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.

A133879 n modulo 9 repeated 9 times.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 9^2=81.

Crossrefs

Programs

  • Mathematica
    Rest[Flatten[Table[Table[Table[n,{9}],{n,0,8}],{3}],1]]//Flatten (* Harvey P. Dale, Apr 15 2018 *)

Formula

a(n)=(1+floor(n/9)) mod 9.
a(n)=1+floor(n/9)-9*floor((n+9)/81).
a(n)=(((n+9) mod 81)-(n mod 9))/9.
a(n)=((n+9-(n mod 9))/9) mod 9.
G.f. g(x)=(1-x^9)(1+2x^9+3x^18+4x^27+5x^36+6x^45+7x^56+8x^63)/((1-x)(1-x^81)).
G.f. g(x)=(1-x^9)*sum{0<=k<8, (k+1)*x^(9*k)}/((1-x)(1-x^81)).
G.f. g(x)=(8x^81-9x^72+1)/((1-x)(1-x^9)(1-x^81)).

A133891 a(n) = binomial(n+p,n) mod p, where p=12.

Original entry on oeis.org

1, 1, 7, 11, 8, 8, 0, 0, 6, 2, 2, 2, 4, 4, 4, 0, 3, 3, 9, 9, 0, 0, 0, 0, 0, 0, 0, 4, 4, 4, 8, 8, 5, 9, 3, 3, 8, 8, 8, 4, 10, 10, 6, 6, 0, 0, 0, 0, 3, 3, 9, 9, 0, 0, 4, 4, 4, 8, 8, 8, 0, 0, 0, 8, 5, 5, 7, 7, 4, 0, 0, 0, 6, 6, 6, 6, 0, 0, 0, 0, 3, 7, 1, 1, 8, 8, 8, 0, 0, 0, 8, 8, 8, 4, 4, 4, 9, 9, 3, 3, 0, 0, 0, 0
Offset: 0

Views

Author

Hieronymus Fischer, Oct 16 2007

Keywords

Comments

Periodic with length 6*12^2 = 864 = A133900(12).

Crossrefs

See A133872, A133873, A133875, A133877, A133884, A133886, A133888, A133889, A133890 for sequences with different values of p.
See A133900 for the respective periods regarding other values of p.

Programs

  • Mathematica
    Table[Mod[Binomial[n+12,n],12],{n,0,110}] (* Harvey P. Dale, Oct 13 2017 *)

Formula

a(n) = binomial(n+12,12) mod 12.

A133899 Numbers m such that binomial(m+9,m) mod 9 = 0.

Original entry on oeis.org

72, 73, 74, 75, 76, 77, 78, 79, 80, 153, 154, 155, 156, 157, 158, 159, 160, 161, 234, 235, 236, 237, 238, 239, 240, 241, 242, 315, 316, 317, 318, 319, 320, 321, 322, 323, 396, 397, 398, 399, 400, 401, 402, 403, 404, 477, 478, 479, 480, 481, 482, 483, 484, 485
Offset: 0

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Also numbers m such that floor(1+(m/9)) mod 9 = 0.
Partial sums of the sequence 72,1,1,1,1,1,1,1,1,73,1,1,1,1,1,1,1,1,73, ... which has period 9.

Crossrefs

Programs

  • Mathematica
    Select[Range[500],Divisible[Binomial[#+9,#],9]&] (* Harvey P. Dale, Apr 03 2011 *)

Formula

a(n)=9n+72-8*(n mod 9).
G.f.: g(x)=(72+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)/((1-x^9)(1-x)).
G.f.: g(x)=(72-71x-x^10) /((1-x^9)(1-x)^2).
Showing 1-3 of 3 results.