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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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

A133873 n modulo 3 repeated 3 times.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 3^2=9.

Crossrefs

Formula

G.f.: (1 + 2*x^3)*(1 - x^3)/((1 - x)*(1 - x^9)).
a(n) = (1 + floor(n/3)) mod 3.
a(n) = A010872(A002264(n+3)).
a(n) = 1+floor(n/3)-3*floor((n+3)/9).
a(n) = (((n+3) mod 9)-(n mod 3))/3.
a(n) = ((n+3-(n mod 3))/3) mod 3.
a(n) = binomial(n+3,n) mod 3 = binomial(n+3,3) mod 3.

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.

Crossrefs

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)).

A133884 a(n) = binomial(n+4,n) mod 4.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 4^2=16.

Examples

			For n=2, binomial(6,2) = 6*5/2 = 15, which is 3 (mod 4) so a(2) = 3. - _Michael B. Porter_, Jul 19 2016
		

Crossrefs

Programs

  • Magma
    [Binomial(n+4,n) mod 4: n in [0..100]]; // Vincenzo Librandi, Jul 15 2016
  • Mathematica
    Table[Mod[Binomial[n + 4, 4], 4], {n, 0, 100}] (* Vincenzo Librandi, Jul 15 2016 *)

Formula

a(n) = binomial(n+4,4) mod 4.
G.f.: (1 + x + 3*x^2 + 3*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + 2*x^7 + 3*x^8 + 3*x^9 + x^10 + x^11)/(1 - x^16) = (1 + 2*x^2 + 2*x^6 + x^8)/((1 - x)*(1 + x^4)*(1 + x^8)).
a(n) = A000505(n+5) mod 4. - John M. Campbell, Jul 14 2016
a(n) = A000506(n+6) mod 4. - John M. Campbell, Jul 15 2016

Extensions

G.f. corrected by Bruno Berselli, Jul 19 2016

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.

Crossrefs

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

A133876 n modulo 6 repeated 6 times.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 0, 0, 0
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 6^2=36.

Crossrefs

Programs

  • Mathematica
    Table[PadRight[{},6,Mod[n,6]],{n,20}]//Flatten (* Harvey P. Dale, Nov 15 2023 *)

Formula

a(n)=(1+floor(n/6)) mod 6.
a(n)=1+floor(n/6)-6*floor((n+6)/36).
a(n)=(((n+6) mod 36)-(n mod 6))/6.
a(n)=((n+6-(n mod 6))/6) mod 6.
G.f. g(x)=(1-x^6)(1+2x^5+3x^12+4x^18+5x^24)/((1-x)(1-x^36)).
G.f. g(x)=(5x^36-6x^30+1)/((1-x)(1-x^6)(1-x^36)).

A133877 n modulo 7 repeated 7 times.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 7^2=49.

Crossrefs

Formula

a(n)=(1+floor(n/7)) mod 7.
a(n)=1+floor(n/7)-7*floor((n+7)/49).
a(n)=(((n+7) mod 49)-(n mod 7))/7.
a(n)=((n+7-(n mod 7))/7) mod 7.
a(n)=binomial(n+7,n) mod 7 =binomial(n+7,7) mod 7.
G.f. g(x)=(1-x^7)(1+2x^7+3x^14+4x^21+5x^28+6x^35)/((1-x)(1-x^49)).
G.f. g(x)=(6x^49-7x^42+1)/((1-x)(1-x^7)(1-x^49)).

A133887 Binomial(n+7,n) mod 7^2.

Original entry on oeis.org

1, 8, 36, 22, 36, 8, 1, 2, 16, 23, 44, 23, 16, 2, 3, 24, 10, 17, 10, 24, 3, 4, 32, 46, 39, 46, 32, 4, 5, 40, 33, 12, 33, 40, 5, 6, 48, 20, 34, 20, 48, 6, 7, 7, 7, 7, 7, 7, 7, 8, 15, 43, 29, 43, 15, 8, 9, 23, 30, 2, 30, 23, 9, 10, 31, 17, 24, 17, 31, 10, 11, 39, 4, 46, 4, 39, 11, 12, 47
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 7^3=343.

Crossrefs

For the sequence regarding binomial(n+7, n) mod 7 see A133877.

Programs

  • Mathematica
    Table[Mod[Binomial[n+7,n],49],{n,0,80}] (* Harvey P. Dale, Apr 08 2018 *)

Formula

a(n)=binomial(n+7,7) mod 7^2.
G.f. g(x)=sum{0<=k<343, a(k)*x^k}/(1-x^343).

A133888 Binomial(n+8,n) mod 8.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 8^2=64.

Crossrefs

Programs

  • Mathematica
    Table[Mod[Binomial[n+8,n],8],{n,0,110}] (* Harvey P. Dale, Aug 08 2011 *)

Formula

a(n)=binomial(n+8,8) mod 8.
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