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a(n) = n*(n+1)*(n+7)*(122+57*n+n^2)/120.
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%I #18 Sep 08 2022 08:45:16

%S 24,108,302,671,1296,2275,3724,5778,8592,12342,17226,23465,31304,

%T 41013,52888,67252,84456,104880,128934,157059,189728,227447,270756,

%U 320230,376480,440154,511938,592557,682776,783401,895280,1019304,1156408,1307572,1473822,1656231

%N a(n) = n*(n+1)*(n+7)*(122+57*n+n^2)/120.

%C Partial sums of A101861.

%C 6th partial summation within series as series accumulate n times from an initial sequence of Euler Triangle's row 4: 1,11,11,1: 6th row of the array in the examples of A101860.

%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*(2-x)*(x^2-12*x+12) / (1-x)^6. - _R. J. Mathar_, Dec 06 2011

%F 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) for n > 6. - _Wesley Ivan Hurt_, Dec 06 2016

%p A101862:=n->n*(n+1)*(n+7)*(122+57*n+n^2)/120: seq(A101862(n), n=1..50); # _Wesley Ivan Hurt_, Dec 06 2016

%t Table[n*(n + 1)*(n + 7)*(122 + 57*n + n^2)/120, {n, 50}] (* _Wesley Ivan Hurt_, Dec 06 2016 *)

%t LinearRecurrence[{6,-15,20,-15,6,-1},{24,108,302,671,1296,2275},50] (* _Harvey P. Dale_, Oct 15 2020 *)

%o (Magma) [n*(n + 1)*(n + 7)*(122 + 57*n + n^2)/120 : n in [1..50]]; // _Wesley Ivan Hurt_, Dec 06 2016

%Y Cf. A101860, A101861.

%K nonn,easy

%O 1,1

%A Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 18 2004