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%I M5394 N2343 #93 Apr 28 2024 11:08:03
%S 0,1,129,2316,18700,96825,376761,1200304,3297456,8080425,18080425,
%T 37567596,73399404,136147921,241561425,412420800,680856256,1091194929,
%U 1703414961,2597286700,3877286700,5678375241,8172733129,11577558576,16164030000,22267545625
%N Sum of 7th powers: 1^7 + 2^7 + ... + n^7.
%C a(n) is divisible by A000537(n) if and only n is congruent to 1 mod 3 (see A016777) - _Artur Jasinski_, Oct 10 2007
%C This sequence is related to A000540 by a(n) = n*A000540(n) - Sum_{i=0..n-1} A000540(i). - _Bruno Berselli_, Apr 26 2010
%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 815.
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, <a href="/A000541/b000541.txt">Table of n, a(n) for n = 0..1000</a>
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H B. Berselli, A description of the recursive method in Comments lines: website <a href="http://www.lanostra-matematica.org/2008/12/sequenze-numeriche-e-procedimenti.html">Matem@ticamente</a> (in Italian).
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).
%F a(n) = n^2*(n+1)^2*(3*n^4 + 6*n^3 - n^2 - 4*n + 2)/24.
%F a(n) = sqrt(Sum_{j=1..n} Sum_{i=1..n} (i*j)^7). - _Alexander Adamchuk_, Oct 26 2004
%F Jacobi formula: a(n) = 2(A000217(n))^4 - A000539(n). - _Artur Jasinski_, Oct 10 2007
%F G.f.: x*(1 + 120*x + 1191*x^2 + 2416*x^3 + 1191*x^4 + 120*x^5 + x^6)/(1-x)^9. - _Colin Barker_, May 25 2012
%F a(n) = 8*a(n-1) - 28* a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) + 5040. - _Ant King_, Sep 24 2013
%F a(n) = -Sum_{j=1..7} j*s(n+1,n+1-j)*S(n+7-j,n), where s(n,k) and S(n,k) are the Stirling numbers of the first kind and the second kind, respectively. - _Mircea Merca_, Jan 25 2014
%F a(n) = 2*A000217(n)^4 - (4/3)*A000217(n)^3 + (1/3)*A000217(n)^2. - _Michael Raney_, Feb 19 2016
%F a(n) = 72*A288876(n-2) + 48*A006542(n+2) + A000537(n). - _Yasser Arath Chavez Reyes_, Apr 27 2024
%p a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^7 od: seq(a[n], n=0..25); # _Zerinvary Lajos_, Feb 22 2008
%t Table[Sum[k^7, {k, 1, n}], {n, 0, 100}] (* _Artur Jasinski_, Oct 10 2007 *)
%t s = 0; lst = {s}; Do[s += n^7; AppendTo[lst, s], {n, 1, 30, 1}]; lst (* _Zerinvary Lajos_, Jul 12 2009 *)
%t LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {0, 1, 129, 2316, 18700, 96825, 376761, 1200304, 3297456}, 35] (* _Vincenzo Librandi_, Feb 20 2016 *)
%o (PARI) a(n)=n^2*(n+1)^2*(3*n^4+6*n^3-n^2-4*n+2)/24 \\ _Edward Jiang_, Sep 10 2014
%o (PARI) a(n) = sum(i=1, n, i^7); \\ _Michel Marcus_, Sep 11 2014
%o (Python)
%o A000541_list, m = [0], [5040, -15120, 16800, -8400, 1806, -126, 1, 0, 0]
%o for _ in range(10**2):
%o for i in range(8):
%o m[i+1] += m[i]
%o A000541_list.append(m[-1]) # _Chai Wah Wu_, Nov 05 2014
%o (Magma) [n^2*(n+1)^2*(3*n^4+6*n^3-n^2-4*n+2)/24: n in [0..30]]; // _Vincenzo Librandi_, Feb 20 2016
%Y Row 7 of array A103438.
%Y Cf. A000217, A000537, A000539.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_
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