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A161703 a(n) = (4*n^3 - 12*n^2 + 14*n + 3)/3.

%I A161703 #28 Oct 04 2024 19:29:37
%S A161703 1,3,5,15,41,91,173,295,465,691,981,1343,1785,2315,2941,3671,4513,
%T A161703 5475,6565,7791,9161,10683,12365,14215,16241,18451,20853,23455,26265,
%U A161703 29291,32541,36023,39745,43715,47941,52431,57193,62235,67565,73191,79121
%N A161703 a(n) = (4*n^3 - 12*n^2 + 14*n + 3)/3.
%C A161703 {a(k): 0 <= k < 4} = divisors of 15:
%C A161703 a(n) = A027750(A006218(14) + k + 1), 0 <= k < A000005(15).
%H A161703 G. C. Greubel, <a href="/A161703/b161703.txt">Table of n, a(n) for n = 0..1000</a>
%H A161703 R. Zumkeller, <a href="/A161700/a161700.txt">Enumerations of Divisors</a>
%H A161703 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A161703 a(n) = C(n,0) + 2*C(n,1) + 8*C(n,3).
%F A161703 G.f.: (1-x-x^2+9*x^3)/(1-x)^4. - _Colin Barker_, Jan 08 2012
%e A161703 Differences of divisors of 15 to compute the coefficients of their interpolating polynomial, see formula:
%e A161703   1     3     5    15
%e A161703      2     2    10
%e A161703         0     8
%e A161703            8
%p A161703 A161703:=n->(4*n^3 - 12*n^2 + 14*n + 3)/3: seq(A161703(n), n=0..100); # _Wesley Ivan Hurt_, Jul 16 2017
%t A161703 CoefficientList[Series[(1 - x - x^2 + 9*x^3)/(1 - x)^4, {x, 0, 50}], x] (* _G. C. Greubel_, Jul 16 2017 *)
%t A161703 LinearRecurrence[{4,-6,4,-1},{1,3,5,15},50] (* _Harvey P. Dale_, Oct 04 2024 *)
%o A161703 (Magma) [(4*n^3 - 12*n^2 + 14*n + 3)/3: n in [0..50]]; // _Vincenzo Librandi_, Dec 27 2010
%o A161703 (PARI) a(n)=n*(4*n^2-12*n+14)/3+1 \\ _Charles R Greathouse IV_, Sep 24 2015
%Y A161703 Cf. A000124, A000125, A000127, A002522, A005408, A006261, A016813, A058331, A080856, A086514, A161701, A161702, A161704, A161706-A161708, A161710, A161711-A161713, A161715.
%K A161703 nonn,easy
%O A161703 0,2
%A A161703 _Reinhard Zumkeller_, Jun 17 2009