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%I #68 Sep 08 2022 08:46:10
%S 1,514,20710,303050,2538515,14851676,67518444,254402940,828707925,
%T 2403012910,6335265586,15427298614,35123831015,75481410200,
%U 154282348760,301802764056,567911055849,1032378638010,1819533917950,3118689197890,5212124524411,8512829068724,13614686274500,21358351020500,32916713032125,49904578722726
%N Second partial sums of ninth powers (A001017).
%C The formula for the second partial sums of m-th powers is: b(n,m) = (n+1)*F(m) - F(m+1), where F(m) are the m-th Faulhaber's formulas.
%H G. C. Greubel, <a href="/A253637/b253637.txt">Table of n, a(n) for n = 1..1000</a>
%H Luciano Ancora, <a href="/A253636/a253636_3.pdf">Recurrence relation for the second partial sums of m-th powers</a>
%H Luciano Ancora, <a href="/A253636/a253636_4.pdf">Second partial sums of the m-th powers</a>
%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).
%F a(n) = n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4 + 24*n^3 + 5*n^2 - 38*n + 25)/660.
%F a(n) = 2*a(n-1) - a(n-2) + n^9.
%F G.f.: x*(1 + 502*x + 14608*x^2 + 88234*x^3 + 156190*x^4 + 88234*x^5 + 14608*x^6 + 502*x^7 + x^8)/(1-x)^12. - _Vincenzo Librandi_, Jan 19 2015
%p seq(n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4+24*n^3+5*n^2-38*n+ 25)/660, n=1..30); # _G. C. Greubel_, Aug 28 2019
%t CoefficientList[Series[(1 +502x +14608x^2 +88234x^3 +156190x^4 +88234x^5 +14608x^6 +502x^7 +x^8)/(1-x)^12, {x, 0, 30}], x] (* _Vincenzo Librandi_, Jan 19 2015 *)
%t Nest[Accumulate,Range[30]^9,2] (* _Harvey P. Dale_, Apr 18 2021 *)
%o (PARI) a(n) = (6*n^11 + 66*n^10 + 275*n^9 + 495*n^8 + 198*n^7 - 462*n^6 - 330*n^5 + 330*n^4 + 231*n^3 - 99*n^2 - 50*n)/660; \\ _Michel Marcus_, Jan 08 2015
%o (Magma) [n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4+24*n^3+5*n^2-38*n+ 25)/660: n in [1..30]]; // _G. C. Greubel_, Aug 28 2019
%o (Sage) [n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4+24*n^3+5*n^2-38*n+ 25)/660 for n in (1..30)] # _G. C. Greubel_, Aug 28 2019
%o (GAP) List([1..30], n-> n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4+24*n^3 +5*n^2-38*n+ 25)/660 ); # _G. C. Greubel_, Aug 28 2019
%Y Cf. A001017.
%K nonn,easy
%O 1,2
%A _Luciano Ancora_, Jan 07 2015
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