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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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.

A133889 a(n) = binomial(n+9,n) mod 9.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 9^2 = 81.

Crossrefs

Programs

  • Mathematica
    Table[Mod[Binomial[n+9,n],9],{n,0,110}] (* Harvey P. Dale, Jan 14 2012 *)

Formula

a(n) = binomial(n+9,9) mod 9.

A133878 n modulo 8 repeated 8 times.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 8^2=64.

Crossrefs

Programs

  • Mathematica
    Flatten[Join[Table[PadRight[{},8,n],{n,7}],Table[PadRight[{},8,n],{n,0,7}]]] (* Harvey P. Dale, Nov 06 2011 *)

Formula

a(n)=(1+floor(n/8)) mod 8.
a(n)=1+floor(n/8)-8*floor((n+8)/64).
a(n)=(((n+8) mod 64)-(n mod 8))/8.
a(n)=((n+8-(n mod 8))/8) mod 8.
G.f. g(x)=(1-x^8)(1+2x^8+3x^16+4x^24+5x^32+6x^40+7x^48)/((1-x)(1-x^64)).
G.f. g(x)=(1-x^8)*sum{0<=k<7, (k+1)*x^(8*k)}/((1-x)(1-x^64)).
G.f. g(x)=(7x^64-8x^56+1)/((1-x)(1-x^8)(1-x^64)).

A133879 n modulo 9 repeated 9 times.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 9^2=81.

Crossrefs

Programs

  • Mathematica
    Rest[Flatten[Table[Table[Table[n,{9}],{n,0,8}],{3}],1]]//Flatten (* Harvey P. Dale, Apr 15 2018 *)

Formula

a(n)=(1+floor(n/9)) mod 9.
a(n)=1+floor(n/9)-9*floor((n+9)/81).
a(n)=(((n+9) mod 81)-(n mod 9))/9.
a(n)=((n+9-(n mod 9))/9) mod 9.
G.f. g(x)=(1-x^9)(1+2x^9+3x^18+4x^27+5x^36+6x^45+7x^56+8x^63)/((1-x)(1-x^81)).
G.f. g(x)=(1-x^9)*sum{0<=k<8, (k+1)*x^(9*k)}/((1-x)(1-x^81)).
G.f. g(x)=(8x^81-9x^72+1)/((1-x)(1-x^9)(1-x^81)).

A133905 Least composite number m such that binomial(n+m,m) mod m = 1.

Original entry on oeis.org

4, 9, 25, 10, 26, 9, 9, 9, 6, 4, 4, 34, 34, 85, 289, 4, 4, 57, 87, 8, 8, 25, 25, 25, 134, 4, 4, 15, 15, 111, 111, 4, 4, 8, 8, 10, 10, 121, 121, 82, 86, 4, 4, 49, 49, 49, 49, 4, 4, 265, 68, 10, 10, 8, 8, 6, 9, 4, 4, 194, 194, 469, 249, 4, 4, 44, 44, 146, 146, 16, 16, 6, 6, 4, 4, 162
Offset: 1

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Examples

			a(1)=4, since binomial(1+4,4) mod 4 = 5 mod 4 = 1 and 4 is the minimal composite number with this property.
a(5)=26 because of binomial(5+26,26)=169911=6535*26+1, but binomial(5+k,k) mod k<>1 for all composite numbers <26.
		

Crossrefs

Programs

  • Mathematica
    lcn[n_]:=Module[{m=4},While[PrimeQ[m]||Mod[Binomial[n+m,m],m]!=1,m++];m]; Array[lcn,80] (* Harvey P. Dale, May 13 2022 *)
  • PARI
    a(n) = { my(m = 4, ok = 0); until (ok, if (! isprime(m) && (binomial(n+m, m) % m == 1), ok = 1, m++);); return (m);} \\ Michel Marcus, Jul 15 2013

A134335 Numbers such that the arithmetic mean of their prime factors (counted with multiplicity) is an integer, but not a prime.

Original entry on oeis.org

15, 35, 39, 42, 50, 51, 55, 65, 77, 78, 87, 91, 92, 95, 110, 111, 114, 115, 119, 123, 140, 141, 143, 155, 159, 161, 164, 170, 183, 185, 186, 187, 189, 201, 203, 204, 209, 215, 219, 221, 222, 225, 230, 235, 236, 242, 247, 258, 259, 264, 267, 284, 285, 287, 290
Offset: 1

Views

Author

Hieronymus Fischer, Oct 23 2007

Keywords

Examples

			a(1) = 15, since 15 = 3*5 and (3+5)/2 = 4 is not prime.
a(5) = 50, since 50 = 2*5*5 and (2+5+5)/3 = 4 is not prime.
		

Crossrefs

Programs

  • Mathematica
    fp[{a_,b_}]:=a*b;s={};Do[If[q=Total[fp/@FactorInteger[n]]/Total[Last/@FactorInteger[n]];IntegerQ[q]&&!PrimeQ[q],AppendTo[s,n]],{n,2,290}];s (* James C. McMahon, Apr 05 2025 *)

Extensions

Definition clarified by the author, May 06 2013

A133881 Even numbers k such that binomial(k+p,k) mod k = 1, where p=10.

Original entry on oeis.org

4, 68, 164, 260, 292, 356, 388, 452, 484, 516, 548, 676, 708, 772, 836, 932, 964, 1028, 1060, 1124, 1156, 1252, 1348, 1412, 1444, 1508, 1572, 1604, 1636, 1732, 1796, 1828, 1892, 1924, 2084, 2116, 2244, 2276, 2308, 2372, 2404, 2468, 2564, 2596, 2692, 2756
Offset: 1

Views

Author

Hieronymus Fischer, Oct 16 2007

Keywords

Crossrefs

Programs

  • Mathematica
    Select[2*Range[1500],Mod[Binomial[10+#,#],#]==1&] (* Harvey P. Dale, Jun 05 2023 *)

A133894 Numbers m such that binomial(m+4,m) mod 4 = 0.

Original entry on oeis.org

12, 13, 14, 15, 28, 29, 30, 31, 44, 45, 46, 47, 60, 61, 62, 63, 76, 77, 78, 79, 92, 93, 94, 95, 108, 109, 110, 111, 124, 125, 126, 127, 140, 141, 142, 143, 156, 157, 158, 159, 172, 173, 174, 175, 188, 189, 190, 191, 204, 205, 206, 207, 220, 221, 222, 223, 236, 237
Offset: 1

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Also numbers m such that floor(1+(m/4)) mod 4 = 0.
Partial sums of the sequence 12,1,1,1,13,1,1,1,13, ... which has period 4.
Numbers congruent to {12,13,14,15} mod 16. Numbers n such that n xor 12 = n - 12. [Brad Clardy, May 06 2013]

Crossrefs

Programs

Formula

a(n) = 4*n + 12 - 3*(n mod 4).
G.f.: 12/(1-x)+x(1+x+x^2+13x^3)/((1-x^4)(1-x)) = (12+x+x^2+x^3+x^4)/((1-x^4)(1-x)) = (12-11x-x^5)/((1-x^4)(1-x)^2).
a(n) = 4*n+3*((1-i)*i^n+(1+i)*(-i)^n+(-1)^n+5)/2, where i=sqrt(-1). - Bruno Berselli, Apr 08 2011

A133908 Least odd prime number m such that binomial(n+m,m) mod m = 1.

Original entry on oeis.org

3, 3, 5, 5, 7, 7, 11, 11, 3, 3, 3, 13, 17, 17, 17, 17, 19, 3, 3, 3, 23, 23, 29, 29, 5, 5, 3, 3, 3, 31, 37, 37, 37, 37, 37, 3, 3, 3, 41, 41, 43, 43, 47, 47, 3, 3, 3, 53, 7, 5, 5, 5, 5, 3, 3, 3, 59, 59, 61, 61, 67, 67, 3, 3, 3, 67, 71, 71, 71, 71, 73, 3, 3, 3, 5, 5, 5, 5, 5, 83, 3, 3, 3, 89, 89
Offset: 1

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Also the least odd prime number m such that m divides floor(n/m) or m>n.

Examples

			a(3)=5, since binomial(3+5,5) mod 5 = 56 mod 5 = 1 and 5 is the minimal odd prime number with this property.
a(8)=11 because of binomial(8+11,11)=75582=6871*11+1, but binomial(8+k,k) mod k<>1 for all odd primes <11.
		

Crossrefs

Programs

  • Mathematica
    With[{oprs=Rest[Prime[Range[100]]]},Flatten[Table[Select[oprs,Mod[ Binomial[ n+#,#],#]==1&,1],{n,90}]]] (* Harvey P. Dale, Jun 27 2012 *)
Previous Showing 41-50 of 58 results. Next