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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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).

A133911 Number of prime factors (counted with multiplicity) of the period numbers defined by A133900.

Original entry on oeis.org

0, 2, 2, 4, 2, 5, 2, 6, 4, 6, 2, 8, 2, 6, 5, 8, 2, 9, 2, 8, 5, 7, 2, 10, 4, 7, 6, 8, 2, 12, 2, 10, 6, 8, 5, 12, 2, 8, 6, 11, 2, 12, 2, 9, 8, 8, 2, 13, 4, 10, 6, 9, 2, 12, 5, 11, 6, 8, 2, 14, 2, 8, 8, 12, 5, 13, 2, 10, 6, 13, 2, 14, 2, 9, 8, 10, 5, 13, 2, 13, 8, 10, 2, 17, 5, 9, 7, 11, 2, 16, 5, 10, 7, 9
Offset: 1

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Examples

			a(6)=5, since A133900(6)=72=2*2*2*3*3.
a(12)=8, since A133900(12)=864=2*2*2*2*2*3*3*3.
		

Crossrefs

Formula

a(n)=A001222(A133900(n)).

A134331 Sum of prime factors (counted with multiplicity) of the period numbers defined by A133900.

Original entry on oeis.org

0, 4, 6, 8, 10, 12, 14, 12, 12, 18, 22, 19, 26, 22, 19, 16, 34, 22, 38, 22, 23, 32, 46, 23, 20, 36, 18, 26, 58, 37, 62, 20, 34, 46, 29, 29, 74, 50, 38, 31, 82, 38, 86, 36, 30, 58, 94, 30, 28, 32, 46, 40, 106, 30, 37, 37, 50, 70, 118, 41, 122, 74, 36, 24, 41, 48, 134, 50, 58, 50
Offset: 1

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Examples

			a(6)=12, since A133900(6)=72=2*2*2*3*3 and 2+2+2+3+3=12.
a(12)=19, since A133900(12)=864=2*2*2*2*2*3*3*3 and 2+2+2+2+2+3+3+3=19.
		

Crossrefs

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

A133883 a(n) = binomial(n+3,n) mod 3^2.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 3^3 = 27.

Crossrefs

For the sequence regarding "Binomial(n+3, n) mod 3" see A133873.

Programs

  • Magma
    [Binomial(n+3,n) mod 9: n in [0..60]]; // Vincenzo Librandi, Jul 20 2016
  • Mathematica
    Table[Mod[Binomial[n + 3, n], 9], {n, 0, 120}] (* or *)
    CoefficientList[Series[(1 + 3 x - 3 x^2 + 2 x^3 + 9 x^4 - 9 x^5 + 3 x^6 + 9 x^7 - 9 x^8 + 4 x^9 + 12 x^10 - 12 x^11 + 5 x^12 + 9 x^13 - 9 x^14 + 6 x^15 + 9 x^16 - 9 x^17 + 7 x^18 + 3 x^19 - 3 x^20 + 8 x^21)/((1 - x) (1 + x^3 + x^6) (1 + x^9 + x^18)), {x, 0, 120}], x] (* Michael De Vlieger, Jul 19 2016 *)
  • PARI
    Vec((1 +3*x -3*x^2 +2*x^3 +9*x^4 -9*x^5 +3*x^6 +9*x^7 -9*x^8 +4*x^9 +12*x^10 -12*x^11 +5*x^12 +9*x^13 -9*x^14 +6*x^15 +9*x^16 -9*x^17 +7*x^18 +3*x^19 -3*x^20 +8*x^21) / ((1 -x)*(1 +x^3 +x^6)*(1 +x^9 +x^18)) + O(x^200)) \\ Colin Barker, Jul 19 2016
    

Formula

a(n) = binomial(n+3,3) mod 3^2.
G.f.: (1 +3*x -3*x^2 +2*x^3 +9*x^4 -9*x^5 +3*x^6 +9*x^7 -9*x^8 +4*x^9 +12*x^10 -12*x^11 +5*x^12 +9*x^13 -9*x^14 +6*x^15 +9*x^16 -9*x^17 +7*x^18 +3*x^19 -3*x^20 +8*x^21) / ((1 -x)*(1 +x^3 +x^6)*(1 +x^9 +x^18)). - Colin Barker, Jul 19 2016

Extensions

Corrected g.f. by Colin Barker, Jul 19 2016

A133886 a(n) = binomial(n+6,n) mod 6.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 2*6^2 = 72.

Crossrefs

Programs

Formula

a(n) = binomial(n+6,6) mod 6.
G.f.: g(x) = (1+x+4*x^2-6*x^9-6*x^56+4*x^63+x^64+x^65+3*x^8*(1+x)(1-x^56)/(1-x^8)+4*x^9(1+x+x^2)(1-x^54)/(1-x^9))/(1-x^72).
a(n) = a(n-1)-a(n-2)+a(n-8)+a(n-11)-a(n-17)-a(n-20)-a(n-24)+a(n-25)+a(n-29)+ a(n-32)- a(n-38)-a(n-41)+a(n-47)-a(n-48)+a(n-49). - Harvey P. Dale, May 04 2013
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