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%I #79 Sep 20 2022 02:37:39
%S 1,17,97,337,881,1921,3697,6497,10657,16561,24641,35377,49297,66977,
%T 89041,116161,149057,188497,235297,290321,354481,428737,514097,611617,
%U 722401,847601,988417,1146097,1321937,1517281,1733521,1972097,2234497,2522257,2836961,3180241
%N 4-dimensional centered cube numbers.
%C Summation of n^4 taken two at a time. - Al Hakanson (hawkuu(AT)gmail.com), May 27 2009
%C The primes in this sequence are given by A152913. - _Jonathan Vos Post_, Aug 17 2011
%H Vincenzo Librandi, <a href="/A008514/b008514.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = n^4 + (n+1)^4.
%F a(n) = 2*n^4 + 4*n^3 + 6*n^2 + 4*n + 1. - Al Hakanson (hawkuu(AT)gmail.com), May 27 2009, corrected _R. J. Mathar_, May 29 2009
%F G.f.: (1+10*x+x^2)*(1+x)^2/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 09 2009
%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with a(0) = 1, a(1) = 17, a(2) = 97, a(3) = 337, a(4) = 881. - _Harvey P. Dale_, Jan 28 2013
%F a(n) = 4*(n+n^2) + 2*(n+n^2)^2 + 1. - _Avi Friedlich_, Mar 31 2015
%F a(n) = 2*A002061(n+1)^2 - 1. - _Bruce J. Nicholson_, Apr 14 2017
%F a(n) = A047838(2*(n^2+n+1)). - _David James Sycamore_, Aug 01 2018
%F E.g.f.: (1 + 16*x + 32*x^2 + 16*x^3 + 2*x^4)*exp(x). - _G. C. Greubel_, Nov 09 2019
%F Sum_{n>=0} 1/a(n) = (tanh((sqrt(2)-1)*Pi/2)*Pi*(2+sqrt(2)) - tanh((sqrt(2)+1)*Pi/2)*Pi*(2-sqrt(2)))/4. - _Amiram Eldar_, Sep 20 2022
%p seq(n^4+(n+1)^4, n=0..40);
%t Total/@Partition[Range[0, 30]^4, 2, 1] (* or *) LinearRecurrence[{5,-10, 10,-5,1}, {1,17,97,337,881}, 30] (* _Harvey P. Dale_, Jan 28 2013 *)
%o (Sage) [i^4+(i+1)^4 for i in range(0,36)] # _Zerinvary Lajos_, Jul 03 2008
%o (Magma) [(n+1)^4+n^4: n in [0..30]]; // _Vincenzo Librandi_, Aug 27 2011
%o (PARI) a(n) = n^4 + (n+1)^4; \\ _Altug Alkan_, Aug 01 2018
%o (GAP) List([0..30],n->n^4+(n+1)^4); # _Muniru A Asiru_, Aug 02 2018
%Y Cf. A002061, A047838, A152913.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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