cp's OEIS Frontend

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 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.

On Jun 17 2010, Zhi-Wei Sun conjectured that a(n) is an integer for every n=1,2,3,... and that a(p) == 3(p/3) (mod p) for any prime p, where (p/3) is the Legendre symbol. He also observed that Sum_{k=0..n-1} (2k+1) T_k*3^{n-1-k} = n * Sum_{k=0..n-1} C(n-1,k)*(-1)^(n-1-k)*(k+1)*C(2k,k).

On July 17, 2010 Zhi-Wei Sun introduced this sequence and made the following conjecture: If p is a prime with (p/7)=1 and p=x^2+7y^2 with x,y integers, then sum_{k=0}^{p-1}(-1)^k*a(k)=4x^2-2p (mod p^2); if p is a prime with (p/7)=-1, then sum_{k=0}^{p-1}(-1)^k*a(k)=0 (mod p^2). He also conjectured that sum_{k=0}^{n-1}(42k+37)(-1)^k*a(k)=0 (mod n) for all n=1,2,3,... and that sum_{k=0}^{p-1}(42k+37)(-1)^k*a(k)=p(21(p/7)+16) (mod p^2) for any prime p.