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A270694
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Alternating sum of centered heptagonal pyramidal numbers.
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1
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0, -1, 8, -23, 51, -94, 157, -242, 354, -495, 670, -881, 1133, -1428, 1771, -2164, 2612, -3117, 3684, -4315, 5015, -5786, 6633, -7558, 8566, -9659, 10842, -12117, 13489, -14960, 16535, -18216, 20008, -21913, 23936, -26079, 28347, -30742, 33269
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OFFSET
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0,3
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COMMENTS
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More generally, the ordinary generating function for the alternating sum of centered k-gonal pyramidal numbers is -x*(1 - (k - 2)*x + x^2)/((1 - x)*(1 + x)^4).
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LINKS
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FORMULA
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G.f.: -x*(1 - 5*x + x^2)/((1 - x)*(1 + x)^4).
a(n) = -3*a(n-1) - 2*a(n-2) + 2*a(n-3) + 3*a(n-4) + a(n-5).
a(n) = ((-1)^n*(2*n + 1)*(14*n^2 + 14*n - 9) + 9)/48.
E.g.f.: (1/48)*(9*exp(x) - (9 + 66*x - 126*x^2 + 28*x^3)*exp(-x)). - G. C. Greubel, Mar 28 2016
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MAPLE
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MATHEMATICA
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LinearRecurrence[{-3, -2, 2, 3, 1}, {0, -1, 8, -23, 51}, 39]
Table[((-1)^n (2 n + 1) (14 n^2 + 14 n - 9) + 9)/48, {n, 0, 38}]
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PROG
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(PARI) my(x='x+O('x^50)); concat(0, Vec(-x*(1-5*x+x^2)/((1-x)*(1+x)^4))) \\ Altug Alkan, Mar 21 2016
(Magma) [((-1)^n*(2*n + 1)*(14*n^2 + 14*n - 9) + 9)/48 : n in [0..40]]; // Wesley Ivan Hurt, Mar 21 2016
(Sage) [((-1)^n*(2*n+1)*(14*n^2+14*n-9) +9)/48 for n in (0..40)] # G. C. Greubel, Apr 02 2021
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CROSSREFS
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Cf. A004126 (centered heptagonal pyramidal numbers).
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KEYWORD
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easy,sign
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AUTHOR
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STATUS
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approved
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