bencze

OQ.4518. There exists n composite such that Ln − 1 is divisible by n, thisnumbers are called Lucas pseudoprimes. The first few705, 2465, 2737, 3745, 4181, 5777, 6721.... Denote LP the set of Lucaspseudoprimes. Compute:

Open questions 1005

1).∑

n∈LP1n2 2).

∑n∈LP

∑1≤i<j≤n

1ij

Mihaly Bencze

OQ.4519. Given a positive integer n, let f (n) be the k power of thenumber of its digits. For example f (2) = 1, f (123) = 3k. ComputeS (n, k) =

∑1≤ii<...<ir≤n

1f(i1)f(i2)...f(ir)

.

Mihaly Bencze

OQ.4520. Find all continuous functions f : R → R such that for allx1, x2, ..., xn ∈ R holdsn∑

i=1f (xi)−

∑1≤i<j≤n

f (xixj) + ...+ (−1)n−1 f (x1x2...xn) = 0.

Mihaly Bencze

OQ.4521. Let p be a prime. Determine all k ∈ N and all prime q for which

the polynomial P (x) =n∏

i=1

(xp + ik

)+ q is irreductible in Z [x] .

Mihaly Bencze

OQ.4522. Determine all ak ∈ N (k = 1, 2, ..., n) , where are in arithmeticalprogression and for which 1 + ai1ai2 ...ain−1 (i1, i2, ..., in−1 ∈ {1, 2, ..., n} ,i1 = i2... = in−1) are all perfect squares.

Mihaly Bencze

OQ.4523. Prove that∞∑k=0

1(nk+1)(nk+2)...(nk+n) =

1(n−1)!

1∫0

(1−x)n−1

1−xn dx.

Compute Sp =∞∑k=0

1(nk+1p)(nk+2p)...(nk+np) .

Mihaly Bencze

OQ.4524. Denote Mk the set of prime numbers p for which exist a, b ∈ Nsuh that p = ak + bk. Compute

1).∑

p∈Mk

1p2

2).∑

p,q∈Mk

1pq

Mihaly Bencze

1006 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

OQ.4525. Denote r (n) the number obtained by reversing the digits of n.Determine all n ∈ N for which rk (n)− nk is a perfect k power of a positiveinteger.

Mihaly Bencze

OQ.4526. Prove that the squars of side ak (k = 1, 2, ..., n) cannot fit inside

a square of side less thann∑

k=1

ak without overlapping.

Mihaly Bencze

OQ.4527. We define magic numbers as follows:

1). all numbers from 0 to 9 are magic2). a number greater than 9 is magic if it is divisible by the number of itsdigits and the number obtained by deleting its final digit is also magic.

Prove that are infinitely magic numbers. If M denote the set of magicnumbers, then compute

∑i1,i2,...,ik∈M

1

(1+i21)(1+i22)...(1+i2k).

Mihaly Bencze

OQ.4528. Let (xk)k≥1 be the positive roots of the equation tgx = x

1). Compute Bp =∞∑k=1

1xpk

2). Compute Cp =∞∑k=1

∑1≤i1<...<ip≤k

1xi1

xi2...xip

3). Compute∞∑p=1

Bp 4). Compute∞∑p=1

Cp

Mihaly Bencze

OQ.4529. Let A1A2...A2n+1 be a regular polygon and B1B2...B2n+1 be the

polygon formed by its diagonals. Compute Area[B1B2...B2n+1]Area[A1A2...A2n+1]

.

Mihaly Bencze

OQ.4530. Let ABC be a triangle with side-lengths a, b, c that satisfyaα + bα = cα, when α ∈ Q and α > 0. Determine all u (α) , (vα) > 0 forwhich πu (α) ≤ c ≤ πv (α) .If α = 2, then C = π

2 and u (2) = v (2) = 12 .

Open questions 1007

If α = 32 then u

(32

)= 1

2 and v(32

)= 3

5 .

Mihaly Bencze

OQ.4531. Solve in Z the equation x2 + y2 + z2 + 6 (ab+ bc+ ca) = 2t2. Ifx = (a1 − b1)u, y = (b1 − c1)u, z = (c1 − a1)u, a = a1u, b = b1u, c = c1u,t = (a1 + b1 + c1)u, where a1, b1, c1, u ∈ Z, then results that the givenequation have infinitely many solutions in Z.Solve the given equation in Q.

Mihaly Bencze

OQ.4532. Solve in Z the equationx4 + y4 + z4 + 12

(a2 + b2 + c2

)(ab+ bc+ ca) + 6 (ab+ bc+ ca)2 = 2t4.

If x = (a1 − b1)u, y = (b1 − c1)u, z = (c1 − a1)u, a = a1u, b = b1u, c = c1u,t = (a1 + b1 + c1)u, when a1, b1, c1, u ∈ Z, then results that the givenequation have infinitely many solutions in Z.Solve in Q the given equation.

Mihaly Bencze

OQ.4533. Compute∞∑k=2

1k2+{logk k!} , where {·} denote the fractional part.

Mihaly Bencze

OQ.4534. Compute∞∑k=1

1k2+{ek} , when {·} denote the fractional part.

Mihaly Bencze

OQ.4535. Compute∞∑k=1

1k2+{πk} , when {·} denote the fractional part.

Mihaly Bencze

OQ.4536. Compute∑

1≤i<,j≤n

1(i+{πi})(j+{ej}) , when {·} denote the fractional

part.

Mihaly Bencze

1008 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

OQ.4537. If ak > 0 (k = 1, 2, ..., n) then

nn∑

k=1a3k

(n−1)

(n∑

k=1ak

)(n∑

k=1a2k

) ≤ ∑cyclic

a1a2+a3+...+an

≤n

n∑k=1

a3k

(n−1)

(n∑

k=1ak

)(∑

cyclica1a2

) .

Mihaly Bencze

OQ.4538. If ak > 0 (k = 1, 2, ..., n) then

∑cyclic

a1 (a2 − a3)2n + 2n

2n+1∏k=1

ak ≥2n

∑cyclic

a2n1 a2n2

(2n+1∑k=1

ak

)2n−1 .

Mihaly Bencze

OQ.4539. Which rational numbers can be written as the sum of finitelymany pairwise distinct Tk number, where Tk = n(n+1)...(n+k−1)

k! .

Mihaly Bencze

OQ.4540. Let ABC be a triangle. Prove that exist infinitely many ceviansca, cb, cc such that 1

4

(∑a2 −∑

ab)≤ ∑

c2a −∑

cacb.

Mihaly Bencze

OQ.4541. A natural number n is called k−faithfull if there exist naturalnumbers a1 < a2 < ... < ak such that a1 |a2, a2| a3, ..., an−1|an. Denote Mk

the set of k-faithfull numbers.

1). Determine Mk

2). Compute∑

n∈N∗\Mk

1n2

3). Determine all n such that n ∈ Mk and n ∈ Mp (k = p) .

Mihaly Bencze

OQ.4542. Show that for any r, p, k, n ∈ N, there exist infinitely manyn-uples (a1, a2, ..., an) of natural numbers, such that[(∑

a1, a2, ..., an)]r =

(ak1 + ak2 + ...+ akn

)np, where [a1, a2, ..., an] denotes the

least common multiple of a1, a2, ..., an.

Mihaly Bencze

Open questions 1009

OQ.4543. Let A1A2...An be a convex n−gon, and B1 ∈ (A1A2) ,B2 ∈ (A2A3) , ..., B3 ∈ (AnA1) such that A1B1 = A2B2 = ... = AnBn. DenoteG1, G2, ..., Gn the centroid of triangles A1B1Bn, A2B2B1, ..., AnBnBn−1....Determine all polygos A1A2...An for which his centroid coincied with thecentroid of B1B2...Bn.

Mihaly Bencze

OQ.4544. Compute∞∫0

cos√x cos 3

√x

cosx dx.

Mihaly Bencze

OQ.4545. Compute Sk =n∑

i=−m(−1)i (km+i)!(kn−i)!

((m+i)!)k((n−i)!)k. We have

S1 =(−1)n+(−1)m

2 and S2 = 1.

Mihaly Bencze

OQ.4546. Compute Sk =∑n≥k

1[nk

k

][nk−1

k

]...[nk−k+1

k

] , where [·] denote the

integer part.

Mihaly Bencze

OQ.4547. Compute1∫0

1∫0

...1∫0

x1

{1

1−x1x2...xn

}dx1dx2...dxn when {·} denote

the fractional part.

Mihaly Bencze

OQ.4548. If f : [0, 1] → R is a continuous function, then compute1∫

a1

1∫a2

...1∫

an

fa1 (x) fa2 (x) ...fan (x) dx1dx2...dxn when 0 < ak < 1

(k = 1, 2, ..., n) .

Mihaly Bencze

OQ.4549. If ζ denote the Riemann zeta function then compute thefollowing sums:

1). Sr =∞∑k=1

1ζ(a1k)ζ(a2k)...ζ(ark)

2). Sk,p =∑

1≤i1<...<ik≤1

ζ(ai1p)...ζ(aikp),

1010 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

where ai > 1 (i = 1, 2, ..., n) and p > 1.

Mihaly Bencze

OQ.4550. Compute the following sums:

1). S1 =∞∑k=1

1[πpk ] 2). S2 =

∞∑k=1

1[epk ]

where pk denote the kth prime, and [·] denote the integer part.

Mihaly Bencze

OQ.4551. Solve in Z the equation

((23 + x1)

2 + (10 + x2)2 + (15 + x3)

2)·

·((23 + x4)

6 + (10 + x5)6 + (15 + x6)

6)=

=((3 + x7)

2 + (19 + x8)2 + (22 + x9)

2)·

·((3 + x10)

6 + (19 + x11)6 + (22 + x12)

6).

Solve in Q the given equation.

Mihaly Bencze

OQ.4552. Solve in Z the equation

((23 + x1)

2 + (10 + x2)2 + (15 + x3)

2)·

·((3 + x4)

6 + (19 + x5)6 + (22 + x6)

6)=

=((23 + x7)

2 + (10 + x8)2 + (15 + x9)

2)·

·((3 + x10)

6 + (19 + x11)6 + (22 + x12)

6).

Solve in Q the given equation.

Mihaly Bencze

OQ.4553. Solve in Z the equation64∑k=1

(29 + xk)4 = (96104 + x65)

2 .

Mihaly Bencze

Open questions 1011

OQ.4554. Solve in Z the equation

((1050 + x1)

4 + (1400 + x2)4 + (1430 + x3)

4+

+(1665 + x4)4 + (1562 + x5)

4)·

·((735 + x6)

4 + (3220 + x7)4 + (3780 + x8)

4+

+(4160 + x9)4 + (5936 + x10)

4)= (3 + x11)

60 .

Solve in Q the given equation.

Mihaly Bencze

OQ.4555. Solve in Z the equation((608 + x1)

4 + (2450 + x2)4 + (11530 + x3)

4+ (34865 + x4)4)·

·((1630 + x5)

4 + (21019 + x6)4 + (22340 + x7)

4 + (33940 + x8)4)=

= (35717 + x9)8 . Solve in Q the given equation.

Mihaly Bencze

OQ.4556. Solve in Z the equation(214 + x1)

6 + (2968 + x2)6 + (6951 + x3)

6++(2046 + x4)

6 + (23457 + x7)6 = (24781 + x8)

6 + (5 + x9)18 . Solve in Q the

given equation.

Mihaly Bencze

OQ.4557. Solve in Z the given equation

((12 + x1)

2 + (16 + x2)2 + (15 + x3)

2)·

·((72 + x4)

2 + (96 + x5)2 + (35 + x6)

2)

((576 + x7)

2 + (168 + x8)2 + (175 + x9)

2)=

= (5 + x10)18 . Solve in Q the given equation.

Mihaly Bencze

OQ.4558. Solve in Z the given equation(5 + x1)

3 + 3 (3 + x2)3 + 4 (2 + x3)

3+

1012 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

+ (1 + x4)3 = 2 (3 + x5)

4 + 4 (2 + x6)4 + 13 (1 + x7)

4 . Solve in Q the givenequation.

Mihaly Bencze

OQ.4559. Solve in Z the given equation (5 + x1)3 + (3 + x2)

4 + (2 + x3)5 =

= 2 (3 + x4)4 + (2 + x5)

6 + 12. Solve in Q the given equation.

Mihaly Bencze

OQ.4560. Solve in Z the given equation((2 + x1)

4 + (2 + x2)4 + (4 + x3)

4+

+(4 + x4)4)(

(2 + x5)4 + (32 + x6)

4 + (34 + x7)4 + (13 + x8)

4 + (84 + x9)4)·

·((16 + x10)

4 + (22 + x11)4 + (38 + x12)

4 + (13 + x13)4 + (84 + x14)

4)=

= (36125 + x15)4 . Solve in Q the given equation.

Mihaly Bencze

OQ.4561. Solve in Z the given equation((23 + x1)

3 + (24 + x2)3 + (25 + x3)

3)·

·((118 + x4)

3 + (119 + x5)3 + (120 + x6)

3 + (121 + x7)3 + (122 + x8)

3)=

= (599760 + x9)2 . Solve in Q the given equation.

Mihaly Bencze

OQ.4562. Solve in Z the given equation((11 + x1)

3 + (12 + x2)3 + (13 + x3)

3+ (14 + x4)3)·

·((31 + x5)

3 + (33 + x6)3 + (35 + x7)

3 + (37 + x8)3 + (39 + x9)

3+

+(41 + x10)3)= (1320 + x11)

3 . Solve in Q the given equation.

Mihaly Bencze

OQ.4563. If aij > 0 (i = 1, 2, ..., n and j = 1, 2, ...,m) andH (x1, x2, ..., xm) = m

1x1

+ 1x2

+...+ 1xm

, then

n∑i=1

1H(ai1,ai2,...,aim) ≥ n2

H

(n∑

i=1ai1,

n∑i=1

ai2,...,n∑

i=1aim,

) .

Mihaly Bencze

Open questions 1013

OQ.4564. Solve in Z the given equationn∑

k=1

x3k = y4.

Mihaly Bencze

OQ.4565. Denote Fk the kth Fibonacci number. Determine all n ∈ N∗ andall p prime for which Fn (mod p) is prime.

Mihaly Bencze

OQ.4566. If ak > 0 (k = 1, 2, ..., n) , then

∑cyclic

a1a2...an−1

an−11 +an−1

2 +...+an−1n−1+λan−1

n≤ n

n+λ−1 , when λ ≥ 1.

Mihaly Bencze

OQ.4567. If ak > 0 (k = 1, 2, ..., n) , thenn∑

k=1

ank + nn∏

k=1

ak ≥

≥ 2n−1

∑cyclic

a1a2...an−1n−1

√(n− 1)n−2 (an−1

1 + an−12 + ...+ an−1

n−1

).

Mihaly Bencze

OQ.4568. If ak > 0 (k = 1, 2, ..., n) , then

n∑k=1

an+1k

∑cyclic

a1a2...an−1+

nn∏

k=1ak

n∑k=1

an−2k

≥ 2n

n∑k=1

a2k.

Mihaly Bencze

OQ.4569. If ak > 0 (k = 1, 2, ..., n) , then

n∑k=1

ank

n∏k=1

ak

+λ

n∏k=1

ak(

n∑k=1

ak

)n ≥ n+ λnn for all

λ ≥ 1.

Mihaly Bencze

OQ.4570. If ak > 0 (k = 1, 2, ..., n) , then∑

cyclic

a21a2

≥ 4

√n3

(n∑

k=1

a4k

).

Mihaly Bencze

1014 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

OQ.4571. If ak > 0 (k = 1, 2, ..., n) , then(n∑

k=1ak

)n∏

k=1ak

n∑k=1

an+1k

+λ

n∑k=1

ank(

n∑k=1

ak

)(n∑

k=1an−1k

) ≥ 1 + λn for all λ ≥ 1.

Mihaly Bencze

OQ.4572. Find suffiecient and necessary conditions on ak ∈ R(k = 1, 2, ..., n) so that the set {({ma1} , {ma2} , ..., {man}) |m ∈ N} is densein ([0, 1])m , when {·} denote the fractional part.

Mihaly Bencze

OQ.4573. If ai > 0 (i = 1, 2, ..., n) , then

n∑i=1

aki(1+a1+a2+...+ai)

k+1 ≤n∑

i=1aki

(1+

n∑i=1

ai

)k , when k ∈ N∗.

Mihaly Bencze

OQ.4574. If a0 = p, a1 = q, a2 = r, b0 = r, b1 = q, b2 = p when p, q, r areprime, and an+3 = x1an+2+ y1an+1+ z1an and bn+3 = x2bn+2+ y2bn+1+ z2bnfor all n ≥ 1. Determine all xk, yk, zk ∈ N (k = 1, 2) for which the sequencesan and bn have in common a finite prime numbers.

Mihaly Bencze

OQ.4575. Determine the best constants a, b, c ∈ R such thatn∑

k=1

kΨ (k) ≥ 12n (n+ 1)

(an2 + bn+ c

).

Mihaly Bencze

OQ.4576. Compute∑

1≤i<j≤n

Φ(i)+Ψ(j)ij+ji

.

Mihaly Bencze

OQ.4577. Computen∑

k=1

∑1≤i1<i2<...<ik≤n

F (i1+i2+...+ik)ir1+ir2+...+irk

when

F ∈ {d,σ,Φ,Ψ,Λ, Fn, Ln} .

Mihaly Bencze

Open questions 1015

OQ.4578. Determine the best constants a, b, c ∈ R such thatn∑

k=1

kφ (k) ≥ 12n (n+ 1)

(an2 + bn+ c

).

Mihaly Bencze

OQ.4579. Determine the best constants a, b, c ∈ R such thatn∑

k=1

Ψ (k) ≥ an2 + bn+ c.

Mihaly Bencze

OQ.4580. Determine the best constants a, b, c ∈ R such thatn∑

k=1

φ (k) ≥ an2 + bn+ c is the best possible. If a = 14 , b = −1

4 , c = 2, then the

inequality holds.

Mihaly Bencze

OQ.4581. Find all n for which there are n consecutive integers whoe sum ofperfect k powers is a prime. If k = 2, then n ∈ {2, 3, 6} .

Mihaly Bencze

OQ.4582. Solve in N the equationn∑

k=1

xmk = yp. The equation have

infinitely many solutions in N .

Mihaly Bencze

OQ.4583. If ak > 0 (k = 1, 2, ..., n) , then

∑cyclic

a1+a2+...+an−1

a1+a2+...+an−1+λan+

∑1≤i<j≤n

aiaj

n∑k=1

a2k

≤ 3n2+(λ−4)n+1−λ2(n−1+λ) , when λ > 0.

Mihaly Bencze

OQ.4584. Solve in N the equationn∏

k=1

(km + 1

m

)=

(ab

)r.

Mihaly Bencze

1016 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

OQ.4585. Solve in N the equationΦ (Φ (aa21 )) + Φ (Φ (aa32 )) + ...+ Φ (Φ (aa1n )) = a1 + a2 + ...+ an.

Mihaly Bencze

OQ.4586. Find all λ ∈ R for which there are functions fk : [0, 1] → R such

thatn∏

k=1

|fk (x)− fk (y)| ≥ |x− y|λ for all x, y ∈ [0, 1] .

Mihaly Bencze

OQ.4587. Solve in N the equation Φ (Φ (nm)) + Φ (Φ (mn)) = m+ n.

Mihaly Bencze

OQ.4588. Solve in N the equation Φ (Φ (nm))Φ (Φ (mn)) = mn.

Mihaly Bencze

OQ.4589. Compute limn→∞

n(√

e−A(n)1

n2 lnn

), when

A (n) =

��������

1 2 ... n1 22 ... n2

− − − −1 2n ... nn

��������.

Mihaly Bencze

OQ.4590. Let f : R → R be a differentiable function and c ∈ R such thatf ′ (x) ≥ 2f (c) for all x ∈ R. Prove that for all a, b ∈ R holds

b∫af (x) dx− 2

a+b2∫c

f (x) dx+b∫cf (x) dx ≥ 1

2

((a− b)2 − c2

)f (c) .

Mihaly Bencze

OQ.4591. Computen∑

k=1

∑1≤i1≤...≤ik≤n

k−(i1+i2+...+ik).

Mihaly Bencze

OQ.4592. Computen∑

k=1

∑1≤i1≤...≤ik≤n

1i1!i2!...ik!

.

Mihaly Bencze

Open questions 1017

OQ.4593. If a1, a2, ..., ak > 0, xi > 0 (i = 1, 2, ..., n) , then determine theminimum of the expression 1

n∑i=1

xi

∑

cyclic

xk+11

(a1x1+a2x2+...+akxk)(a1x2+a2x3+...+akx1)...(a1xk+a2x1+...+akxk−1).

Mihaly Bencze

OQ.4594. Solve in Z the equation kn∑

i=1xk+1i = (k + 1)

(n∑

i=1xi

)k

, when

k ∈ N.

Mihaly Bencze

OQ.4595. If ak > 0 (k = 1, 2, ..., n) , then

nn−1

∑cyclic

a21 (a2 + a3 + ...+ an) ≤(

n∑k=1

ak

)(n∑

k=1

a2k

).

Mihaly Bencze

OQ.4596. Solve in Z the equationn∏

k=1

(a2k − b2k

)= 2013m.

Mihaly Bencze

OQ.4597. Solve in Z the equationn∑

i=1xk+1i = (k + 1)

n∑i=1

xki , when k ∈ N.

Mihaly Bencze

OQ.4598. Determine the best constants ak, bk, ck, dk, ek ∈ R (k = 1, 2) for

which a1x+b1x3

c1+d1x2+e1x4 < sinx < a2x+b2x3

c2+d2x2+e2x4 for all x ∈(0, π2

).

Mihaly Bencze

OQ.4599. Determine the best constants a, b ∈ R such thatb2 ln

1b+x2 < cosx < a

2 ln1

a+x2 for all x ∈(0, π2

).

Mihaly Bencze

OQ.4600. Determine the best constants a, b ∈ R such thata(a−x2)(a+x2)2

< cosx <b(b−x2)(b+x2)2

, for all x ∈(0, π2

).

Mihaly Bencze

1018 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

OQ.4601. Let ak > 0 (k = 1, 2, ..., n) and λ =

(n∑

k=1

ak

)(n∑

k=1

1ak

).

Determine the best polynomial P ∈ Z [x] of degree m such that(n∑

k=1

amk

)(n∑

k=1

1amk

)≥ 1

m+1P (λ) .

Mihaly Bencze

OQ.4602. If Tk = 1+ 2 + ...+ k is a triangular number, then solve in N the

equationn∑

i=1T 2ki

= T 2m.

Mihaly Bencze

OQ.4603. If Tk = 1 + 2 + ...+ k is a triangular number, the solve in N the

equation∑

1≤i<j≤nTkiTkj =

(n∑

i=1Ti

)2

.

Mihaly Bencze

OQ.4604. Find the best constants ak, bk, ck, dk ∈ R (k = 1, 2) for whicha1−b1x

c1−d1x+e1x2 < cosx < a2−b2xc2−d2x+e2x2 for all x ∈

(0, π2

).

Mihaly Bencze

OQ.4605. Determine all ak, bk ∈ N (k = 1, 2, ..., n) for which

n∑k=1

(ak − bk) = (a1, a2, ..., an)

n∑k=1

(ak + bk) = [a1, a2, ..., an].

Mihaly Bencze

OQ.4606. Solve the following equationn∏

k=1

(an−1k + x

)=

∏cyclic

(a1a2...an−1 + x) , when ak ∈ R (k = 1, 2, ..., n) .

Mihaly Bencze

OQ.4607. Determine all prime p which can be written in the formxk1 (x2 − x3) + xk2 (x3 − x4) + ...+ xkn (x1 − x2) when xi ∈ Z (i = 1, 2, ..., n)and k ∈ N.

Mihaly Bencze

Open questions 1019

OQ.4608. If ak > 0 (k = 1, 2, ..., n) , thenn∑

k=1

ank + nn∏

k=1

ak ≥ n+1n n−1√n−1

∑cyclic

a1a2...an−1n−1

√an−11 + an−1

2 + ...+ an−1n−1.

Mihaly Bencze

OQ.4609. If ak > 0 (k = 1, 2, ..., n) , then∑

cyclic

a21a2+a3+...+an

≥n

n∑k=1

a3k

(n−1)n∑

k=1a2k

.

Mihaly Bencze

OQ.4610. If ak > 0 (k = 1, 2, ..., n) , then∑

cyclic

an1+na1a2...an

(a2+a3+...+an)n−1 ≥ n+1

(n−1)n−1

n∑k=1

ak.

Mihaly Bencze

OQ.4611. If ak > 0 (k = 1, 2, ..., n) , thenn∑

k=1

ak +∑

cyclic

a21a2

≥ 2n

n∑k=1

a2k

n∑k=1

ak

.

Mihaly Bencze

OQ.4612. If ai > 0 (i = 1, 2, ..., n) , then determine all k ∈ N for which∑cyclic

√a1

ka1+(n2−k)a2≤ 1.

Mihaly Bencze

OQ.4613. If ak > 0 (k = 1, 2, ..., n) , then∑

cyclic

(a2+a3+...+an)n−1

an−11 +a2a3...an

≥ n(n−1)n−1

2 .

Mihaly Bencze

OQ.4614. If ak > 0 (k = 1, 2, ..., n) , then∑cyclic

(a2+a3+...+an)n−2

(n−1)an−11 +a2a3...an

≥ n(n−1)n−2

n∑k=1

ak

.

Mihaly Bencze

OQ.4615. If ak > 0 (k = 1, 2, ..., n) , then

1).∑

cyclic

an1+nn∏

i=1ai

a2+a3+...+an≥ n+1

n−1

∑cyclic

a1a2...an−1

1020 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

2).∑

cyclic

an1+nn∏

i=1ai

(a2+a3+...+an)n ≥ n(n+1)

(n−1)n

Mihaly Bencze

OQ.4616. If ak > 0 (k = 1, 2, ..., n) , then

1).∑

cyclic

an1+(n−1)n∏

i=1ai

a2+a3+...+an≥ 1

(n−1)nn−3

(n∑

k=1

ak

)n−1

2).∑

cyclic

an−11 +(n−1)a2a3...an

a2+a3+...+an≥ 1

(n−1)nn−4

(n∑

k=1

ak

)n−2

Mihaly Bencze

OQ.4617. If ai > 0 (i = 1, 2, ..., n) and

x1 =2k

√a2k1 − a2k−1

1 a2 + a2k−21 a22 − ...+ a2k2 , ..., xn =

= 2k

√a2kn − a2k−1

n a1 + a2k−2n a21 − ...+ a2k1 , then

∑cyclic

x1x2...xn−1 ≥n∑

i=1a2ki .

Mihaly Bencze

OQ.4618. Determine the best constants ck > 0 (k = 1, 2, ..., n) , for which(n∑

k=1

1xk

)2

≥ c1x21+ c2

x21+x2

2+ ...+ cn

x21+x2

2+...+x2n, for all xk > 0 (k = 1, 2, ..., n). A

solution is ck = k2 (k = 1, 2, ..., n) .

Mihaly Bencze

OQ.4619. If ai > 0 (i = 1, 2, ..., n) and

x1 =2k

√a2k1 + a2k−1

1 a2 + a2k−21 a22 − ...+ a2k2 , ..., xn =

= 2k

√a2kn + a2k−1

n a1 + a2k−2n a21 − ...+ a2k1 , then

∑cyclic

x1x2...xn−1 ≥(

n∑i=1

ai

)2k

.

Mihaly Bencze

OQ.4620. If ai > 0 (i = 1, 2, ..., n) and λ ∈ [0, 2] , then∑

cyclic

an−11 −a2a3...an

λan−11 +an−1

2 +...+an−1n

≥ 0.

Mihaly Bencze

Open questions 1021

OQ.4621. If ak > 0 (k = 1, 2, ..., n) , then∑

cyclic

n−1

√an−11 + (nn−1 − 1) a2a3...an ≤ n

n∑k=1

ak

Mihaly Bencze

OQ.4622. If ak > 0 (k = 1, 2, ..., n) , then∑

cyclic

a1+a2+...+an−1

a2+a3+...+an≤

(n∑

k=1ak

)2

∑cyclic

a1a2.

Mihaly Bencze

OQ.4623. If ak > 0 (k = 1, 2, ..., n) , then∑

cyclic

(a1

a1+a2+...+an−1

)2≥ n

(n−1)2.

Mihaly Bencze

OQ.4624. Solve in Z the equation xy+z + y

z+x + zx+y = x

1+yz + y1+zx + z

1+xy .

Mihaly Bencze

OQ.4625. If ak > 0 (k = 1, 2, ..., n) , then∑

cyclic

an−11

n−1√(ax2+ax3+...+axn)(a

y2+ay3+...+ayn)

≥ nn−1√

(n−1)2, when x, y > 0 and

x+ y = (n− 1)2 .

Mihaly Bencze

OQ.4626. If ai > 0 (i = 1, 2, ..., n) and k ∈ N , thenn∑

i=1ak+1i + n (n− 2)

n∏i=1

ak+1n

i ≥ ∑cyclic

ak1 (a2 + a3 + ...+ an) .

Mihaly Bencze

OQ.4627. Determine the value of∞∑n=1

FnLn

(n!)2.

Mihaly Bencze

OQ.4628. Compute Sk =∞∑n=1

1nk|sin(nπΦ)| , when k ≥ 2 and Φ = 1+

√5

2 .

Mihaly Bencze

1022 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

OQ.4629. The concatenation of possitive integers a and b is defined asa ⊥ b = a

(10[log10 b]+1

)+ b, where [·] denote the integer part. Determine all

a, b, c ∈ N∗ for which (a ⊥ b) ⊥ c is prime. Exist infinitely manya1, a2, ..., an ∈ N∗ for which (((a1 ⊥ a2) ⊥ a3) ... ⊥ an) is prime?

Mihaly Bencze

OQ.4630. The Jacobsthal numbers Jn are defined recursively by J0, J1 = 1and Jn = Jn−1 + 2Jn−2 for all n ≥ 2. The Jacobsthal-Lucas numbers kn aredefined recursively by k0 = 2, k1 = 1 and kn = kn−1 + 2kn−2 for all n ≥ 2.

1). Compute A (n, r) =n∑

k=0

Jrk and B (n, r) =

n∑p=0

krp. We have

A (n, 1) = Jn+1 − 1+(−1)n

2 and B (n, 1) = kn+1 +3(−1)n+1

2 .

2). Compute∞∑n=1

11+J2

nand

∞∑n=1

11+k2n

3). Compute C (n, r) =n∑

p=0Jrpk

rp and D (n, r, s) =

n∑p=0

Jrpk

sp

Mihaly Bencze

OQ.4631. A real number is called p−Cantor number if it can be written isbase p notation without the use of the digit 1. For example 1

3 is 3−Cantornumber, because 1

3 = 0, 1 = 0, 02. Prove that every real number is the sum ofone p−Cantor and one q−Cantor numbers.

Mihaly Bencze

OQ.4632. If ak ∈ (0, 1] (k = 1, 2, ..., n) , then∑

cyclic

1a21+a22

≥ n2

2

(n−1+

n∏k=1

ak

) .

Mihaly Bencze

OQ.4633. If ai ∈ (0, 1] (i = 1, 2, ..., n) and k ∈ {1, 2, ..., n} , then∑cyclic

1a21+a22+...+a2k

≥ n2

k

(n−1+

n∏i=1

ai

) .

Mihaly Bencze

Open questions 1023

OQ.4634. Compute the integrals Ik (n) =π∫0

(sinx)k sin (nx) dx and

Jk (n) =π∫0

(cosx)k cos (nx) dx.

Mihaly Bencze

OQ.4635. If xk > 0 (k = 1, 2, ..., n) andn∑

k=1

xk = 1, then

n∑k=1

1xk

≥ n+ (n+ 1)

(n∏

k=1

xxk−1

k

) 1n−1

.

Mihaly Bencze

OQ.4636. Denote T (n, k) =(n+k−1

k

). Determine all k ∈ N∗ such that the

sequence (T (n, k))n≥1 does not contain any infinite subsequence with allterms in arithmetic progression.

Mihaly Bencze

OQ.4637. Prove that exist infinitely many primes p ≥ 3 for whichp(p−1)n − (p− 1) and pp(p−1)n + p− 1 are both composite for infinitely manyn ∈ N.

Mihaly Bencze

OQ.4638. Let pn + pn+1 − pn+2 = q where (pn) is the sequence of primesand q is a prime. Examine if the double-inequalitypn−1 ≤ pn + pn+1 − pn+2 ≤ pn−1 holds true for all integers n ≥ 4.

George Miliakos

OQ.4639. Find all triples (pn, pn+1, pn+2) for which pn + pn+1 − pn+2 = 1,where (pn) is the sequence of primes.Remark. p2 + p3 − p4 = 1 since p2 = 3, p3 = 5, p4 = 7. Also for p3 = 5,p4 = 7, p5 = 11 one has p3 + p4 − p5 = 1.

George Miliakos

OQ.4640. Examine if there exists an even integer greater than 2, whichdoes not take the form pn+2 − pn where (pn) is the sequence of primes.

George Miliakos

1024 Octogon Mathematical Magazine, Vol. 21, No.2, October 2013

OQ.4641. Examine if every even integer greater than 4 takes the formpn+2 − pn for infinite values of n, where (pn) is the sequence of primes.

George Miliakos

OQ.4642. Let min(pn+K − pn) = 2λ, where (pn) is the sequence of primesand K ≥ 1 is given, then examine, if there exists an even integer greater than2λ, which does not take the form pn+K − pn.

George Miliakos

OQ.4643. Compute Sk =∞∑n=0

1(2n2+1)(3n2+2)...(kn2+k−1)

.

Mihaly Bencze

OQ.4644. Solve in Z the equation x21 + (y21 + y22 + y23 − y24 + y25−−y26 − y27 + y28)

2 +(z21 + z22 + z23 + z24 − z25

)2= 14t.

A solution is t = 1, x1 = 1, y1 = 1, y2 = 3, y4 = 5, y4 = 7, y5 = 9, y6 = 11,y7 = 13, y8 = 15, z1 = 1, z2 = 3, z3 = 5, z4 = 7, z5 = 9.

Mihaly Bencze

OQ.4645. Denote Ck the kth composite number. Compute

Sk =∞∑n=0

1(C2n+C1)(C3n+C2)...(Ckn+Ck−1)

.

Mihaly Bencze

OQ.4646. Denote Ck and pk the kth composite, respectively the prime

numbers. Compute Sk =∞∑n=0

1(p1n+c1)(p2n+c2)...(pkn+ck)

.

Mihaly Bencze

OQ.4647. Denote pk the kth prime number. Compute

Sk =∞∑n=0

1(p2n+p1)(p3n+p2)...(pkn+pk−1)

.

Mihaly Bencze