cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A131372 Period 7: repeat [1, -1, 0, 1, 0, -1, 1].

Original entry on oeis.org

1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1, 1, -1, 0, 1, 0, -1, 1
Offset: 0

Views

Author

Paul Curtz, Oct 01 2007

Keywords

Examples

			G.f. = 1 - x + x^3 - x^5 + x^6 + x^7 - x^8 + x^10 - x^12 + x^13 + x^14 + ...
		

Crossrefs

Programs

Formula

From Wesley Ivan Hurt, Dec 23 2016: (Start)
G.f.: (1 - x + x^3 - x^5 + x^6)/(1 - x^7).
a(n) = a(n-7) for all n in Z.
a(n) = 1 - floor((1-n)/7) + floor((3-n)/7) - floor((5-n)/7) + floor((6-n)/7) + floor((7-n)/7) + floor((n-7)/7) + floor((n-6)/7) - floor((n-5)/7) + floor((n-3)/7) - floor((n-1)/7). (End)
G.f.: 1 / (1 + x / (1 - x / (1 + x / (1 - x / (1 - x^2 / (1 + x^4 / (1 + x^3))))))). - Michael Somos, Dec 26 2016
a(n) = (-1)^(mod(mod(n, 7), 3)>0) * A098457(n+1). - Michael Somos, Dec 26 2016

A271842 Positive numbers m such that m^2 - 1 divides 4^m - 1.

Original entry on oeis.org

2, 4, 6, 16, 36, 52, 66, 256, 378, 456, 1296, 1470, 1548, 1800, 2002, 2556, 4356, 6480, 8008, 11952, 23580, 26320, 33930, 36636, 37170, 43290, 44100, 47520, 47880, 49680, 57240, 65536, 74448, 84420, 97812, 101920, 127050, 134946, 139860, 141156, 157080, 164880, 165600, 209220, 225456
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Apr 15 2016

Keywords

Examples

			2 is in this sequence because (4^2 - 1)/(2^2 - 1) = 5.
		

Crossrefs

Cf. positive numbers m such that m^2 - 1 divides (2^k)^m - 1:
A247219 (k=1), this sequence (k=2), A242062 (k=3).

Programs

  • Magma
    [0] cat [n: n in [2..240000] | Denominator((4^n-1)/(n^2-1)) eq 1];
    
  • Maple
    A271842:=n->`if`((4^n-1) mod (n^2-1) = 0, n, NULL): seq(A271842(n), n=2..10^4); # Wesley Ivan Hurt, Apr 18 2016
  • Mathematica
    Select[Range[1, 100], IntegerQ[(4^# - 1)/(#^2 - 1)] &] (* G. C. Greubel, Apr 15 2016 *)
  • PARI
    is(n)=Mod(4,n^2-1)^n==1 \\ Charles R Greathouse IV, Apr 15 2016
Showing 1-2 of 2 results.