%I #30 Jul 10 2022 09:42:44
%S 1,3,7,21,53,111,203,337,521,763,1071,1453,1917,2471,3123,3881,4753,
%T 5747,6871,8133,9541,11103,12827,14721,16793,19051,21503,24157,27021,
%U 30103,33411,36953,40737,44771,49063,53621,58453,63567,68971,74673
%N a(n) = (4*n^3 - 9*n^2 + 11*n + 3)/3.
%C {a(k): 0 <= k < 4} = divisors of 21:
%C a(n) = A027750(A006218(20) + k + 1), 0 <= k < A000005(21).
%H G. C. Greubel, <a href="/A161707/b161707.txt">Table of n, a(n) for n = 0..1000</a>
%H Reinhard Zumkeller, <a href="/A161700/a161700.txt">Enumerations of Divisors</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = C(n,0) + 2*C(n,1) + 2*C(n,2) + 8*C(n,3).
%F G.f.: (7*x^3 + x^2 - x + 1)/(x-1)^4. - _Harvey P. Dale_, Mar 28 2011
%F E.g.f.: (1/3)*(4*x^3 + 3*x^2 + 6*x + 3)*exp(x). - _G. C. Greubel_, Jul 16 2017
%e Differences of divisors of 21 to compute the coefficients of their interpolating polynomial, see formula:
%e 1 3 7 21
%e 2 4 14
%e 2 10
%e 8
%p A161707:=n->(4*n^3 - 9*n^2 + 11*n + 3)/3: seq(A161707(n), n=0..100); # _Wesley Ivan Hurt_, Jan 19 2017
%t Table[(4n^3-9n^2+11n+3)/3,{n,0,40}] (* or *)
%t CoefficientList[Series[(7x^3+x^2-x+1)/(x-1)^4, {x,0,60}], x] (* _Harvey P. Dale_, Mar 28 2011 *)
%o (Magma) [(4*n^3 - 9*n^2 + 11*n + 3)/3: n in [0..50]]; // _Vincenzo Librandi_, Dec 27 2010
%o (PARI) a(n)=(4*n^3-9*n^2+11*n)/3+1 \\ _Charles R Greathouse IV_, Jul 16 2011
%Y Cf. A000005, A006218, A027750.
%K nonn,easy
%O 0,2
%A _Reinhard Zumkeller_, Jun 17 2009