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 1-5 of 5 results.

A133634 Nonprime numbers k such that binomial(k+p,k) mod k = 1, where p=4.

Original entry on oeis.org

10, 25, 26, 34, 35, 49, 50, 55, 58, 65, 74, 77, 82, 85, 91, 95, 98, 106, 115, 119, 121, 122, 125, 130, 133, 143, 145, 146, 154, 155, 161, 169, 170, 175, 178, 185, 187, 194, 202, 203, 205, 209, 215, 217, 218, 221, 226, 235, 242, 245, 247, 250, 253, 259, 265, 266
Offset: 1

Views

Author

Hieronymus Fischer, Sep 30 2007

Keywords

Links

Crossrefs

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.
Cf. A133874, A133884, A133880, A133890, A133900, A133910.

Programs

  • Mathematica
    Select[Range[300],!PrimeQ[#]&&Mod[Binomial[#+4,#],#]==1&] (* Harvey P. Dale, Oct 09 2011 *)

A133874 n modulo 4 repeated 4 times.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 4^2 = 16.

Links

Crossrefs

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.
Cf. A133884, A133880, A133890, A133900, A133910.

Programs

Formula

a(n) = (1 + floor(n/4)) mod 4.
a(n) = A010873(A002265(n+4)).
a(n) = 1 + floor(n/4) - 4*floor((n+4)/16).
a(n) = (((n+4) mod 16) - (n mod 4))/4.
a(n) = ((n + 4 - (n mod 4))/4) mod 4.
G.f. g(x) = (1 + x + x^2 + x^3 + 2x^4 + 2x^5 + 2x^6 + 2x^7 + 3x^8 + 3x^9 + 3x^10 + 3x^11)/(1-x^16).
G.f. g(x) = ((1-x^4)*(1+2x^4+3x^8))/((1-x)*(1-x^16)).
G.f. g(x) = (3x^16-4x^12+1)/((1-x)*(1-x^4)*(1-x^16)).
G.f. g(x) = (1+2x^4+3x^8)/((1-x)*(1+x^4)*(1+x^8)).

A133624 Binomial(n+p, n) mod n, where p=4.

Original entry on oeis.org

0, 1, 2, 2, 1, 0, 1, 7, 4, 1, 1, 8, 1, 8, 6, 13, 1, 7, 1, 6, 8, 12, 1, 3, 1, 1, 10, 8, 1, 26, 1, 25, 12, 1, 1, 22, 1, 20, 14, 31, 1, 15, 1, 12, 16, 24, 1, 5, 1, 1, 18, 14, 1, 46, 1, 43, 20, 1, 1, 36, 1, 32, 22, 49, 1, 23, 1, 18, 24, 36, 1, 7, 1, 1, 26, 20, 1, 66, 1, 61, 28, 1, 1, 50, 1, 44, 30
Offset: 1

Views

Author

Hieronymus Fischer, Sep 30 2007

Keywords

Comments

Let d(m)...d(2)d(1)d(0) be the base-n representation of n+p. The relation a(n)=d(1) holds, if n is a prime index. For this reason there are infinitely many terms which are equal to 1.

Links

Crossrefs

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636, A133874, A133884, A133880, A133890, A133900, A133910, A362686, A362687, A362688, A362689.

Programs

  • Magma
    [Binomial(n+4,4) mod n: n in [1..100]]; // Vincenzo Librandi, Apr 27 2014
  • Mathematica
    Table[Mod[Binomial[n+4,n],n],{n,90}] (* Harvey P. Dale, Apr 26 2014 *)

Formula

a(n) = binomial(n+4,4) mod n.
a(n)=1 if n is a prime > 4, since binomial(n+4,n) == (1+floor(4/n))(mod n), provided n is a prime.
From Chai Wah Wu, May 26 2016: (Start)
a(n) = (n^4 + 10*n^3 + 11*n^2 + 2*n + 24)/24 mod n.
For n > 6:
if n mod 24 == 0, then a(n) = n/12 + 1.
if n mod 24 is in {1, 2, 5, 7, 10, 11, 13, 17, 19, 23}, then a(n) = 1.
if n mod 24 is in {3, 9, 15, 18, 21}, then a(n) = n/3 + 1.
if n mod 24 is in {4, 20}, then a(n) = n/4 + 1.
if n mod 24 == 6, then a(n) = 5*n/6 + 1.
if n mod 24 is in {8, 16}, then a(n) = 3*n/4 + 1.
if n mod 24 == 12, then a(n) = 7*n/12 + 1.
if n mod 24 is in {14, 22}, then a(n) = n/2 + 1.
(End)
For n > 54, a(n) = 2*a(n-24) - a(n-48). - Ray Chandler, Apr 23 2023

A133894 Numbers m such that binomial(m+4,m) mod 4 = 0.

Original entry on oeis.org

12, 13, 14, 15, 28, 29, 30, 31, 44, 45, 46, 47, 60, 61, 62, 63, 76, 77, 78, 79, 92, 93, 94, 95, 108, 109, 110, 111, 124, 125, 126, 127, 140, 141, 142, 143, 156, 157, 158, 159, 172, 173, 174, 175, 188, 189, 190, 191, 204, 205, 206, 207, 220, 221, 222, 223, 236, 237
Offset: 1

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Also numbers m such that floor(1+(m/4)) mod 4 = 0.
Partial sums of the sequence 12,1,1,1,13,1,1,1,13, ... which has period 4.
Numbers congruent to {12,13,14,15} mod 16. Numbers n such that n xor 12 = n - 12. [Brad Clardy, May 06 2013]

Links

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635, A133874, A133884, A133890, A133900, A133910.

Programs

Formula

a(n) = 4*n + 12 - 3*(n mod 4).
G.f.: 12/(1-x)+x(1+x+x^2+13x^3)/((1-x^4)(1-x)) = (12+x+x^2+x^3+x^4)/((1-x^4)(1-x)) = (12-11x-x^5)/((1-x^4)(1-x)^2).
a(n) = 4*n+3*((1-i)*i^n+(1+i)*(-i)^n+(-1)^n+5)/2, where i=sqrt(-1). - Bruno Berselli, Apr 08 2011

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

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636, A133910.
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.
Showing 1-5 of 5 results.

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