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%I #32 Sep 08 2022 08:45:45
%S 1,2,3,4,6,12,28,64,135,262,473,804,1300,2016,3018,4384,6205,8586,
%T 11647,15524,20370,26356,33672,42528,53155,65806,80757,98308,118784,
%U 142536,169942,201408,237369,278290,324667,377028,435934,501980,575796,658048
%N a(n) = (n^5 - 5*n^4 + 5*n^3 + 5*n^2 + 114*n + 120)/120.
%C {a(k): 0 <= k < 6} = divisors of 12:
%C a(n) = A027750(A006218(11) + k + 1), 0 <= k < A000005(12).
%H Vincenzo Librandi, <a href="/A161701/b161701.txt">Table of n, a(n) for n = 0..10000</a>
%H R. Zumkeller, <a href="/A161700/a161700.txt">Enumerations of Divisors</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 a(n) = C(n,0) + C(n,1) + C(n,4) + C(n,5).
%F G.f.: (1-4*x+6*x^2-4*x^3+2*x^4)/(1-x)^6. - _Colin Barker_, Aug 20 2012
%e Differences of divisors of 12 to compute the coefficients of their interpolating polynomial, see formula:
%e 1 2 3 4 6 12
%e 1 1 1 2 6
%e 0 0 1 4
%e 0 1 3
%e 1 2
%e 1
%p A161701:=n->(n^5 - 5*n^4 + 5*n^3 + 5*n^2 + 114*n + 120)/120: seq(A161701(n), n=0..60); # _Wesley Ivan Hurt_, Jul 16 2017
%t CoefficientList[Series[(1-4*x+6*x^2-4*x^3+2*x^4)/(1-x)^6, {x, 0, 50}], x] (* _G. C. Greubel_, Jul 16 2017 *)
%o (Magma) [(n^5 - 5*n^4 + 5*n^3 + 5*n^2 + 114*n + 120)/120: n in [0..50]]; // _Vincenzo Librandi_, Dec 27 2010
%o (PARI) a(n)=(n^5-5*n^4+5*n^3+5*n^2+114*n+120)/120 \\ _Charles R Greathouse IV_, Sep 24 2015
%Y Cf. A000124, A000125, A000127, A002522, A005408, A006261, A016813, A058331, A080856, A086514, A161702, A161703, A161704, A161706, A161707, A161708, A161710, A161711, A161712, A161713, A161715.
%K nonn,easy
%O 0,2
%A _Reinhard Zumkeller_, Jun 17 2009
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