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Conjecture: the number a(n) = n^(-1)*Sum_{k=0..n-1}(2*k+1)*A_k is always an integer.

We can prove that for any prime p>3 we have a(p)=p (mod p^4).

Conjecture: If p=5,7 (mod 8) is a prime then sum_{k=0}^{p-1}A_k=0 (mod p^2); if p=1,3 (mod 8) is a prime greater than 3 and p=x^2+2y^2 with x,y integers then sum_{k=0}^{p-1}A_k=4x^2-2p (mod p^2).

a(n) is always an integer. The detailed proof can be found in the latest version of arXiv:1006.2776 . - Zhi-Wei Sun, Jun 17 2010

On Jun 14 2010, Zhi-Wei Sun conjectured that the number a(n) = n^{-1}*Sum_{k=0..n-1} (2*k+1)*(-1)^k*A_k is always an integer and that a(p) = p(p/3) (mod p^3) for any prime p>3. He also formulated the following conjecture: If p=1 (mod 3) is a prime and p = x^2 + 3y^2 with x,y integers then Sum_{k=0..p-1} (-1)^k*A_k = 4*x^2 - 2*p (mod p^2); if p=2 (mod 3) is a prime then Sum_{k=0..p-1 }(-1)^k*A_k=0 (mod p^2).

On Jun 14 2010, Zhi-Wei Sun conjectured that a(n) = (1/n^2) * Sum_{k = 0..n-1} (2*k+1)*(D_k)^2 is always an integer and that p^2*a(p) = p^2 - 4*p^3*q_p(2) - 2*p^4*q_p(2)^2 (mod p^5) for any prime p > 3, where q_p(2) denotes the Fermat quotient (2^(p-1) - 1)/p (see Sun, Remark 4.3, p. 26, 2014). He also conjectured that Sum_{k = 0..n-1} (2*k+1)*(-1)^k*(D_k)^2 == 0 (mod n*D_n/(3,D_n)) for all n = 1,2,3,....

The fact that a(n) is an integer follows directly from the formulas for a(n) in the formula section below. - Mark van Hoeij, Nov 13 2022

On Apr 04 2010, Zhi-Wei Sun introduced this sequence and conjectured that each term a(n) is a positive integer. He also guessed that a(n) is odd if and only if n is a power of two. It is easy to see that 8*n^2*binomial(2*n,n)^2*a(n) equals Sum_{k=0..n-1} (205*k^2 + 160*k + 32)*(-1)^(n-1-k)*binomial(2*k,k)^5. Sun also conjectured that for any prime p > 5 we have Sum_{k=0..p-1} (205*k^2 + 160*k + 32)(-1)^k*binomial(2*k,k)^5 == 32*p^2 + 64*p^3*Sum_{k=1..p-1} 1/k (mod p^7) and Sum_{k=0..(p-1)/2} (205*k^2 + 160*k + 32)(-1)^k*binomial(2*k,k)^5 == 32*p^2 + 896/3*p^5*B_{p-3} (mod p^6), where B_0, B_1, B_2, ... are Bernoulli numbers. It is also remarkable that Sum_{n>0} (-1)^n(205*n^2 - 160*n + 32)/(n^5*binomial(2*n,n)^5) = -2*zeta(3) as proved by T. Amdeberhan and D. Zeilberger via the WZ method.

During April 26-28, 2010, Zhi-Wei Sun introduced this new sequence and proved that a(n) = binomial(6n,3n)*binomial(3n,n)/(2*(2n+1)*binomial(2n,n)) is a positive integer for every n=1,2,3,... He also observed that a(n) is odd if and only if n is a power of two, and that 3a(n)=0 (mod 2n+3). By Stirling's formula, we have lim_n (8n*sqrt(n*Pi)a(n)/108^n) = 1. It is interesting to find a combinatorial interpretation or recursion for the sequence.

From Tatiana Hessami Pilehrood, Dec 01 2015: (Start)

Zhi-Wei Sun formulated two conjectures concerning a(n) (see Conjectures 1.1 and 1.2 in Z.-W. Sun, "Products and sums divisible by central binomial coefficients" and Conjecture A89 in "Open conjectures on congruences"). The first conjecture states that Sum_{n=1..p-1} a(n)/(108^n) is congruent to 0 or -1 modulo a prime p > 3 depending on whether p is congruent to +-1 or +-5 modulo 12, respectively.

The second conjecture asks about an exact formula for a companion sequence of a(n). Both conjectures as well as many numerical congruences involving a(n) and (2n+1)a(n) were solved by Kh. Hessami Pilehrood and T. Hessami Pilehrood, see the link below. (End)

This is the even bisection of A078531 divided by 2. The odd bisection divided by 2 is A281733. - Akiva Weinberger, Dec 09 2024

On Jun 28 2010, Zhi-Wei Sun conjectured that a(n) is an integer for every n=1,2,3,... and that a(p) = (1+p/3)/2 (mod p) for any prime p, where (p/3) is the Legendre symbol. In contrast, he showed that Sum_{k=0..n-1} (2k+1)*T_k^2*3^(n-1-k) = n*T_n*T_{n-1} for all n=1,2,3,...

The formula for a(n) in the formula section implies that a(n) is an integer. - Mark van Hoeij, Nov 13 2022

On July 1, 2010 Zhi-Wei Sun introduced this sequence and made the following conjecture: If p is a prime with p=1,9 (mod 20) and p=x^2+5y^2 with x,y integers, then sum_{k=0}^{p-1}a(k)=4x^2-2p (mod p^2); if p is a prime with p=3,7 (mod 20) and 2p=x^2+5y^2 with x,y integers, then sum_{k=0}^{p-1}a(k)=2x^2-2p (mod p^2); if p is a prime with p=11,13,17,19 (mod 20), then sum_{k=0}^{p-1}w_k=0 (mod p^2). He also conjectured that sum_{k=0}^{n-1}(20k+17)w_k=0 (mod n) for all n=1,2,3,... and that sum_{k=0}^{p-1}(20k+17)w_k=p(10(-1/p)+7) (mod p^2) for any odd prime p. Sun also formulated similar conjectures for some sequences similar to a(n).

On Apr 06 2010, Zhi-Wei Sun introduced this sequence and conjectured that each term a(n) is a positive integer. He also guessed that a(n) is odd if and only if n = 2, 2^2, 2^3, .... It is easy to see that 16*(2n+1)^3*binomial(2n,n)^3*a(n) equals Sum_{k=0..n} (21k^3 + 22k^2 + 8k + 1)*256^(n-k)*binomial(2k,k)^7. Sun also conjectured that for any prime p > 5 we have Sum_{k=0..p-1} (21k^3 + 22k^2 + 8k + 1)*binomial(2k,k)^7/256^k == p^3 (mod p^8). It is also remarkable that Sum_{n>=1} 256^n*(21n^3 - 22n^2 + 8n - 1)/(n^7*binomial(2n,n)^7) = Pi^4/8 as conjectured by J. Guillera.

On Jul 17 2010, Zhi-Wei Sun introduced this sequence and made the following conjecture: If p is a prime with p=1,9,11,19 (mod 40) and p = x^2+10y^2 with x,y integers, then Sum_{k=0..p-1} a(k) == 4x^2-2p (mod p^2); if p is a prime with p == 7,13,23,37 (mod 40) and 2p = x^2 + 10y^2 with x,y integers, then Sum_{k=0..p-1} a(k) == 2p - 2x^2 (mod p^2); if p is an odd prime with (-10/p)=-1, then Sum_{k=0..p-1} a(k) == 0 (mod p^2). He also conjectured that Sum_{k=0..n-1} (10k+9)*a(k) == 0 (mod n) for all n=1,2,3,... and that Sum_{k=0..p-1} (10k+9)*a(k) == p(4(-2/p)+5) (mod p^2) for any prime p > 3.