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A045894 4-fold convolution of A001700(n), n >= 0.

%I A045894 #20 May 22 2025 10:20:06
%S A045894 1,12,94,608,3525,19044,97954,486000,2345930,11081880,51447036,
%T A045894 235454848,1064832173,4767347796,21160397050,93223960784,408037319262,
%U A045894 1775744775592,7688699122724,33140226601920,142262721338146
%N A045894 4-fold convolution of A001700(n), n >= 0.
%H A045894 Indranil Ghosh, <a href="/A045894/b045894.txt">Table of n, a(n) for n = 0..1500</a>
%H A045894 José Agapito, Ângela Mestre, Maria M. Torres, and Pasquale Petrullo, <a href="http://cs.uwaterloo.ca/journals/JIS/VOL18/Agapito/agapito2.html">On One-Parameter Catalan Arrays</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.1.
%H A045894 Milan Janjić, <a href="https://www.emis.de/journals/JIS/VOL21/Janjic2/janjic103.html">Pascal Matrices and Restricted Words</a>, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
%F A045894 a(n) = (n+11)*4^(n+2) - (n+5)*binomial(2*(n+4), n+4)/2;
%F A045894 G.f.: c(x)^4/(1-4*x)^2, where c(x) = g.f. for Catalan numbers A000108;
%F A045894 recursion: a(n)= (2*(2*n+10)/(n+4))*a(n-1) + (4/(n+4))*A045720(n), a(0)=1.
%t A045894 Table[(n + 11)*4^(n + 2) - (n + 5) Binomial[2 (n + 4), n + 4]/2, {n, 0, 20}] (* _Michael De Vlieger_, Feb 18 2017 *)
%o A045894 (Python)
%o A045894 import math
%o A045894 def C(n,r):
%o A045894     f=math.factorial
%o A045894     return f(n)/f(r)/f(n-r)
%o A045894 def A045894(n):
%o A045894     return (n+11)*4**(n+2)-(n+5)*C(2*(n+4),(n+4))/2 # _Indranil Ghosh_, Feb 18 2017
%Y A045894 Cf. A001700, A000108, A045720.
%K A045894 easy,nonn
%O A045894 0,2
%A A045894 _Wolfdieter Lang_