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 1-6 of 6 results.

A133620 Binomial(n+p,n) mod n where p=10.

Original entry on oeis.org

0, 0, 1, 1, 3, 4, 2, 6, 2, 6, 1, 2, 1, 10, 5, 7, 1, 12, 1, 15, 18, 12, 1, 12, 21, 14, 4, 12, 1, 28, 1, 29, 1, 18, 6, 5, 1, 20, 14, 10, 1, 14, 1, 34, 15, 24, 1, 3, 8, 16, 18, 27, 1, 34, 23, 16, 1, 30, 1, 16, 1, 32, 17, 57, 40, 56, 1, 1, 47, 60, 1, 54, 1, 38, 36, 58, 12, 66, 1, 63, 10, 42, 1
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, A133621-A133625, A133630, A038509, A133634, A133635, A133636.
Cf. A133880, A133890, A133900, A133910, A362686, A362687, A362688, A362689.

Programs

  • Mathematica
    Table[Mod[Binomial[n + 10, n], n], {n, 90}] (* Harvey P. Dale, Apr 04 2015 *)
  • PARI
    a(n) = binomial(n+10, n) % n \\ Michel Marcus, Jul 15 2013

Formula

a(n) = binomial(n+p,p) mod n.
a(n) = 1 if n is a prime > p, since binomial(n+p,n)==(1+floor(p/n))(mod n), provided n is a prime.
a(n) = A001287(n+10) mod n. - Michel Marcus, Jul 15 2013; corrected by Michel Marcus, Jan 27 2020
For n > 58060802, a(n) = 2*a(n-29030400) - a(n-58060800). - Ray Chandler, Apr 29 2023

A133625 Binomial(n+p, n) mod n where p=5.

Original entry on oeis.org

0, 1, 2, 2, 2, 0, 1, 7, 4, 3, 1, 8, 1, 8, 9, 13, 1, 7, 1, 10, 8, 12, 1, 3, 6, 1, 10, 8, 1, 2, 1, 25, 12, 1, 8, 22, 1, 20, 14, 39, 1, 15, 1, 12, 25, 24, 1, 5, 1, 11, 18, 14, 1, 46, 12, 43, 20, 1, 1, 48, 1, 32, 22, 49, 14, 23, 1, 18, 24, 50, 1, 7, 1, 1, 41, 20, 1, 66, 1, 77, 28, 1, 1, 50, 18, 44
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.
Cf. A133875, A133885, A133880, A133890, A133900, A133910, A362686, A362687, A362688, A362689.

Programs

  • Mathematica
    Table[Mod[Binomial[n+5,n],n],{n,90}] (* Harvey P. Dale, Oct 02 2015 *)

Formula

a(n)=binomial(n+5,5) mod n.
a(n)=1 if n is a prime > 5, since binomial(n+5,n)==(1+floor(5/n))(mod n), provided n is a prime.
From Chai Wah Wu, May 26 2016: (Start)
a(n) = (n^5 + 15*n^4 + 85*n^3 + 105*n^2 + 34*n + 120)/120 mod n.
For n > 6:
if n mod 120 == 0, then a(n) = 17*n/60 + 1.
if n mod 120 is in {1, 2, 7, 11, 13, 17, 19, 23, 26, 29, 31, 34, 37, 41, 43, 47, 49, 53, 58, 59, 61, 67, 71, 73, 74, 77, 79, 82, 83, 89, 91, 97, 98, 101, 103, 106, 107, 109, 113, 119}, then a(n) = 1.
if n mod 120 is in {3, 9, 18, 21, 27, 33, 39, 42, 51, 57, 63, 66, 69, 81, 87, 93, 99, 111, 114, 117}, then a(n) = n/3 + 1.
if n mod 120 is in {4, 28, 44, 52, 68, 76, 92, 116}, then a(n) = n/4 + 1.
if n mod 120 is in {5, 10, 25, 35, 50, 55, 65, 85, 95, 115}, then a(n) = n/5 + 1.
if n mod 120 is in {6, 54, 78, 102}, then a(n) = 5*n/6 + 1.
if n mod 120 is in {8, 16, 32, 56, 64, 88, 104, 112}, then a(n) = 3*n/4 + 1.
if n mod 120 is in {12, 36, 84, 108}, then a(n) = 7*n/12 + 1.
if n mod 120 is in {14, 22, 38, 46, 62, 86, 94, 118}, then a(n) = n/2 + 1.
if n mod 120 is in {15, 45, 75, 90, 105}, then a(n) = 8*n/15 + 1.
if n mod 120 is in {20, 100}, then a(n) = 9*n/20 + 1.
if n mod 120 is in {24, 48, 72, 96}, then a(n) = n/12 + 1.
if n mod 120 == 30, then a(n) = n/30 + 1.
if n mod 120 is in {40, 80}, then a(n) = 19*n/20 + 1.
if n mod 120 == 60, then a(n) = 47*n/60 + 1.
if n mod 120 is in {70, 110}, then a(n) = 7*n/10 + 1.
(End)
For n > 246, a(n) = 2*a(n-120) - a(n-240). - Ray Chandler, Apr 23 2023

A362687 Binomial(n+p, n) mod n where p=7.

Original entry on oeis.org

0, 0, 0, 2, 2, 0, 2, 3, 1, 8, 1, 0, 1, 10, 9, 5, 1, 10, 1, 10, 18, 12, 1, 15, 6, 14, 1, 12, 1, 12, 1, 9, 12, 18, 13, 10, 1, 20, 27, 19, 1, 0, 1, 12, 10, 24, 1, 45, 8, 36, 18, 14, 1, 28, 12, 23, 39, 30, 1, 48, 1, 32, 10, 17, 14, 12, 1, 18, 24, 60, 1, 19, 1
Offset: 1

Views

Author

Ray Chandler, Apr 29 2023

Keywords

Links

Crossrefs

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.
Cf. A133875, A133885, A133880, A133890, A133900, A133910, A362686, A362688, A362689.

Programs

  • Mathematica
    Table[Mod[Binomial[n+7,n],n],{n,90}]

Formula

a(n)=binomial(n+7,n) mod n.
For n > 10122, a(n) = 2*a(n-5040) - a(n-10080).

A362688 Binomial(n+p, n) mod n where p=8.

Original entry on oeis.org

0, 1, 0, 3, 2, 3, 2, 6, 1, 8, 1, 6, 1, 10, 9, 15, 1, 1, 1, 5, 18, 1, 1, 12, 6, 14, 1, 12, 1, 12, 1, 13, 12, 1, 13, 19, 1, 1, 27, 34, 1, 0, 1, 34, 10, 24, 1, 27, 8, 11, 18, 1, 1, 1, 12, 16, 39, 30, 1, 48, 1, 32, 10, 25, 14, 45, 1, 35, 24, 25, 1, 46, 1, 38, 66
Offset: 1

Views

Author

Ray Chandler, Apr 29 2023

Keywords

Links

Crossrefs

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.
Cf. A133875, A133885, A133880, A133890, A133900, A133910, A362686, A362687, A362689.

Programs

  • Mathematica
    Table[Mod[Binomial[n+8,n],n],{n,90}]

Formula

a(n)=binomial(n+8,n) mod n.
For n > 645240, a(n) = 2*a(n-322560) - a(n-645120).

A362689 Binomial(n+p, n) mod n where p=9.

Original entry on oeis.org

0, 1, 1, 3, 2, 1, 2, 6, 2, 8, 1, 2, 1, 10, 14, 15, 1, 3, 1, 5, 18, 1, 1, 12, 6, 14, 4, 12, 1, 22, 1, 13, 1, 1, 13, 23, 1, 1, 14, 34, 1, 14, 1, 34, 15, 24, 1, 27, 8, 11, 18, 1, 1, 7, 12, 16, 1, 30, 1, 28, 1, 32, 17, 25, 14, 23, 1, 35, 47, 25, 1, 54, 1, 38, 66
Offset: 1

Views

Author

Ray Chandler, Apr 29 2023

Keywords

Links

Crossrefs

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.
Cf. A133875, A133885, A133880, A133890, A133900, A133910, A362686, A362687, A362688.

Programs

  • Mathematica
    Table[Mod[Binomial[n+9,n],n],{n,90}]

Formula

a(n)=binomial(n+9,n) mod n.
For n > 5806081, a(n) = 2*a(n-2903040) - a(n-5806080).

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
Showing 1-6 of 6 results.

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