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 1-3 of 3 results.

A165657 Numerators of A002110 divided by A102647, starting from the second term of both.

Original entry on oeis.org

2, 3, 15, 105, 385, 5005, 17017, 323323, 1062347, 30808063, 434113615, 35336848261, 1448810778701, 33545541876077, 266186053068611, 5426100312552455, 9156001667401012567, 42962777054727828199
Offset: 1

Views

Author

Creighton Dement, Sep 24 2009

Keywords

Crossrefs

Programs

  • PARI
    a1(n)=prod(i=1, n, prime(i)); b1(n)=prod(i=1, n, prime(n+1)%prime(i));
    a(n)=if(n<0, 0, numerator(a1(n)/b1(n))); for(n=1, 20, print1(a(n) ", "))

A165658 Denominators of A002110 divided by A102647, starting from the second term of both.

Original entry on oeis.org

1, 1, 1, 4, 6, 48, 64, 96, 576, 1728, 13824, 165888, 1036800, 9953280, 119439360, 297271296, 134369280000, 222953472000, 75246796800, 32105299968000, 229323571200, 568865783808000000, 125150472437760000, 6129819058176000
Offset: 1

Views

Author

Creighton Dement, Sep 24 2009

Keywords

Crossrefs

Programs

  • PARI
    a1(n)=prod(i=1, n, prime(i));
    b1(n)=prod(i=1, n, prime(n+1)%prime(i));
    a(n)=if(n<0, 0, denominator(a1(n)/b1(n)));
    for(n=1, 25, print1(a(n) ", "))

A165660 Denominators of A007504 divided by A033955, starting from the second term of A033955.

Original entry on oeis.org

1, 3, 2, 8, 13, 18, 27, 29, 23, 56, 7, 74, 44, 98, 67, 49, 171, 200, 217, 28, 137, 309, 17, 116, 209, 448, 471, 174, 571, 629, 137, 739, 111, 793, 853, 318, 997, 1002, 560, 164, 610, 446, 1419, 1466, 385, 1615, 1573, 1633, 1707, 1825, 946, 662, 2221, 781, 1198
Offset: 1

Views

Author

Creighton Dement, Sep 24 2009

Keywords

Comments

Conjecture: with the exception of the second term, 2 <= A165659(n)/a(n) < 3.

Crossrefs

Programs

  • PARI
    a1(n)=sum(i=1, n, prime(i)); b1(n)=sum(i=1, n, prime(n+1)%prime(i)); a(n)=if(n<0, 0, denominator(a1(n)/b1(n))); for(n=1, 50, print1(a(n) ", "))

Extensions

Terms corrected by Creighton Dement, Oct 03 2009
Removed a conjecture - R. J. Mathar, Oct 09 2009
Typo in definition corrected by Creighton Dement, Oct 09 2009
Showing 1-3 of 3 results.