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 1-9 of 9 results.

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

A133635 Nonprime numbers k such that binomial(k+p,k) mod k = 1, where p=5.

Original entry on oeis.org

26, 34, 49, 58, 74, 77, 82, 91, 98, 106, 119, 121, 122, 133, 143, 146, 154, 161, 169, 178, 187, 194, 202, 203, 209, 217, 218, 221, 226, 242, 247, 253, 259, 266, 274, 287, 289, 298, 299, 301, 314, 319, 322, 323, 329, 338, 341, 343, 346, 361, 362, 371, 377, 386
Offset: 1

Views

Author

Hieronymus Fischer, Sep 30 2007

Keywords

Links

Crossrefs

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

Programs

  • Mathematica
    nn=400;With[{c=Complement[Range[nn],Prime[Range[PrimePi[nn]]]]}, Select[ c,Mod[Binomial[#+5,#],#]==1&]] (* Harvey P. Dale, Sep 24 2012 *)
  • Python
    from itertools import count, islice
from math import comb
from sympy import isprime
def A133635_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda k:comb(k+5,k)%k==1 and not isprime(k),count(max(startvalue,1)))
A133635_list = list(islice(A133635_gen(),30)) # Chai Wah Wu, Feb 22 2023

A133885 Binomial(n+5,n) mod 5^2.

Original entry on oeis.org

1, 6, 21, 6, 1, 2, 12, 17, 12, 2, 3, 18, 13, 18, 3, 4, 24, 9, 24, 4, 5, 5, 5, 5, 5, 6, 11, 1, 11, 6, 7, 17, 22, 17, 7, 8, 23, 18, 23, 8, 9, 4, 14, 4, 9, 10, 10, 10, 10, 10, 11, 16, 6, 16, 11, 12, 22, 2, 22, 12, 13, 3, 23, 3, 13, 14, 9, 19, 9, 14, 15, 15, 15, 15, 15, 16, 21, 11, 21, 16, 17
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 5^3=125.

Links

  • Ray Chandler, Table of n, a(n) for n = 0..1000
  • Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1).

Crossrefs

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.
Cf. A133875, A133880, A133890, A133900, A133910.
For the sequence regarding binomial(n+5, n) mod 5 see A133875.

Programs

  • Mathematica
    Table[Mod[Binomial[n+5,n],25],{n,0,90}] (* Harvey P. Dale, Jan 12 2023 *)

Formula

a(n)=binomial(n+5,5) mod 5^2.
G.f. g(x)=sum{0<=k<125, a(k)*x^k}/(1-x^125).

A362686 Binomial(n+p, n) mod n where p=6.

Original entry on oeis.org

0, 0, 0, 2, 2, 0, 1, 3, 1, 8, 1, 0, 1, 8, 9, 5, 1, 10, 1, 10, 15, 12, 1, 15, 6, 14, 1, 8, 1, 12, 1, 9, 12, 18, 8, 10, 1, 20, 27, 19, 1, 36, 1, 12, 10, 24, 1, 45, 1, 36, 18, 14, 1, 28, 12, 15, 39, 30, 1, 48, 1, 32, 1, 17, 14, 12, 1, 18, 24, 50, 1, 19, 1, 38
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, A362687, A362688, A362689.

Programs

  • Mathematica
    Table[Mod[Binomial[n+6, n], n], {n, 90}]

Formula

a(n)=binomial(n+6,n) mod n.
For n > 1452, a(n) = 2*a(n-720) - a(n-1440).

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).

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.

A133895 Numbers m such that binomial(m+5,m) mod 5 = 0.

Original entry on oeis.org

20, 21, 22, 23, 24, 45, 46, 47, 48, 49, 70, 71, 72, 73, 74, 95, 96, 97, 98, 99, 120, 121, 122, 123, 124, 145, 146, 147, 148, 149, 170, 171, 172, 173, 174, 195, 196, 197, 198, 199, 220, 221, 222, 223, 224, 245, 246, 247, 248, 249, 270, 271, 272, 273, 274, 295
Offset: 0

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Also numbers m such that floor(1+(m/5)) mod 5 = 0.
Partial sums of the sequence 20,1,1,1,1,21,1,1,1,1, 21, ... which has period 5.

Links

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133875, A133885, A133890, A133900, A133910.

Formula

a(n)=5n+20-4*(n mod 5).
G.f.: g(x)=(20+x+x^2+x^3+x^4+x^5)/((1-x^5)(1-x)).
G.f.: g(x)=(20-19x-x^6) /((1-x^5)(1-x)^2).
Showing 1-9 of 9 results.

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