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a(n) = (1/5)*(4 + sum of C(5k,k)) for k = 0,1,2,...,n.
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%I #9 Jul 27 2021 17:35:03

%S 1,2,11,102,1071,11697,130452,1475356,16856293,194088920,2248544554,

%T 26179257724,306051026719,3590265729775,42241017111607,

%U 498243554454823,5889887780556528,69761293355975193,827690921897694948

%N a(n) = (1/5)*(4 + sum of C(5k,k)) for k = 0,1,2,...,n.

%H Harvey P. Dale, <a href="/A024721/b024721.txt">Table of n, a(n) for n = 0..922</a>

%t (Accumulate[Table[Binomial[5n,n],{n,0,20}]]+4)/5 (* _Harvey P. Dale_, Jul 27 2021 *)

%K nonn

%O 0,2

%A _Clark Kimberling_

%E More terms from _James A. Sellers_, May 01 2000