A183069 L.g.f.: Sum_{n>=1,k>=0} CATALAN(n,k)^2 * x^(n+k)/n = Sum_{n>=1} a(n)*x^n/n, where CATALAN(n,k) = n*C(n+2*k-1,k)/(n+k) is the coefficient of x^k in C(x)^n and C(x) is the g.f. of the Catalan numbers.

1, 3, 19, 163, 1626, 17769, 206487, 2508195, 31504240, 406214878, 5349255726, 71672186953, 974311431094, 13408623649893, 186491860191519, 2617716792257955, 37040913147928380, 527875569932002608, 7570657419156212256, 109194783587953243038

Offset: 1

Author

Paul D. Hanna, Dec 23 2010

Keywords

• nonn

Comments

Logarithmic derivative of A183070.

A bisection of A003162.

We conjecture that the sequence satisfies the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k. - Peter Bala, Mar 20 2023

Examples

			L.g.f.: L(x) = x + 3*x^2/2 + 19*x^3/3 + 163*x^4/4 + 1626*x^5/5 + ...
L(x) = (1 + x + 2^2*x^2 + 5^2*x^3 + 14^2*x^4 + ...)*x
+ (1 + 2^2*x + 5^2*x^2 + 14^2*x^3 + 42^2*x^4 + ...)*x^2/2
+ (1 + 3^2*x + 9^2*x^2 + 28^2*x^3 + 90^2*x^4 + ...)*x^3/3
+ (1 + 4^2*x + 14^2*x^2 + 48^2*x^3 + 165^2*x^4 + ...)*x^4/4
+ (1 + 5^2*x + 20^2*x^2 + 75^2*x^3 + 275^2*x^4 + ...)*x^5/5
+ (1 + 6^2*x + 27^2*x^2 + 110^2*x^3 + 429^2*x^4 + ...)*x^6/6 + ...
which consists of the squares of coefficients in the powers of C(x),
where C(x) = 1 + x*C(x)^2 is g.f. of the Catalan numbers.
...
Exponentiation yields the g.f. of A183070:
exp(L(x)) = 1 + x + 2*x^2 + 8*x^3 + 49*x^4 + 380*x^5 + 3400*x^6 + ...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..500
• Pedro J. Miana, Hideyuki Ohtsuka, and Natalia Romero, Sums of powers of Catalan triangle numbers, arXiv:1602.04347 [math.NT], 2016.

Crossrefs

Cf. A183070, A003162, A000108, A009766.

Programs

Magma

[&+[(n-k)*Binomial(n+k-1, k)^2/n: k in [0..n]]: n in [1..30]]; // Vincenzo Librandi, Feb 16 2016

Mathematica

Table[Sum[(n-k)*Binomial[n+k-1,k]^2/n,{k,0,n}],{n,1,20}] (* Vaclav Kotesovec, Mar 06 2014 *)

PARI

{a(n)=if(n<1,0,sum(k=0,n,(n-k)*binomial(n+k-1,k)^2/n))}

PARI

{a(n)=sum(k=0,n,(binomial(2*n+1,k)-binomial(2*n+1,k-1))^3)/binomial(2*n+1, n)}

Formula

a(n) = Sum_{k=0..n} (n-k)*C(n+k-1,k)^2/n for n>=1.

a(n) = Sum_{k=0..n-1} (C(2*n-1,k) - C(2*n-1,k-1))^3/C(2*n-1,n). [From a formula given in A003162 by Michael Somos.]

Recurrence: 2*n^2*(2*n-1)*(7*n^2 - 20*n + 14)*a(n) = (455*n^5 - 2427*n^4 + 4850*n^3 - 4406*n^2 + 1728*n - 216)*a(n-1) - 4*(n-2)*(2*n - 3)^2*(7*n^2 - 6*n + 1)*a(n-2). - Vaclav Kotesovec, Mar 06 2014

a(n) ~ 16^n/(9*Pi*n^2). - Vaclav Kotesovec, Mar 06 2014

Extensions

a(19)-a(20) from Vincenzo Librandi, Feb 16 2016