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a(n) = (-1)^n*(n^8 + 4*n^7 - 14*n^5 + 28*n^3 - 17*n)/2.
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%I #34 Sep 08 2022 08:46:13

%S 0,-1,255,-6306,59230,-331395,1348221,-4416580,12360636,-30686085,

%T 69313915,-145044966,284936730,-530793991,944995065,-1617895560,

%U 2677071736,-4298685705,6721274871,-10262288170,15337711830,-22485147531,32390726005,-45920259276,64155054900,-88432835725

%N a(n) = (-1)^n*(n^8 + 4*n^7 - 14*n^5 + 28*n^3 - 17*n)/2.

%C Alternating sum of eighth powers (A001016).

%C For n>0, a(n) is divisible by A000217(n).

%H Robert Israel, <a href="/A261032/b261032.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (-9,-36,-84,-126,-126,-84,-36,-9,-1 ).

%F G.f.: -x*(1 - 246*x + 4047*x^2 - 11572*x^3 + 4047*x^4 - 246*x^5 + x^6)/(1 + x)^9.

%F a(n) = Sum_{k = 0..n} (-1)^k*k^8.

%F a(n) = (-1)^n*n*(n + 1)*(n^6 + 3*n^5 - 3*n^4 - 11*n^3 + 11*n^2 + 17*n - 17)/2.

%F Sum_{n>0} 1/a(n) = -0.9962225712723456482...

%F Sum_{j=0..9} binomial(9,j)*a(n-j) = 0. - _Robert Israel_, Nov 18 2015

%F E.g.f.: (x/2)*(-2 +253*x -1848*x^2 +2961*x^3 -1596*x^4 +350*x^5 -32*x^6 +x^7)*exp(-x). - _G. C. Greubel_, Apr 02 2021

%e a(0) = 0^8 = 0,

%e a(1) = 0^8 -1^8 = -1,

%e a(2) = 0^8 -1^8 + 2^8 = 255,

%e a(3) = 0^8 -1^8 + 2^8 - 3^8 = -6306,

%e a(4) = 0^8 -1^8 + 2^8 - 3^8 + 4^8 = 59230,

%e a(5) = 0^8 -1^8 + 2^8 - 3^8 + 4^8 - 5^8 = -331395, etc.

%p seq((-1)^n*(n^8 + 4*n^7 - 14*n^5 + 28*n^3 - 17*n)/2, n = 0 .. 100); # _Robert Israel_, Nov 18 2015

%t Table[(1/2) (-1)^n n (n + 1) (n^6 + 3 n^5 - 3 n^4 - 11 n^3 + 11 n^2 + 17 n - 17), {n, 0, 25}]

%o (PARI) vector(100, n, n--; (-1)^n*(n^8+4*n^7-14*n^5+28*n^3-17*n)/2) \\ _Altug Alkan_, Nov 18 2015

%o (Magma) [(-1)^n*(n^8+4*n^7-14*n^5+28*n^3-17*n)/2: n in [0..30]]; // _Vincenzo Librandi_, Nov 20 2015

%o (Sage) [(-1)^n*(n^8 +4*n^7 -14*n^5 +28*n^3 -17*n)/2 for n in (0..40)] # _G. C. Greubel_, Apr 02 2021

%Y Cf. A000217, A001016, A000542, A089594, A232599, A062392, A062393, A152725, A152726.

%K sign,easy

%O 0,3

%A _Ilya Gutkovskiy_, Nov 18 2015