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%I #19 Sep 08 2022 08:45:45
%S 1,2,3,6,9,18,59,190,513,1186,2435,4566,7977,13170,20763,31502,46273,
%T 66114,92227,125990,168969,222930,289851,371934,471617,591586,734787,
%U 904438,1104041,1337394,1608603,1922094,2282625,2695298,3165571,3699270
%N a(n) = (3*n^5 - 35*n^4 + 145*n^3 - 235*n^2 + 152*n + 30)/30.
%C {a(k): 0 <= k < 6} = divisors of 18:
%C a(n) = A027750(A006218(17) + k + 1), 0 <= k < A000005(18).
%H G. C. Greubel, <a href="/A161704/b161704.txt">Table of n, a(n) for n = 0..1000</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) + 2*C(n,3) - 4*C(n,4) + 12*C(n,5).
%F G.f.: ( 1-4*x+6*x^2-2*x^3-7*x^4+18*x^5 ) / (x-1)^6. - _R. J. Mathar_, Jul 12 2016
%e Differences of divisors of 18 to compute the coefficients of their interpolating polynomial, see formula:
%e 1 2 3 6 9 18
%e 1 1 3 3 9
%e 0 2 0 6
%e 2 -2 6
%e -4 8
%e 12
%p A161704:=n->(3*n^5 - 35*n^4 + 145*n^3 - 235*n^2 + 152*n + 30)/30: seq(A161704(n), n=0..50); # _Wesley Ivan Hurt_, Jul 16 2017
%t CoefficientList[Series[(1 - 4*x + 6*x^2 - 2*x^3 - 7*x^4 + 18*x^5)/(x - 1)^6, {x, 0, 50}], x] (* _G. C. Greubel_, Jul 16 2017 *)
%o (Magma) [(3*n^5 - 35*n^4 + 145*n^3 - 235*n^2 + 152*n + 30)/30: n in [0..50]]; // _Vincenzo Librandi_, Dec 27 2010
%o (PARI) a(n)=n*(3*n^4-35*n^3+145*n^2-235*n+152)/30+1
%Y Cf. A000124, A000125, A000127, A002522, A005408, A006261, A016813, A018251, A058331, A080856, A086514, A161701, A161702, A161703, A161706, A161707, A161708, A161710, A161711, A161712, A161713, A161715, A161856.
%K nonn,easy
%O 0,2
%A _Reinhard Zumkeller_, Jun 17 2009
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