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.

Showing 1-7 of 7 results.

A225003 Duplicate of A047332.

Original entry on oeis.org

23, 2, 3, 5, 6, 7, 9, 10, 12, 13, 14, 16, 17, 19, 20, 21, 23, 24, 26, 27, 28, 30, 31, 33, 34, 35, 37, 38, 40, 41, 42, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 58, 59, 61, 62, 63, 65, 66, 68, 69, 70, 72, 73, 75, 76, 77, 79, 80, 82, 83, 84, 86, 87, 89, 90, 91, 93
Offset: 1

Views

Author

T. D. Noe, Apr 26 2013

Keywords

Programs

  • Mathematica
    th = E^(5/7); t = Table[Floor[1/FractionalPart[th^(1/n)]], {n, 100}]

A085269 Integer part of the conversion from centimeters to inches.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 14, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 20, 20, 20, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 27, 27, 27, 28, 28, 29, 29, 29, 30
Offset: 0

Views

Author

Cino Hilliard, Aug 12 2003

Keywords

Comments

2.54 = 127/50. - Eric Desbiaux, Nov 16 2008

Crossrefs

Programs

  • Mathematica
    Floor[Range[0, 100]*100/254] (* Paolo Xausa, Jul 16 2025 *)
  • PARI
    a(n) = floor(n*100/254);

Formula

a(n) = floor(n/2.54).
Eric Desbiaux suggested (Apr 19 2008) that A047332(n)-A080686(n) =? a(n-1), but R. J. Mathar points out that this is only true for the first 23 terms and so is nothing more than a coincidence. - N. J. A. Sloane, Apr 26 2008

Extensions

Corrected by T. D. Noe, Nov 02 2006

A224996 a(n) = floor(1/f(x^(1/n))) for x = 2, where f computes the fractional part.

Original entry on oeis.org

2, 3, 5, 6, 8, 9, 11, 12, 13, 15, 16, 18, 19, 21, 22, 24, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 39, 41, 42, 44, 45, 47, 48, 49, 51, 52, 54, 55, 57, 58, 60, 61, 62, 64, 65, 67, 68, 70, 71, 73, 74, 75, 77, 78, 80, 81, 83, 84, 86, 87, 88, 90, 91, 93, 94, 96, 97
Offset: 2

Views

Author

T. D. Noe, Apr 26 2013

Keywords

Comments

First denominator of continued fraction representing 2^(1/n): [1,a(n),....] so that 1+1/a(n) is first convergent for 2^(1/n). - Carmine Suriano, Apr 29 2014
a(n) is the largest integer y that satisfies (y+1)^n - y^n >= y^n, or equivalently (y+1)^n >= 2*y^n. - Charles Kusniec, Jan 19 2025

Crossrefs

Cf. A078607 (the smallest integer y that satisfies (y+1)^n - y^n < y^n).

Programs

  • Mathematica
    th = 2; t = Table[Floor[1/FractionalPart[th^(1/n)]], {n, 2, 100}]

Formula

a(n) = floor(n/log(2)-1/2). - Andrey Zabolotskiy, Dec 01 2017

A224995 Floor(1/f(x^(1/n))) for x = 3/2, where f computes the fractional part.

Original entry on oeis.org

2, 4, 6, 9, 11, 14, 16, 19, 21, 24, 26, 29, 31, 34, 36, 38, 41, 43, 46, 48, 51, 53, 56, 58, 61, 63, 66, 68, 71, 73, 75, 78, 80, 83, 85, 88, 90, 93, 95, 98, 100, 103, 105, 108, 110, 112, 115, 117, 120, 122, 125, 127, 130, 132, 135, 137, 140, 142, 145, 147, 149
Offset: 1

Views

Author

T. D. Noe, Apr 26 2013

Keywords

Crossrefs

Programs

  • Mathematica
    th = 3/2; t = Table[Floor[1/FractionalPart[th^(1/n)]], {n, 100}]

Formula

a(n) = floor(n/log(3/2)-1/2) for n>1. - Andrey Zabolotskiy, Dec 01 2017

A224997 Floor(1/f(x^(1/n))) for x = 17, where f computes the fractional part.

Original entry on oeis.org

8, 1, 32, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 25
Offset: 2

Views

Author

T. D. Noe, Apr 26 2013

Keywords

Crossrefs

Programs

  • Mathematica
    th = 17; t = Table[Floor[1/FractionalPart[th^(1/n)]], {n, 2, 100}]

Formula

a(n) = floor(n/log(17)-1/2) for n>7. - Andrey Zabolotskiy, Dec 01 2017

A224998 Floor(1/f(x^(1/n))) for x = Pi, where f computes the fractional part.

Original entry on oeis.org

7, 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 9, 10, 11, 12, 13, 14, 15, 16, 16, 17, 18, 19, 20, 21, 22, 23, 23, 24, 25, 26, 27, 28, 29, 30, 30, 31, 32, 33, 34, 35, 36, 37, 37, 38, 39, 40, 41, 42, 43, 44, 44, 45, 46, 47, 48, 49, 50, 51, 51, 52, 53, 54, 55, 56, 57, 58, 58
Offset: 1

Views

Author

T. D. Noe, Apr 26 2013

Keywords

Crossrefs

Programs

  • Mathematica
    th = Pi; t = Table[Floor[1/FractionalPart[th^(1/n)]], {n, 100}]

Formula

a(n) = floor(n/log(Pi)-1/2) for n>4. - Andrey Zabolotskiy, Dec 01 2017

A382657 Number of minimum total dominating sets in the n-Goldberg graph.

Original entry on oeis.org

16, 277, 10, 2386, 28, 33301, 360, 10, 4334, 60, 67288, 728, 10, 9856, 102, 150750, 1292, 10, 19222, 154, 299368, 2124, 10, 34112, 216, 549276, 3306, 10, 56730, 288, 951456, 4930, 10, 89916, 370, 1576202, 7098, 10, 137268, 462, 2518596, 9922, 10, 203274, 564, 3905148, 13524, 10
Offset: 3

Views

Author

Eric W. Weisstein, Apr 02 2025

Keywords

Comments

For n > 3, the total domination number is given by 2*floor((7*n+4)/5) = 2*A047332(n+1). For n = 3, the total domination number is 9. - Andrew Howroyd, May 24 2025

Crossrefs

Cf. A382431.

Programs

  • PARI
    a(n) = { my(m=n\5+1,r=-n%5); if(n<=8, [16, 277, 10, 2386, 28, 33301][n-2], if(r<=2, if(r==0, 10, n*if(r==1, (m + 13)*(m^2 + 2*m + 24)/12, (m^7 + 70*m^6 + 1750*m^5 + 39340*m^4 + 525889*m^3 + 2944270*m^2 + 15922920*m + 12216960)/20160)), n*if(r==3, (m + 2), (m^5 + 35*m^4 + 365*m^3 + 5485*m^2 + 21114*m + 16200)/360) )) } \\ Andrew Howroyd, May 24 2025

Formula

a(5*n) = 10.
From Andrew Howroyd, May 24 2025: (Start)
a(5*n-1) = (5*n-1)*(n + 13)*(n^2 + 2*n + 24)/12 for n >= 2;
a(5*n-2) = (5*n-2)*(n^7 + 70*n^6 + 1750*n^5 + 39340*n^4 + 525889*n^3 + 2944270*n^2 + 15922920*n + 12216960)/20160 for n >= 3;
a(5*n-3) = (5*n-3)*(n + 2) for n >= 2;
a(5*n-4) = (5*n-4)*(n^5 + 35*n^4 + 365*n^3 + 5485*n^2 + 21114*n + 16200)/360 for n >= 3. (End)

Extensions

a(8)-a(12) from Eric W. Weisstein, May 11 2025
a(13) onwards from Andrew Howroyd, May 24 2025
Showing 1-7 of 7 results.