The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A245480

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A245480

Numbers n such that the n-th cyclotomic polynomial has a root mod 11.
(history; published version)

#21 by Harvey P. Dale at Wed Aug 04 11:13:23 EDT 2021

STATUS

editing

approved

#20 by Harvey P. Dale at Wed Aug 04 11:13:19 EDT 2021

MATHEMATICA

LinearRecurrence[{0, 0, 0, 11}, {1, 2, 5, 10}, 40] (* Harvey P. Dale, Aug 04 2021 *)

STATUS

approved

editing

#19 by Charles R Greathouse IV at Tue Aug 02 09:12:59 EDT 2016

STATUS

proposed

approved

#18 by Michel Marcus at Mon Aug 01 17:10:15 EDT 2016

STATUS

editing

proposed

#17 by Michel Marcus at Mon Aug 01 17:10:01 EDT 2016

FORMULA

For example, n=31=2^0+2^1+2^2+2^3+2^4, then a(31+1)=a(2)*a(3)*a(5)*a(9)*a(17) i.e. 194871710=2*5*11*121*14641.

n=31=2^0+2^1+2^2+2^3+2^4, then a(31+1)=a(2)*a(3)*a(5)*a(9)*a(17) i.e. 194871710=2*5*11*121*14641.

STATUS

proposed

editing

#16 by G. C. Greubel at Sat Jul 30 00:40:07 EDT 2016

STATUS

editing

proposed

#15 by G. C. Greubel at Sat Jul 30 00:39:54 EDT 2016

LINKS

<a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,11).

FORMULA

a(n) = 11*a(n-4).

a(n) = 11*a(n-4). G.f.: -x*(1 + 2*x+)*(1)*( + 5*x^2+)/(1)/( - 11*x^4-1).

a(n) appears to satisfy x*Prod_{n>=0} (1 + a(2^n+1)x^(2^n)) = Sum_{n>=1} a(n)*x^n.

Then a(n+1) = a(2^x+1)*a(2^y+1)*a(2^z+1)..., where n=2^x+2^y+2^z+... .

MATHEMATICA

CoefficientList[Series[-x(2x+1)(5x^2+1)/(1-11x^4-1), {x, 0, 20}], x] (* Benedict W. J. Irwin, Jul 24 2016 *)

STATUS

proposed

editing

#14 by Benedict W. J. Irwin at Fri Jul 29 05:32:27 EDT 2016

STATUS

editing

proposed

#13 by Benedict W. J. Irwin at Fri Jul 29 05:32:01 EDT 2016

FORMULA

From Benedict W. J. Irwin, Jul 29 2016: (Start)

a(n) = 11*a(n-4). G.f.: -x*(2*x+1)*(5*x^2+1)/(11*x^4-1). _Benedict W. J. Irwin_, Jul 24 2016

a(n) appears to satisfy x*Prod_{n>=0} (1 + a(2^n+1)x^(2^n)) = Sum_{n>=1} a(n)x^n.

Then a(n+1)=a(2^x+1)a(2^y+1)a(2^z+1)..., where n=2^x+2^y+2^z+... .

For example,

n=31=2^0+2^1+2^2+2^3+2^4, then a(31+1)=a(2)*a(3)*a(5)*a(9)*a(17) i.e. 194871710=2*5*11*121*14641.

(End)

#12 by Benedict W. J. Irwin at Sun Jul 24 12:51:30 EDT 2016

FORMULA

a(n) = 11*a(n-4). G.f.: -x*(2*x+1)*(5*x^2+1)/(11*x^4-1). Benedict W. J. Irwin, Jul 24 2016

MATHEMATICA

CoefficientList[Series[-x(2x+1)(5x^2+1)/(11x^4-1), {x, 0, 20}], x] (* Benedict W. J. Irwin, Jul 24 2016 *)

STATUS

approved

editing