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.

A224677 Number of compositions of n*(n+1)/2 into sums of positive triangular numbers.

Original entry on oeis.org

1, 1, 2, 7, 40, 351, 4876, 104748, 3487153, 179921982, 14387581923, 1783124902639, 342504341570010, 101962565961894431, 47044167891731682278, 33640402686770010577421, 37282664267078280296013183, 64038780633654058635677191329, 170478465430659361252118580217675
Offset: 0

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Author

Paul D. Hanna, Apr 14 2013

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n) option remember; local i; if n=0 then 1 else 0;
          for i while i*(i+1)/2<=n do %+b(n-i*(i+1)/2) od; %  fi
        end:
    a:= n-> b(n*(n+1)/2):
    seq(a(n), n=0..20);  # Alois P. Heinz, Feb 05 2018
  • Mathematica
    b[n_] := b[n] = If[n==0, 1, Sum[If[IntegerQ[Sqrt[8j+1]], b[n-j], 0], {j, 1, n}]];
    a[n_] := b[n(n+1)/2];
    a /@ Range[0, 20] (* Jean-François Alcover, Oct 31 2020, after Alois P. Heinz in A023361 *)
  • PARI
    {a(n)=polcoeff(1/(1-sum(r=1,n+1, x^(r*(r+1)/2)+x*O(x^(n*(n+1)/2)))), n*(n+1)/2)}
    for(n=0, 20, print1(a(n), ", "))

Formula

a(n) = A023361(n*(n+1)/2), where A023361(n) is the number of compositions of n into positive triangular numbers.
a(n) = [x^(n*(n+1)/2)] 1/(1 - Sum_{k>=1} x^(k*(k+1)/2)).

A299032 Number of ordered ways of writing n-th triangular number as a sum of n squares of positive integers.

Original entry on oeis.org

1, 1, 0, 3, 6, 0, 12, 106, 420, 2718, 18240, 120879, 694320, 5430438, 40668264, 300401818, 2369504386, 19928714475, 174151735920, 1543284732218, 14224347438876, 135649243229688, 1331658133954940, 13369350846412794, 138122850643702056, 1462610254141337590
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 01 2018

Keywords

Examples

			a(4) = 6 because fourth triangular number is 10 and we have [4, 4, 1, 1], [4, 1, 4, 1], [4, 1, 1, 4], [1, 4, 4, 1], [1, 4, 1, 4] and [1, 1, 4, 4].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, t) option remember; local i; if n=0 then
          `if`(t=0, 1, 0) elif t<1 then 0 else 0;
          for i while i^2<=n do %+b(n-i^2, t-1) od; % fi
        end:
    a:= n-> b(n*(n+1)/2, n):
    seq(a(n), n=0..25);  # Alois P. Heinz, Feb 05 2018
  • Mathematica
    Table[SeriesCoefficient[(-1 + EllipticTheta[3, 0, x])^n/2^n, {x, 0, n (n + 1)/2}], {n, 0, 25}]

Formula

a(n) = [x^(n*(n+1)/2)] (Sum_{k>=1} x^(k^2))^n.

A299031 Number of ordered ways of writing n-th triangular number as a sum of n squares of nonnegative integers.

Original entry on oeis.org

1, 1, 0, 3, 18, 60, 252, 1576, 10494, 64152, 458400, 3407019, 27713928, 225193982, 1980444648, 17626414158, 165796077562, 1593587604441, 15985672426992, 163422639872978, 1729188245991060, 18743981599820280, 208963405365941380, 2378065667103672024, 27742569814633730608
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 01 2018

Keywords

Examples

			a(3) = 3 because third triangular number is 6 and we have [4, 1, 1], [1, 4, 1] and [1, 1, 4].
		

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[(1 + EllipticTheta[3, 0, x])^n/2^n, {x, 0, n (n + 1)/2}], {n, 0, 24}]

Formula

a(n) = [x^(n*(n+1)/2)] (Sum_{k>=0} x^(k^2))^n.
Showing 1-3 of 3 results.