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.

A369438 a(n) = [x^(n^3)] Product_{k=1..n} (x^(k^3) + 1 + 1/x^(k^3)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 11, 24, 46, 106, 238, 537, 1318, 3007, 7027, 18199, 43202, 105900, 279860, 688474, 1741235, 4641670, 11790546, 30529486, 82306963, 213852619, 563866091, 1531711961, 4047719392, 10835966180, 29624064007, 79423421277, 215083283638
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 23 2024

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n>(i*(i+1)/2)^2, 0,
         `if`(i=0, 1, b(n, i-1)+b(n+i^3, i-1)+b(abs(n-i^3), i-1)))
        end:
    a:= n-> b(n^3, n):
    seq(a(n), n=0..35);  # Alois P. Heinz, Jan 23 2024
  • Mathematica
    Table[Coefficient[Product[(x^(k^3) + 1 + 1/x^(k^3)), {k, 1, n}], x, n^3], {n, 0, 34}]

A367416 Triangle read by rows: T(n,k) = number of solutions to +- 1^k +- 2^k +- 3^k +- ... +- n^k is a k-th power, n >= 2.

Original entry on oeis.org

4, 8, 1, 16, 1, 32, 0, 2, 64, 6, 128, 8, 256, 16, 4, 512, 26, 1024, 17, 10, 2048, 67, 4, 3, 4096, 100, 10, 8192, 137, 34, 6, 16384, 426, 28, 1, 32768, 661, 96, 6, 65536, 1351, 146, 16, 8, 131072, 2637, 230, 15, 262144, 3831, 258, 40, 524288, 8095, 1130, 50
Offset: 2

Views

Author

Jean-Marc Rebert, Jan 26 2024

Keywords

Comments

In the case of n = 1, there are solutions for all k. In particular, 1^k is always a k-th power and -(1^k) is a k-th power for odd k. As a formula: T(1,k) = 1 + (k mod 2). This row is not included in the sequence.

Examples

			Triangle begins:
            k = 1      2     3   4  5
  n= 2:         4;
  n= 3:         8,     1;
  n= 4:        16,     1;
  n= 5:        32,     0,    2;
  n= 6:        64,     6;
  n= 7:       128,     8;
  n= 8:       256,    16,    4;
  n= 9:       512,    26;
  n=10:      1024,    17,   10;
  n=11:      2048,    67,    4,  3;
  n=12:      4096,   100,   10;
  n=13:      8192,   137,   34,  6;
  n=14:     16384,   426,   28,  1;
  n=15:     32768,   661,   96,  6;
  n=16:     65536,  1351,  146, 16, 8;
  n=17:    131072,  2637,  230, 15;
  n=18:    262144,  3831,  258, 40;
  n=19:    524288,  8095, 1130, 50;
  n=20:   1048576, 15241,  854, 77, 6;
  ...
The T(6,2) = 6 solutions are:
  - 1^2 - 2^2 + 3^2 - 4^2 + 5^2 + 6^2 = 49 = 7^2,
  - 1^2 - 2^2 + 3^2 + 4^2 + 5^2 - 6^2 =  9 = 3^2,
  - 1^2 - 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 81 = 9^2,
  + 1^2 - 2^2 + 3^2 - 4^2 - 5^2 + 6^2 =  1 = 1^2,
  + 1^2 + 2^2 - 3^2 + 4^2 + 5^2 - 6^2 =  1 = 1^2,
  + 1^2 + 2^2 + 3^2 - 4^2 - 5^2 + 6^2 =  9 = 3^2.
		

Crossrefs

Programs

  • PARI
    f(k,u)=my(x=0,v=vector(#u));for(i=1,#u,u[i]=if(u[i]==0,-1,1);v[i]=i^k);u*v~
    is(k,u)=my(x=f(k,u));ispower(x,k)
    T(n,k)=my(u=vector(n,i,[0,1]),nbsol=0);if(k%2==1,u[1]=[1,1]);forvec(X=u,if(is(k,X),nbsol++));if(k%2==1,nbsol*=2);nbsol

A368348 a(n) = [x^(n^4)] Product_{k=1..n} (x^(k^4) + 1/x^(k^4)).

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 0, 0, 19, 26, 0, 0, 40, 129, 0, 0, 616, 785, 0, 0, 4080, 9309, 0, 0, 44775, 72659, 0, 0, 430297, 781505, 0, 0, 3934457, 7765047, 0, 0, 44740433, 78818429, 0, 0, 463089552, 900950811, 0, 0, 5344766190, 9806206864, 0, 0
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 25 2024

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; (m-> `if`(n>m, 0, `if`(n=m, 1,
          b(abs(n-i^4), i-1)+b(n+i^4, i-1))))(i*(i+1)*(2*i+1)*(3*i^2+3*i-1)/30)
        end:
    a:= n-> `if`(irem(n, 4)>1, 0, b(n^4, n)):
    seq(a(n), n=0..43);  # Alois P. Heinz, Jan 25 2024
  • Mathematica
    b[n_, i_] := b[n, i] = Function[m, If[n > m, 0, If[n == m, 1, b[Abs[n-i^4], i-1] + b[n+i^4, i-1]]]][i*(i+1)*(2*i+1)*(3*i^2+3*i-1)/30];
    a[n_] := If[Mod[n, 4] > 1, 0, b[n^4, n]];
    Table[a[n], {n, 0, 55}] (* Jean-François Alcover, Feb 14 2025, after Alois P. Heinz *)

Extensions

a(46)-a(59) from Alois P. Heinz, Jan 25 2024
Showing 1-3 of 3 results.