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Revision History for A152104

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Showing entries 1-10 | older changes
Quartic product sequence: a(n) = 2^n*Product_{k=1..(n-1)/2} (1 + m*cos(k*Pi/n)^2 + q*cos(k*Pi/n)^4), with m=6, q=4.
(history; published version)
#17 by Charles R Greathouse IV at Thu Sep 08 08:45:39 EDT 2022
PROG

(MAGMAMagma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!( 1+2*(1-x)*x*(1+x)/(1-2*x-8*x^2-2*x^3+x^4) )); // G. C. Greubel, May 15 2019

Discussion
Thu Sep 08
08:45
OEIS Server: https://oeis.org/edit/global/2944
#16 by Giovanni Resta at Wed May 15 15:22:23 EDT 2019
STATUS

reviewed

approved

#15 by Michel Marcus at Wed May 15 12:37:00 EDT 2019
STATUS

proposed

reviewed

#14 by G. C. Greubel at Wed May 15 12:35:20 EDT 2019
STATUS

editing

proposed

#13 by G. C. Greubel at Wed May 15 12:34:44 EDT 2019
NAME

Quartic product sequence: a(n) = 2^n*productProduct_{k=1..(n-1)/2} (1 + m*cos(k*Pi/n)^2 + lq*cos(k*Pi/n)^4, k=1..(n-1)/2), with m=6, lq=4.

COMMENTS

a(n+1)/a(n) tends to 1/2*(1 + sqrt(11) + sqrt(2*(4+sqrt(11)))) /2 = 4.070983928708143809... - Vaclav Kotesovec, Nov 30 2012

FORMULA

G.f.: 1 + 2*(1-x)*x*(1+x)/(1-2*x-8*x^2-2*x^3+x^4). - Vaclav Kotesovec, Nov 30 2012

MATHEMATICA

With[{m = 6; l , q = 4; b = }, Table[2^n*Round[Product[1 + m*Cos[k*Pi/n]^2 + lq*Cos[k*Pi/n]^4, {k, 1, (n - 1)/2}], ], {n, 0, 30}]; FullSimplify[ExpandAll[%]] Round[b] (* modified by _G. C. Greubel_, May 15 2019 *)

PROG

(PARI) my(x='x+O('x^30)); Vec(1+2*(1-x)*x*(1+x)/(1-2*x-8*x^2-2*x^3+x^4)) \\ G. C. Greubel, May 15 2019

(MAGMA) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!( 1+2*(1-x)*x*(1+x)/(1-2*x-8*x^2-2*x^3+x^4) )); // G. C. Greubel, May 15 2019

(Sage) (1+2*(1-x)*x*(1+x)/(1-2*x-8*x^2-2*x^3+x^4)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 15 2019

CROSSREFS
STATUS

approved

editing

#12 by Bruno Berselli at Sun May 26 04:20:39 EDT 2013
STATUS

editing

approved

#11 by Bruno Berselli at Sun May 26 04:19:16 EDT 2013
NAME

Quartic product sequence: m = 6; l = 4; a(n) = 2^n*Product[product(1 + m*Cos[cos(k*Pi/n])^2 + l*Cos[cos(k*Pi/n])^4, {k, =1, ..(n - 1)/2}]), with m=6, l=4.

FORMULA

m = 6; l = 4; a(n)=2^n*Product[1 + m*Cos[k*Pi/n]^2 + l*Cos[k*Pi/n]^4, {k, 1, (n - 1)/2}].

KEYWORD

nonn,easy,changed

STATUS

approved

editing

#10 by Bruno Berselli at Sun May 26 04:14:31 EDT 2013
STATUS

reviewed

approved

#9 by Giovanni Resta at Sun May 26 04:11:29 EDT 2013
STATUS

proposed

reviewed

#8 by Vincenzo Librandi at Sun May 26 03:11:26 EDT 2013
STATUS

editing

proposed