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%I #28 Nov 12 2023 05:33:18
%S 1,2,11,22,29,26,7,-34,-103,-206,-349,-538,-779,-1078,-1441,-1874,
%T -2383,-2974,-3653,-4426,-5299,-6278,-7369,-8578,-9911,-11374,-12973,
%U -14714,-16603,-18646,-20849,-23218,-25759,-28478,-31381,-34474,-37763,-41254
%N a(n) = -n^3 + 7*n^2 - 5*n + 1.
%C {a(k): 0 <= k < 4} = divisors of 22:
%C a(n) = A027750(A006218(21) + k + 1), 0 <= k < A000005(22).
%H Vincenzo Librandi, <a href="/A161708/b161708.txt">Table of n, a(n) for n = 0..10000</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) + C(n,1) + 8*C(n,2) - 6*C(n,3).
%F G.f.: -(-1+2*x-9*x^2+14*x^3)/(-1+x)^4. - _R. J. Mathar_, Jun 18 2009
%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) with a(0)=1, a(1)=2, a(2)=11, a(3)=22. - _Harvey P. Dale_, Nov 12 2013
%F E.g.f.: (-x^3 + 4*x^2 + x + 1)*exp(x). - _G. C. Greubel_, Jul 16 2017
%e Differences of divisors of 22 to compute the coefficients of their interpolating polynomial, see formula:
%e 1 2 11 22
%e 1 9 11
%e 8 2
%e -6
%t Table[-n^3+7n^2-5n+1,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,2,11,22},40] (* _Harvey P. Dale_, Nov 12 2013 *)
%o (Magma) [-n^3 + 7*n^2 - 5*n + 1: n in [0..40]]; // _Vincenzo Librandi_, Jul 17 2011
%o (PARI) a(n)=-n^3+7*n^2-5*n+1 \\ _Charles R Greathouse IV_, Sep 24 2015
%Y Cf. A005408, A000124, A016813, A086514, A000125, A058331, A002522, A161701, A161702, A161703, A000127, A161704, A161706, A161707, A161710, A080856, A161711, A161712, A161713, A161715, A006261.
%K sign,easy
%O 0,2
%A _Reinhard Zumkeller_, Jun 17 2009
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