A165248 Quintisection A061037(5*n+2).
0, 45, 35, 285, 30, 725, 255, 1365, 110, 2205, 675, 3245, 240, 4485, 1295, 5925, 420, 7565, 2115, 9405, 650, 11445, 3135, 13685, 930, 16125, 4355, 18765, 1260, 21605, 5775, 24645, 1640, 27885, 7395, 31325, 2070, 34965, 9215, 38805, 2550, 42845, 11235, 47085
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..5000
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).
Programs
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Magma
m:=25; R
:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 + 4*x^7 + 33*x^8 + 3*x^9 + x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))); // G. C. Greubel, Sep 19 2018 -
Mathematica
CoefficientList[Series[5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 +4*x^7 + 33*x^8 + 3*x^9 +x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 19 2018 *) LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,45,35,285,30,725,255,1365,110,2205,675,3245},50] (* Harvey P. Dale, Sep 18 2021 *) -
PARI
a(n) = numerator(1/4 - 1/(5*n+2)^2); \\ Altug Alkan, Apr 19 2016
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PARI
x='x+O('x^50); concat([0], Vec(5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 + 4*x^7 + 33*x^8 + 3*x^9 + x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))) \\ G. C. Greubel, Sep 19 2018
Formula
Conjecture: a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>11. - R. J. Mathar, Mar 02 2010
The conjecture is equivalent to a(4n) = 5n*(5n+1), a(4n+1) = 5*(20n+9)*(4n+1), a(4n+2) = 5*(10n+7)*(2n+1) and a(4n+3) = 5*(20n+19)*(4n+3). - R. J. Mathar, Feb 13 2011
The conjectures can be proved by taking the closed form of A061037, and writing up the quadrisections case by case. - Bruno Berselli, Feb 20 2011
From Ilya Gutkovskiy, Apr 19 2016: (Start)
G.f.: 5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 + 4*x^7 + 33*x^8 + 3*x^9 + x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = -5*n (5*n + 4)*(27*(-1)^n + 6*cos((Pi*n)/2) - 37)/64. (End)
Extensions
Extended by R. J. Mathar, Mar 02 2010
Comments