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a(n) = (Sum_{k=0..n-1} A001850(k)*A001003(k+1))/n.
3

%I #44 Nov 28 2024 18:40:56

%S 1,5,51,747,13245,264329,5721415,131425079,3159389817,78729848397,

%T 2019910325499,53087981674275,1423867359013749,38855956977763857,

%U 1076297858301372687,30203970496501504239,857377825323716359665,24586286492003180067989,711463902659879056604995,20756358426519694831851227

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

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

%C (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.

%C 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

%H Zhi-Wei Sun, <a href="/A268138/b268138.txt">Table of n, a(n)for n = 1..100</a>

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2011.06.005">On Delannoy numbers and Schroder numbers</a>, J. Number Theory 131(2011), no.12, 2387-2397.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1602.00574">Arithmetic properties of Delannoy numbers and Schröder numbers</a>, preprint, arXiv:1602.00574 [math.CO], 2016.

%F 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

%F 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

%e 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.

%p 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

%p # Alternative (which also gives an integer for n = 0):

%p f := n -> hypergeom([-n, -n], [1], 2): # A001850

%p h := n -> hypergeom([-n, n], [1], 2): # A182626

%p g := n -> hypergeom([-n, n, 1/2], [1, 1], -8): # A358388

%p a := n -> (f(n)*((3*n + 1)*f(n) - (-1)^n*(6*n + 3)*h(n)) - n*g(n))/(2*n + 2):

%p seq(simplify(a(n)), n = 1..20); # _Peter Luschny_, Nov 13 2022

%t d[n_]:=Sum[Binomial[n,k]Binomial[n+k,k],{k,0,n}]

%t s[n_]:=Sum[Binomial[n,k]Binomial[n,k-1]/n*2^(k-1),{k,1,n}]

%t a[n_]:=Sum[d[k]s[k+1],{k,0,n-1}]/n

%t Table[a[n],{n,1,20}]

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

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Jan 26 2016