A268138 a(n) = (Sum_{k=0..n-1} A001850(k)*A001003(k+1))/n.

1, 5, 51, 747, 13245, 264329, 5721415, 131425079, 3159389817, 78729848397, 2019910325499, 53087981674275, 1423867359013749, 38855956977763857, 1076297858301372687, 30203970496501504239, 857377825323716359665, 24586286492003180067989, 711463902659879056604995, 20756358426519694831851227

Offset: 1

Author

Zhi-Wei Sun, Jan 26 2016

Keywords

• nonn

Comments

Conjecture: (i) All the terms are odd integers. Also, p | a(p) for any odd prime p.

(ii) Let D_n(x) = Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k = Sum_{k=0..n} binomial(n,k)^2*x^k*(x+1)^(n-k) for n >= 0, and s_n(x) = Sum_{k=1..n} (binomial(n,k)*binomial(n,k-1)/n)*x^(k-1)*(x+1)^(n-k) = (Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k/(k+1))/(x+1) for n > 0. Then, for any positive integer n, all the coefficients of the polynomial (1/n)*Sum_{k=0..n-1} D_k(x)*s_{k+1}(x) are integral and the polynomial is irreducible over the field of rational numbers.

The conjecture was essentially proved by the author in arXiv:1602.00574, except for the irreducibility of (Sum_{k=0..n-1} D_k(x)*s_{k+1}(x))/n. - Zhi-Wei Sun, Feb 01 2016

Examples

			a(3) = 51 since (A001850(0)*A001003(1) + A001850(1)*A001003(2) + A001850(2)*A001003(3))/3 = (1*1 + 3*3 + 13*11)/3 = 153/3 = 51.
		

Links

• Zhi-Wei Sun, Table of n, a(n)for n = 1..100
• Zhi-Wei Sun, On Delannoy numbers and Schroder numbers, J. Number Theory 131(2011), no.12, 2387-2397.
• Zhi-Wei Sun, Arithmetic properties of Delannoy numbers and Schröder numbers, preprint, arXiv:1602.00574 [math.CO], 2016.

Crossrefs

Cf. A001003, A001850, A006318, A268136, A268137, A182626, A358388.

Programs

Maple

A001850 := n -> LegendreP(n, 3); seq(((3*(2*n+1)*A001850(n)*A001850(n-1)-n*A001850(n-1)^2)/(n+1) - A001850(n)^2)/4, n=1..20); # Mark van Hoeij, Nov 12 2022
# Alternative (which also gives an integer for n = 0):
f := n -> hypergeom([-n, -n], [1], 2):          # A001850
h := n -> hypergeom([-n,  n], [1], 2):          # A182626
g := n -> hypergeom([-n,  n, 1/2], [1, 1], -8): # A358388
a := n -> (f(n)*((3*n + 1)*f(n) - (-1)^n*(6*n + 3)*h(n)) - n*g(n))/(2*n + 2):
seq(simplify(a(n)), n = 1..20); # Peter Luschny, Nov 13 2022

Mathematica

d[n_]:=Sum[Binomial[n,k]Binomial[n+k,k],{k,0,n}]
s[n_]:=Sum[Binomial[n,k]Binomial[n,k-1]/n*2^(k-1),{k,1,n}]
a[n_]:=Sum[d[k]s[k+1],{k,0,n-1}]/n
Table[a[n],{n,1,20}]

Formula

a(n) = ((3*(2*n+1)*A001850(n)*A001850(n-1) - n*A001850(n-1)^2)/(n+1) - A001850(n)^2)/4. - Mark van Hoeij, Nov 12 2022

G.f.: (1-(1+1/x)*Int((1-34*x+x^2)^(1/2) * hypergeom([-1/2,1/2],[1], -32*x/(1-34*x+x^2))/((1-x)*(1+x)^2),x))/4. - Mark van Hoeij, Nov 28 2024