A266087 Alternating sum of 11-gonal (or hendecagonal) numbers.
0, -1, 10, -20, 38, -57, 84, -112, 148, -185, 230, -276, 330, -385, 448, -512, 584, -657, 738, -820, 910, -1001, 1100, -1200, 1308, -1417, 1534, -1652, 1778, -1905, 2040, -2176, 2320, -2465, 2618, -2772, 2934, -3097, 3268, -3440, 3620, -3801, 3990, -4180
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..5000
- OEIS Wiki, Figurate numbers
- Index entries for linear recurrences with constant coefficients, signature (-2,0,2,1).
Programs
-
Magma
[(18*(-1)^n*n^2 + 4*(-1)^n*n - 7*(-1)^n + 7)/8: n in [0..50]]; // Vincenzo Librandi, Dec 21 2015
-
Mathematica
Table[((18 n^2 + 4 n - 7) (-1)^n + 7)/8, {n, 0, 43}] CoefficientList[Series[(x - 8 x^2)/(x^4 + 2 x^3 - 2 x - 1), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 21 2015 *) Accumulate[Times@@@Partition[Riffle[PolygonalNumber[11,Range[0,50]],{1,-1},{2,-1,2}],2]] (* Requires Mathematica version 10 or later *) (* or *) LinearRecurrence[{-2,0,2,1},{0,-1,10,-20},50] (* Harvey P. Dale, Aug 27 2019 *) -
PARI
x='x+O('x^100); concat(0, Vec(-x*(1-8*x)/((1-x)*(1+x)^3))) \\ Altug Alkan, Dec 21 2015
Formula
G.f.: -x*(1 - 8*x)/((1 - x)*(1 + x)^3).
a(n) = ((18*n^2 + 4*n - 7)*(-1)^n + 7)/8.
a(n) = Sum_{k = 0..n} (-1)^k*A051682(k).
Lim_{n -> infinity} a(n + 1)/a(n) = -1.
E.g.f.: (1/4)*(9*x^2 - 11*x)*cosh(x) - (1/4)*(9*x^2 - 11*x - 7)*sinh(x). - G. C. Greubel, Jan 27 2016
Comments