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%I #53 Sep 08 2022 08:45:17
%S 1,20,81,208,425,756,1225,1856,2673,3700,4961,6480,8281,10388,12825,
%T 15616,18785,22356,26353,30800,35721,41140,47081,53568,60625,68276,
%U 76545,85456,95033,105300,116281,128000,140481,153748,167825,182736
%N Number of divisors of 240^n.
%C Geometric interpretation: Take a simple cubical grid of size (2n+1). Number the coordinates along each axis from 1 to (2n+1). Select only the cells that have at least two odd coordinates, and discard the rest. The number of selected cells is a(n). - _Arun Giridhar_, Mar 27 2015
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F From _R. J. Mathar_ and _Stefan Steinerberger_, Aug 31 2008: (Start)
%F a(n) = (4*n+1)*(n+1)^2.
%F G.f.: (1+16x+7x^2)/(1-x)^4.
%F Inverse binomial transform: 1, 19, 42, 24, 0 (0 continued). (End)
%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3. - _Harvey P. Dale_, Jan 21 2013
%F a(n) = (n+1)*A001107(n+1), where A001107 are the partial sums of A017007. - _J. M. Bergot_, Jul 08 2013
%F a(n) = Sum_{i=0..n} (n+1)*(8*i+1). [_Bruno Berselli_, Sep 08 2015]
%F Sum_{n>=0} 1/a(n) = 2*Pi/9 - Pi^2/18 + 4*log(2)/3 = 1.07401658592825... . - _Vaclav Kotesovec_, Oct 04 2016
%e a(2) = 81 because 240^2 has 81 divisors.
%e a(2) = 81 because a 5 X 5 X 5 grid has 81 cells with at least two odd coordinates each, coordinate numbering starting at 1.
%p A103532 := proc(n) (4*n+1)*(n+1)^2 ; end proc: # _R. J. Mathar_, Aug 31 2008
%t Table[(4 n + 1) (n + 1)^2, {n, 0, 40}] (* _Stefan Steinerberger_, Aug 31 2008 *)
%t DivisorSigma[0,240^Range[0,40]] (* or *) LinearRecurrence[{4,-6,4,-1},{1,20,81,208},40] (* _Harvey P. Dale_, Jan 21 2013 *)
%o (Magma) [(4*n+1)*(n+1)^2: n in [0..45]]; // _Vincenzo Librandi_, Feb 10 2016
%Y Cf. similar sequences, with the formula (k*n-k+2)*n^2/2, listed in A262000.
%K nonn,easy
%O 0,2
%A _J. Lowell_, Aug 30 2008
%E More terms from _Stefan Steinerberger_ and _R. J. Mathar_, Aug 31 2008
%E Example corrected by _Harvey P. Dale_, Jan 21 2013
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