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.

Previous Showing 11-18 of 18 results.

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.

Links

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133872, A133880, A133890, A133900, A133910.

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 *)

A133893 Numbers m such that binomial(m+3,m) mod 3 = 0.

Original entry on oeis.org

6, 7, 8, 15, 16, 17, 24, 25, 26, 33, 34, 35, 42, 43, 44, 51, 52, 53, 60, 61, 62, 69, 70, 71, 78, 79, 80, 87, 88, 89, 96, 97, 98, 105, 106, 107, 114, 115, 116, 123, 124, 125, 132, 133, 134, 141, 142, 143, 150, 151, 152, 159, 160, 161, 168, 169, 170, 177, 178, 179, 186
Offset: 0

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Also numbers m such that floor(1+(m/3)) mod 3 = 0.
Partial sums of the sequence 6,1,1,7,1,1,7,1,1,7, ... which has period 3.

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133873, A133883, A133890, A133900, A133910.

Programs

  • Mathematica
    Select[Range[200],Mod[Binomial[#+3,#],3]==0&] (* Harvey P. Dale, Aug 27 2023 *)

Formula

a(n)=3n+6-2*(n mod 3).
G.f.: g(x)=6/(1-x)+x(1+x+7x^2)/((1-x^3)(1-x)) = (6+x+x^2+x^3)/((1-x^3)(1-x)).
G.f.: g(x)=(6-5x-x^4)/((1-x^3)(1-x)^2).

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).

A133896 Numbers m such that binomial(m+6,m) mod 6 = 0.

Original entry on oeis.org

3, 4, 5, 6, 7, 12, 13, 14, 15, 21, 22, 23, 26, 30, 31, 34, 35, 39, 42, 43, 44, 50, 51, 52, 53, 58, 59, 60, 61, 62, 66, 67, 68, 69, 70, 71, 75, 76, 77, 78, 79, 84, 85, 86, 87, 93, 94, 95, 98, 102, 103, 106, 107, 111, 114, 115, 116, 122, 123, 124, 125, 130, 131, 132, 133, 134
Offset: 0

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Partial sums of the sequence 3,1,1,1,1,5,1,1,1,6,1,1,3,4,1,3,1,4,3,1,1,6,1,1,1,5,1,1,1,1,4,1,1,1,1,1,4, ... which has period 36.

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133876, A133886, A133890, A133900, A133910.

Programs

  • Mathematica
    Select[Range[140], Mod[Binomial[# + 6, #], 6] == 0&] (* Jean-François Alcover, Nov 12 2017 *)
  • PARI
    isok(n) = !(binomial(n+6, n) % 6); \\ Michel Marcus, Nov 12 2017

Formula

G.f.: g(x)=3/(1-x)+ x/(1-x)^2+(4x^5+5x^9+2x^12+3x^13+2x^15+3x^17+2x^18+5x^21+3x^26+3x^32) /((1-x^36)(1-x)).
G.f.: g(x)=(3-2x+4x^5+5x^9+2x^12+3x^13+2x^15+3x^17+2x^18+5x^21+3x^26+3x^32-x^37) /((1-x^36)(1-x)^2).

A133897 Numbers m such that binomial(m+7,m) mod 7 = 0.

Original entry on oeis.org

42, 43, 44, 45, 46, 47, 48, 91, 92, 93, 94, 95, 96, 97, 140, 141, 142, 143, 144, 145, 146, 189, 190, 191, 192, 193, 194, 195, 238, 239, 240, 241, 242, 243, 244, 287, 288, 289, 290, 291, 292, 293, 336, 337, 338, 339, 340, 341, 342, 385, 386, 387, 388, 389, 390
Offset: 0

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Also numbers m such that floor(1+(m/7)) mod 7 = 0.
Partial sums of the sequence 42,1,1,1,1,1,1,43,1,1,1,1,1,1,43,... which has period 7.

Links

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133877, A133887, A133890, A133900, A133910.

Programs

  • Mathematica
    Select[Range[390],Mod[Binomial[#+7,#],7]==0&] (* or *) LinearRecurrence[{1,0,0,0,0,0,1,-1},{42, 43, 44, 45, 46, 47, 48, 91},55] (* James C. McMahon, Mar 30 2025 *)

Formula

a(n) = 7*n + 42 - 6*(n mod 7).
G.f.: (42+x+x^2+x^3+x^4+x^5+x^6+x^7)/((1-x^7)(1-x)).
G.f.: (42-41x-x^8) /((1-x^7)(1-x)^2).

A133898 Numbers m such that binomial(m+8,m) mod 8 = 0.

Original entry on oeis.org

56, 57, 58, 59, 60, 61, 62, 63, 120, 121, 122, 123, 124, 125, 126, 127, 184, 185, 186, 187, 188, 189, 190, 191, 248, 249, 250, 251, 252, 253, 254, 255, 312, 313, 314, 315, 316, 317, 318, 319, 376, 377, 378, 379, 380, 381, 382, 383, 440, 441, 442, 443, 444
Offset: 0

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Partial sums of the sequence 56,1,1,1,1,1,1,1,57,1,1,1,1,1,1,1,57, ... which has period 8.

Links

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133878, A133888, A133890, A133900, A133910.

Programs

  • Mathematica
    Select[Range[500],Mod[Binomial[#+8,#],8]==0&] (* or *) LinearRecurrence[{1,0,0,0,0,0,0,1,-1},{56,57,58,59,60,61,62,63,120},60] (* Harvey P. Dale, Apr 07 2025 *)
  • PARI
    a(n)=8*n+56-n%8*7 \\ Charles R Greathouse IV, Oct 13 2022

Formula

a(n)=8n+56-7*(n mod 8). [Corrected by Charles R Greathouse IV, Oct 13 2022]
G.f.: g(x)=(56+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8)/((1-x^8)(1-x)).
G.f.: g(x)=(56-55x-x^9) /((1-x^8)(1-x)^2).

A133899 Numbers m such that binomial(m+9,m) mod 9 = 0.

Original entry on oeis.org

72, 73, 74, 75, 76, 77, 78, 79, 80, 153, 154, 155, 156, 157, 158, 159, 160, 161, 234, 235, 236, 237, 238, 239, 240, 241, 242, 315, 316, 317, 318, 319, 320, 321, 322, 323, 396, 397, 398, 399, 400, 401, 402, 403, 404, 477, 478, 479, 480, 481, 482, 483, 484, 485
Offset: 0

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Comments

Also numbers m such that floor(1+(m/9)) mod 9 = 0.
Partial sums of the sequence 72,1,1,1,1,1,1,1,1,73,1,1,1,1,1,1,1,1,73, ... which has period 9.

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133879, A133889, A133890, A133900, A133910.

Programs

  • Mathematica
    Select[Range[500],Divisible[Binomial[#+9,#],9]&] (* Harvey P. Dale, Apr 03 2011 *)

Formula

a(n)=9n+72-8*(n mod 9).
G.f.: g(x)=(72+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)/((1-x^9)(1-x)).
G.f.: g(x)=(72-71x-x^10) /((1-x^9)(1-x)^2).

A133909 Least odd number m such that binomial(n+m,m) mod m = 1.

Original entry on oeis.org

3, 3, 5, 5, 7, 7, 9, 9, 3, 3, 3, 13, 17, 17, 17, 17, 19, 3, 3, 3, 23, 23, 25, 25, 5, 5, 3, 3, 3, 31, 37, 37, 37, 37, 37, 3, 3, 3, 41, 41, 43, 43, 47, 47, 3, 3, 3, 49, 7, 5, 5, 5, 5, 3, 3, 3, 9, 9, 9, 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, 89, 9, 9
Offset: 1

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Examples

			a(3)=5, since binomial(3+5,5) mod 5 = 56 mod 5 = 1 and 5 is the minimal odd number with this property.
a(8)=9 because of binomial(8+9,9)=24310=2701*9+1, but binomial(8+k,k) mod k<>1 for all odd numbers <9.

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133872, A133880, A133890, A133900, A133910.
Previous Showing 11-18 of 18 results.

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