A295113 a(n) = (1/n)*Sum_{k=0..n-1} (8*k + 9)*A295112(k)^2.

9, 13, 17, 219, 1561, 8169, 39321, 191389, 985201, 5430789, 31943961, 198471183, 1288665177, 8665236121, 59922226809, 423935337411, 3056528058577, 22392246851973, 166311049602681, 1250027314777795, 9494339129623329, 72784922204637153, 562626619763553217, 4381665416129531961, 34354964747652467697

Offset: 1

Author

Zhi-Wei Sun, Nov 14 2017

Keywords

• nonn

Comments

Conjecture: a(n) is always integral for every n = 1,2,3,.... Moreover, for any odd prime p we have a(p) == 24 + 10*Leg(-1,p) - 18*Leg(3,p) - 9*Leg(p,3) (mod p^2), where Leg(m,p) denotes the Legendre symbol (m/p).

We also observe that Sum_{k=0}^{p-1}A295112(k)^2 == 2 (mod p) for any prime p > 3.

Examples

			a(2) = 13 since (1/2)*Sum_{k=0..1} (8k + 9)*A295112(k)^2 = (1/2)*((8*0 + 9)*A295112(0)^2 + (8 + 9)*A295112(1)^2) = (1/2)*(9*(-1)^2 + 17*(-1)^2) = 13.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..200
• Zhi-Wei Sun, On Motzkin numbers and central trinomial coefficients, arXiv:1801.08905 [math.CO], 2018. (See Conjecture 5.1.)

Crossrefs

Cf. A001006, A295112.

Programs

Mathematica

W[n_]:=W[n]=Sum[Binomial[n,2k]Binomial[2k,k]/(2k-1),{k,0,n/2}];
a[n_]:=a[n]=1/n*Sum[(8k+9)W[k]^2,{k,0,n-1}];
Table[a[n],{n,1,25}]