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%I #13 Jun 17 2017 03:53:51
%S 6,30,91,216,441,812,1386,2232,3432,5082,7293,10192,13923,18648,24548,
%T 31824,40698,51414,64239,79464,97405,118404,142830,171080,203580,
%U 240786,283185,331296,385671,446896,515592,592416,678062,773262,878787,995448
%N a(n) = n*(n+1)*(n+2)*(n+4)*(n+23)/120.
%C 6th partial summation within series as series accumulate n times from an initial sequence of Euler Triangle's row 3: 1,4,1. 6th row of the array shown in A101853. Partial sums of A101854.
%H C. Rossiter, <a href="http://noticingnumbers.net/300SeriesCube.htm">Depictions, Explorations and Formulas of the Euler/Pascal Cube</a>.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F G.f.: x*(6-6*x+x^2) / (x-1)^6. - _R. J. Mathar_, Dec 06 2011
%F a(1)=6, a(2)=30, a(3)=91, a(4)=216, a(5)=441, a(6)=812, a(n)=6*a(n-1)- 15*a(n-2)+ 20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - _Harvey P. Dale_, Feb 07 2013
%t Table[n(n+1)(n+2)(n+4)(n+23)/120,{n,40}] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{6,30,91,216,441,812},40](* _Harvey P. Dale_, Feb 07 2013 *)
%K easy,nonn
%O 1,1
%A Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 18 2004
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