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4-fold convolution of A001700(n), n >= 0.
5

%I #18 Sep 04 2018 11:40:01

%S 1,12,94,608,3525,19044,97954,486000,2345930,11081880,51447036,

%T 235454848,1064832173,4767347796,21160397050,93223960784,408037319262,

%U 1775744775592,7688699122724,33140226601920,142262721338146

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

%H Indranil Ghosh, <a href="/A045894/b045894.txt">Table of n, a(n) for n = 0..1500</a>

%H 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 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 a(n) = (n+11)*4^(n+2) - (n+5)*binomial(2*(n+4), n+4)/2;

%F G.f.: c(x)^4/(1-4*x)^2, where c(x) = g.f. for Catalan numbers A000108;

%F recursion: a(n)= (2*(2*n+10)/(n+4))*a(n-1) + (4/(n+4))*A045720(n), a(0)=1.

%t 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 (Python)

%o import math

%o def C(n,r):

%o ....f=math.factorial

%o ....return f(n)/f(r)/f(n-r)

%o def A045894(n):

%o ....return (n+11)*4**(n+2)-(n+5)*C(2*(n+4),(n+4))/2 # _Indranil Ghosh_, Feb 18 2017

%Y Cf. A001700, A000108, A045720.

%K easy,nonn

%O 0,2

%A _Wolfdieter Lang_