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.

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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

Crossrefs

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

A133623 Binomial(n+p, n) mod n where p=3.

Original entry on oeis.org

0, 0, 2, 3, 1, 0, 1, 5, 4, 6, 1, 11, 1, 8, 6, 9, 1, 16, 1, 11, 8, 12, 1, 21, 1, 14, 10, 15, 1, 26, 1, 17, 12, 18, 1, 31, 1, 20, 14, 21, 1, 36, 1, 23, 16, 24, 1, 41, 1, 26, 18, 27, 1, 46, 1, 29, 20, 30, 1, 51, 1, 32, 22, 33, 1, 56, 1, 35, 24, 36, 1, 61, 1, 38, 26, 39, 1, 66, 1, 41, 28, 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.
Appears to satisfy the recurrence: a(n) = -2*a(n-1) - a(n-2) + 2*a(n-3) + 4*a(n-4) + 2*a(n-5) - a(n-6) - 2*a(n-7) - a(n-8) for n > 14. - Chai Wah Wu, May 25 2016

Crossrefs

Programs

  • Mathematica
    Table[Mod[Binomial[n+3,n],n],{n,90}] (* Harvey P. Dale, Nov 22 2011 *)

Formula

a(n) = binomial(n+3,3) mod n.
a(n)=1 if n is a prime > 3, since binomial(n+3,n)==(1+floor(3/n))(mod n), provided n is a prime.
From Chai Wah Wu, May 26 2016: (Start)
a(n) = (n^3 + 5*n + 6)/6 mod n.
For n > 6:
if n mod 6 == 0, then a(n) = 5*n/6 + 1.
if n mod 6 is in {1, 5}, then a(n) = 1.
if n mod 6 is in {2, 4}, then a(n) = n/2 + 1.
if n mod 6 == 3, then a(n) = n/3 + 1.
(End)

A133622 a(n) = 1 if n is odd, a(n) = n/2+1 if n is even.

Original entry on oeis.org

1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 1, 12, 1, 13, 1, 14, 1, 15, 1, 16, 1, 17, 1, 18, 1, 19, 1, 20, 1, 21, 1, 22, 1, 23, 1, 24, 1, 25, 1, 26, 1, 27, 1, 28, 1, 29, 1, 30, 1, 31, 1, 32, 1, 33, 1, 34, 1, 35, 1, 36, 1, 37, 1, 38, 1, 39, 1, 40, 1, 41, 1, 42, 1, 43, 1, 44, 1
Offset: 1

Views

Author

Hieronymus Fischer, Sep 30 2007

Keywords

Comments

a(n) is the count of terms a(n+1) present so far in the sequence, with a(n+1) included in the count; example: a(1) = 1 "says" that there is 1 term "2" so far in the sequence; a(2) = 2 "says" that there are 2 terms "1" so far in the sequence... etc. This comment was inspired by A039617. - Eric Angelini, Mar 03 2020

Crossrefs

Programs

  • Haskell
    import Data.List (transpose)
    a133622 n = (1 - m) * n' + 1 where (n', m) = divMod n 2
    a133622_list = concat $ transpose [[1, 1 ..], [2 ..]]
    -- Reinhard Zumkeller, Feb 20 2015
    
  • Maple
    seq([1,n][],n=2..100); # Robert Israel, May 27 2016
  • Mathematica
    Riffle[Range[2,50],1,{1,-1,2}] (* Harvey P. Dale, Jan 19 2013 *)
  • PARI
    a(n)=if(n%2,1,n/2+1) \\ Charles R Greathouse IV, Sep 02 2015

Formula

a(n)=1+(binomial(n+1,2)mod n)=1+(binomial(n+1,n-1)mod n).
a(n)=binomial(n+2,2) mod n = binomial(n+2,n) mod n for n>2.
a(n)=1+(1+(-1)^n)*n/4.
a(n)=1+(A000217(n) mod n).
a(n)=a(n-2)+1, if n is even, a(n)=a(n-2) if n is odd.
a(n)=a(n-2)+1-(n mod 2)=a(n-2)+(1+(-1)^n)/2 for n>2.
a(n)=(a(n-3)+a(n-2))/a(n-1) for n>3.
G.f.: g(x)=x(1+2x-x^2-x^3)/(1-x^2)^2.
G.f.: (Q(0)-1-x)/x^2, where Q(k)= 1 + (k+1)*x/(1 - x/(x + (k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 23 2013
a(n) = 2*a(n-2)-a(n-4) for n > 4. - Chai Wah Wu, May 26 2016
E.g.f.: exp(x) - 1 + x*sinh(x)/2. - Robert Israel, May 27 2016

A133875 n modulo 5 repeated 5 times.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 5^2 = 25.

Crossrefs

Programs

  • Magma
    [(1 + Floor(n/5)) mod 5 : n in [0..50]]; // Wesley Ivan Hurt, Jun 06 2014
  • Maple
    A133875:=n->((1+floor(n/5)) mod 5); seq(A133875(n), n=0..100); # Wesley Ivan Hurt, Jun 06 2014
  • Mathematica
    Table[Mod[1 + Floor[n/5], 5], {n, 0, 100}] (* Wesley Ivan Hurt, Jun 06 2014 *)
    LinearRecurrence[{1,0,0,0,-1,1,0,0,0,-1,1,0,0,0,-1,1,0,0,0,-1,1},{1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,0},120] (* Harvey P. Dale, Dec 14 2017 *)

Formula

a(n) = (1 + floor(n/5)) mod 5.
a(n) = A010874(A002266(n+5)).
a(n) = 1 + floor(n/5) - 5*floor((n+5)/25).
a(n) = (((n+5) mod 25) - (n mod 5)) / 5.
a(n) = ((n + 5 - (n mod 5)) / 5) mod 5.
a(n) = A010874((n + 5 - A010874(n))/5).
a(n) = binomial(n+5, n) mod 5 = binomial(n+5, 5) mod 5.
a(n) = +a(n-1) -a(n-5) +a(n-6) -a(n-10) +a(n-11) -a(n-15) +a(n-16) -a(n-20) +a(n-21). - R. J. Mathar, Sep 03 2011
G.f.: ( 1+2*x^5+3*x^10+4*x^15 ) / ( (1-x)*(x^20+x^15+x^10+x^5+1) ). - R. J. Mathar, Sep 03 2011

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.

Crossrefs

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

Crossrefs

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

Crossrefs

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

Crossrefs

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

Crossrefs

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).

A133882 a(n) = binomial(n+2,n) mod 2^2.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 2^3 = 8.

Crossrefs

For the sequence regarding "binomial(n+2, n) mod 2" see A133872.
A105198 shifted once left.

Programs

Formula

a(n) = binomial(n+2,2) mod 2^2.
G.f.: (1 + 3*x + 2*x^2 + 2*x^3 + 3*x^4 + x^5)/(1-x^8).
G.f.: (1+x)*(1+2*x+2*x^3+x^4)/(1-x^8) = (1+2*x+2*x^3+x^4)/((1-x)*(1+x^2)*(1+x^4)).
a(n) = A105198(n+1). - R. J. Mathar, Jun 08 2008
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