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.

A096022 Numbers that are congruent to {15, 27, 39, 51} mod 60.

Original entry on oeis.org

15, 27, 39, 51, 75, 87, 99, 111, 135, 147, 159, 171, 195, 207, 219, 231, 255, 267, 279, 291, 315, 327, 339, 351, 375, 387, 399, 411, 435, 447, 459, 471, 495, 507, 519, 531, 555, 567, 579, 591, 615, 627, 639, 651, 675, 687, 699, 711, 735, 747, 759, 771, 795
Offset: 1

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Author

Klaus Brockhaus, Jun 15 2004

Keywords

Comments

Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to 2 and (n+3) mod 5 <> 1.
This is one of a family of sequences which are defined (or could be defined) according to the same scheme: Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to k-1 and (n+k) mod (2+k) <> 1. We have A007310 for k = 1, A017629 for k = 2, this one (A096022) for k = 3, A096023 for k = 5, A096024 for k = 6, A096025 for k = 7, A096026 for k = 9, A096027 for k = 11. Remarkably these sequences are empty for k = 4, 8, 10, ... (i.e., if k+1 is a term of A080765).
Numbers n such that n mod 12 = 3 and n mod 60 <> 3.
Subsequence of A017557: 12n+3.

Examples

			51 mod 2 = 52 mod 3 = 53 mod 4 = 1 and 54 mod 5 = 4, hence 51 is in the sequence; 3 mod 2 = 4 mod 3 = 5 mod 4 = 6 mod 5 = 1, hence 3 is not in the sequence.
		

Crossrefs

Programs

  • Magma
    [ n : n in [1..1500] | n mod 60 in [15, 27, 39, 51] ] // Vincenzo Librandi, Mar 24 2011
  • Maple
    A096022:=n->3*(10*n-3-I^(2*n)-(1-I)*I^(-n)-(1+I)*I^n)/2: seq(A096022(n), n=1..80); # Wesley Ivan Hurt, Jun 04 2016
  • Mathematica
    Table[3*(10n-3-I^(2n)-(1-I)*I^(-n)-(1+I)*I^n)/2, {n, 80}] (* Wesley Ivan Hurt, Jun 04 2016 *)
  • PARI
    {k=3;m=800;for(n=1,m,j=0;b=1;while(b&&j
    				

Formula

G.f.: 3*x*(5+4*x+4*x^2+4*x^3+3*x^4) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jun 04 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = 3*(10*n-3-i^(2*n)-(1-i)*i^(-n)-(1+i)*i^n)/2 where i=sqrt(-1). (End)
E.g.f.: 3*(3 + sin(x) - cos(x) + (5*x - 1)*sinh(x) - (2 - 5*x)*cosh(x)). - Ilya Gutkovskiy, Jun 05 2016

Extensions

New definition from Ralf Stephan, Dec 01 2004