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.

A186674 Total number of n-digit numbers requiring 14 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 6, 60, 632, 6135, 60132, 600115, 6000118, 60000129, 600000127, 6000000136
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Comments

A102831(n) + A186650(n) + A186652(n) + A186654(n) + A186656(n) + A186658(n) + A186660(n) + A186662(n) + A186664(n) + A186666(n) + A186668(n) + A186670(n) + A186672(n) + a(n) + A186676(n) + A186678(n) + A186681(n) + A186683(n) + A186685(n) = A052268(n), for n>1.

Crossrefs

Formula

a(n) = A186673(n) - A186673(n-1).

Extensions

a(5)-a(11) from Giovanni Resta, Apr 29 2016

A186676 Total number of n-digit numbers requiring 15 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 6, 60, 624, 6071, 60073, 600069, 6000069, 60000069, 600000061, 6000000071
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Comments

A102831(n) + A186650(n) + A186652(n) + A186654(n) + A186656(n) + A186658(n) + A186660(n) + A186662(n) + A186664(n) + A186666(n) + A186668(n) + A186670(n) + A186672(n) + A186674(n) + a(n) + A186678(n) + A186681(n) + A186683(n) + A186685(n) = A052268(n), for n>1.

Crossrefs

Formula

a(n) = A186675(n) - A186675(n-1).

Extensions

a(5)-a(11) from Giovanni Resta, Apr 29 2016

A186678 Total number of n-digit numbers requiring 16 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 4, 43, 241, 299, 287, 304, 309, 316, 286, 299
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Comments

A102831(n) + A186650(n) + A186652(n) + A186654(n) + A186656(n) + A186658(n) + A186660(n) + A186662(n) + A186664(n) + A186666(n) + A186668(n) + A186670(n) + A186672(n) + A186674(n) + A186676(n) + a(n) + A186681(n) + A186683(n) + A186685(n) = A052268(n), for n>1.

Crossrefs

Formula

a(n) = A186677(n) - A186677(n-1).

Extensions

a(5)-a(11) from Giovanni Resta, Apr 29 2016

A186685 Total number of n-digit numbers requiring 19 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 1, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Crossrefs

Programs

  • Mathematica
    PadRight[{0, 1, 6}, 100] (* Paolo Xausa, Jul 26 2024 *)

Formula

a(n) = A186684(n) - A186684(n-1).
A161905(n) + A186650(n) + A186652(n) + A186654(n) + A186656(n) + A186658(n) + A186660(n) + A186662(n) + A186664(n) + A186666(n) + A186668(n) + A186670(n) + A186672(n) + A186674(n) + A186676(n) + A186678(n) + A186681(n) + A186683(n) + a(n) = A052268(n).
a(n) = 0 for n >= 4. - Nathaniel Johnston, May 09 2011

A186650 Total number of n-digit numbers requiring 2 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

1, 4, 9, 29, 100, 317, 1007, 3146, 10016, 31712, 100204, 316799, 1002314, 3169309, 10022310, 31693094
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Comments

A102831(n) + a(n) + A186652(n) + A186654(n) + A186656(n) + A186658(n) + A186660(n) + A186662(n) + A186664(n) + A186666(n) + A186668(n) + A186670(n) + A186672(n) + A186674(n) + A186676(n) + A186678(n) + A186681(n) + A186683(n) + A186685(n) = A052268(n), for n>1.

Crossrefs

Programs

  • Maple
    isbiquadrate:=proc(n) type(root(n,4),posint); end:
    isA003336:=proc(n) local x,y4; if isbiquadrate(n) then false; else for x from 1 do y4:=n-x^4; if y4A003336(k) then i:=i+1; fi; od: return(i); end: for n from 1 do print(a(n)); od;

Formula

a(n) = A186649(n)-A186649(n-1).

Extensions

a(6) from Martin Renner, Feb 26 2011
a(7)-a(16) from Giovanni Resta, Apr 29 2016

A186652 Total number of n-digit numbers requiring 3 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

1, 5, 23, 112, 648, 3564, 19820, 110506, 622268, 3501263, 19699896
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Comments

A102831(n) + A186650(n) + a(n) + A186654(n) + A186656(n) + A186658(n) + A186660(n) + A186662(n) + A186664(n) + A186666(n) + A186668(n) + A186670(n) + A186672(n) + A186674(n) + A186676(n) + A186678(n) + A186681(n) + A186683(n) + A186685(n) = A052268(n), for n>1.

Crossrefs

Formula

a(n) = A186651(n) - A186651(n-1).

Extensions

a(5)-a(11) from Giovanni Resta, Apr 29 2016

A186681 Total number of n-digit numbers requiring 17 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 3, 30, 30, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Comments

A161905(n) + A186650(n) + A186652(n) + A186654(n) + A186656(n) + A186658(n) + A186660(n) + A186662(n) + A186664(n) + A186666(n) + A186668(n) + A186670(n) + A186672(n) + A186674(n) + A186676(n) + A186678(n) + a(n) + A186683(n) + A186685(n) = A052268(n)
a(n) = 0 for n >= 6. - Nathaniel Johnston, May 09 2011

Crossrefs

Formula

a(n) = A186680(n) - A186680(n-1).

A186683 Total number of n-digit numbers requiring 18 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 2, 17, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Comments

A161905(n) + A186650(n) + A186652(n) + A186654(n) + A186656(n) + A186658(n) + A186660(n) + A186662(n) + A186664(n) + A186666(n) + A186668(n) + A186670(n) + A186672(n) + A186674(n) + A186676(n) + A186678(n) + A186681(n) + a(n) + A186685(n) = A052268(n)
a(n) = 0 for n >= 5. - Nathaniel Johnston, May 09 2011

Crossrefs

Programs

  • Mathematica
    PadRight[{0, 2, 17, 5}, 100] (* Paolo Xausa, Jul 30 2024 *)

Formula

a(n) = A186682(n) - A186682(n-1).
Previous Showing 11-18 of 18 results.