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 1-3 of 3 results.

A010691 Period 2: repeat (1,10).

Original entry on oeis.org

1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10
Offset: 0

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Author

Keywords

Comments

Regular continued fraction of (5+sqrt 35)/10. - R. J. Mathar, Nov 21 2011
Sequence is an infinite palindrome in two ways (numbers and English names): ONE, TEN, ONE, TEN, ONE, TEN, ONE, ... . - Eric Angelini, Sep 16 2023

Crossrefs

Programs

  • Magma
    [10^n mod 11: n in [0..80]]; // Vincenzo Librandi, Aug 24 2011
  • Maple
    g:=(1+10*z)/((1-z^2)): gser:=series(g, z=0, 66): seq((coeff(gser, z, n)), n=0..65); # Zerinvary Lajos, Feb 25 2009
  • Mathematica
    PadRight[{},100,{1,10}] (* Harvey P. Dale, Aug 27 2013 *)

Formula

a(n) = -9/2*(-1)^n + 11/2.
G.f.: (1+10*z)/(1-z^2). - Zerinvary Lajos, Feb 25 2009
a(n) = 10^n mod 11. - M. F. Hasler, Mar 10 2011
From Nicolas Bělohoubek, Nov 11 2021: (Start)
a(n) = 10/a(n-1). See also A010695.
a(n) = 11 - a(n-1). See also A010712. (End)

A141281 Primes p such that p-6^4, p-6^3, p-6^2, p-6, p, p+6, p+6^2, p+6^3 and p+6^4 are primes.

Original entry on oeis.org

11459317, 18726137, 73718633, 181975727, 361471043, 374195537, 419533753, 420522673, 428739323, 429198703, 456975157, 483576523, 544795393, 653578573, 682118777, 703313623, 753422317, 764967257, 797492477, 960985037, 1059913073
Offset: 1

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Author

Rick L. Shepherd, Jun 22 2008

Keywords

Comments

Subsequence of A006489, A141279 and A141280. Each term is congruent to 1 or 10 mod 11 so for no prime p can this pattern be extended also to include primes p-6^5 and p+6^5 (one of them is divisible by 11). See A070392 for residues mod 11 of powers of 6. As each term of A006489 greater than 11 is congruent to 3 or 7 mod 10, combining results gives that a(n) is congruent to 23, 43, 67, or 87 mod 110.

Crossrefs

Programs

  • Mathematica
    Select[Prime[Range[53734400]],AllTrue[#+{1296,216,36,6,-6,-36,-216,-1296},PrimeQ]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Apr 01 2021 *)

A187466 a(n) = 9^n mod 11.

Original entry on oeis.org

1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1
Offset: 0

Views

Author

M. F. Hasler, Mar 10 2011

Keywords

Comments

Period 5: repeat [1, 9, 4, 3, 5].

Crossrefs

Programs

Formula

G.f.: (5*x^4 + 3*x^3 + 4*x^2 + 9*x + 1)/(1 - x^5). - Chai Wah Wu, Jun 04 2016
a(n) = a(n-5) for n>4. - Wesley Ivan Hurt, Jun 11 2016
Showing 1-3 of 3 results.