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%I #10 Apr 30 2014 01:26:48
%S 1,19,135,605,2054,5778,14178,31350,63855,121693,219505,378027,625820,
%T 1001300,1555092,2352732,3477741,5035095,7155115,9997801,13757634,
%U 18668870,25011350,33116850,43375995,56245761,72257589,92026135
%N Third partial sums of fourth powers (A000583).
%C In general, the r-th successive summation of the fourth powers from 1 to n = (2*n+r)*(12*n^2+12*n*r+r^2-5*r)*(r+n)!/((r+4)!*(n-1)!). Here r = 3. - _Gary Detlefs_, Mar 01 2013
%H C. Rossiter, <a href="http://noticingnumbers.net/300SeriesCube.htm">Depictions, Explorations and Formulas of the Euler/Pascal Cube</a>.
%F a(n) = (n*(1+n)*(2+n)*(3+n)*(3+2*n)*(-1+2*n*(3+n)))/840.
%F G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^8. [Colin Barker, Apr 04 2012]
%F a(n)= (2*n+3)*(12*n^2+36*n-6)*(n+3)!/(5040*(n-1)!), n>0 - _Gary Detlefs_, Mar 01 2013
%Y Cf. A101091, A101089.
%K easy,nonn
%O 1,2
%A Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004
%E Edited by _Ralf Stephan_, Dec 16 2004
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