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 1-5 of 5 results.

A066655 Number of partitions of n*(n-1)/2.

Original entry on oeis.org

1, 1, 3, 11, 42, 176, 792, 3718, 17977, 89134, 451276, 2323520, 12132164, 64112359, 342325709, 1844349560, 10015581680, 54770336324, 301384802048, 1667727404093, 9275102575355, 51820051838712, 290726957916112, 1637293969337171, 9253082936723602
Offset: 1

Views

Author

Roberto E. Martinez II, Jan 10 2002

Keywords

Comments

Number of partitions of the number of edges of the complete graph of order n, K_n.

Examples

			a(4) = p(6) = 11.
		

Crossrefs

Programs

  • Mathematica
    Table[PartitionsP[n(n-1)/2], {n, 1, 30}]
  • MuPAD
    combinat::partitions::count(binomial(n+2,n)) $n=-1..40 // Zerinvary Lajos, Apr 16 2007
    
  • PARI
    a(n) = numbpart(n*(n-1)/2); \\ Michel Marcus, Dec 18 2017

Formula

a(n) = p(n*(n-1)/2) = A000041(n*(n-1)/2).
a(n) ~ exp(Pi*sqrt(n*(n-1)/3))/(2*sqrt(3)*n*(n - 1)). - Ilya Gutkovskiy, Jan 13 2017
a(n) ~ exp(Pi*(n - 1/2) / sqrt(3)) / (2*sqrt(3)*n^2). - Vaclav Kotesovec, May 17 2018

Extensions

More terms from Vladeta Jovovic, Jan 12 2002
Edited by Dean Hickerson, Jan 14 2002

A104382 Triangle, read by rows, where T(n,k) equals number of distinct partitions of triangular number n*(n+1)/2 into k different summands for n>=k>=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 7, 12, 6, 1, 1, 10, 27, 27, 10, 1, 1, 13, 52, 84, 57, 14, 1, 1, 17, 91, 206, 221, 110, 21, 1, 1, 22, 147, 441, 674, 532, 201, 29, 1, 1, 27, 225, 864, 1747, 1945, 1175, 352, 41, 1, 1, 32, 331, 1575, 4033, 5942, 5102, 2462, 598, 55, 1, 1, 38, 469
Offset: 1

Views

Author

Paul D. Hanna, Mar 04 2005

Keywords

Comments

Secondary diagonal equals partitions of n - 1 (A000065).
Third diagonal is A104384.
Third column is A104385.
Row sums are A104383 where limit_{n --> inf} A104383(n+1)/A104383(n) = exp(sqrt(Pi^2/6)) = 3.605822247984...

Examples

			Rows begin:
1;
1, 1;
1, 2, 1;
1, 4, 4, 1;
1, 7, 12, 6, 1;
1, 10, 27, 27, 10, 1;
1, 13, 52, 84, 57, 14, 1;
1, 17, 91, 206, 221, 110, 21, 1;
1, 22, 147, 441, 674, 532, 201, 29, 1;
1, 27, 225, 864, 1747, 1945, 1175, 352, 41, 1;
1, 32, 331, 1575, 4033, 5942, 5102, 2462, 598, 55, 1; ...
		

References

  • Abramowitz, M. and Stegun, I. A. (Editors). "Partitions into Distinct Parts." S24.2.2 in Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables, 9th printing. New York: Dover, pp. 825-826, 1972.

Crossrefs

Programs

  • PARI
    T(n,k)=if(n
    				

Formula

T(n, 1) = T(n, n) = 1.
T(n, n-1) = A000065(n).
T(n, 2) = [(n*(n+1)/2-1)/2].
From Álvar Ibeas, Jul 23 2020: (Start)
T(n, k) = A008284((n-k+1)*(n+k)/2, k).
T(n, k) = A026820((n-k)*(n+k+1)/2, k), with A026820(0, k) = 1. (End)

A126683 Number of partitions of the n-th triangular number n(n+1)/2 into distinct odd parts.

Original entry on oeis.org

1, 1, 1, 1, 2, 4, 8, 16, 33, 68, 144, 312, 686, 1523, 3405, 7652, 17284, 39246, 89552, 205253, 472297, 1090544, 2525904, 5867037, 13663248, 31896309, 74628130, 174972341, 411032475, 967307190, 2280248312, 5383723722, 12729879673, 30141755384, 71462883813
Offset: 0

Views

Author

Moshe Shmuel Newman, Feb 15 2007

Keywords

Comments

Also the number of self-conjugate partitions of the n-th triangular number.

Examples

			The 5th triangular number is 15. Writing this as a sum of distinct odd numbers: 15 = 11 + 3 + 1 = 9 + 5 + 1 = 7 + 5 + 3 are all the possibilities. So a(5) = 4.
		

Crossrefs

Sequences A066655 and A104383 do the same thing for triangular numbers, with partitions or distinct partitions. Sequences A072213 and A072243 are analogs for squares rather than triangular numbers.
Cf. A000217.

Programs

  • Maple
    g:= mul(1+x^(2*j+1),j=0..900): seq(coeff(g,x,n*(n+1)/2),n=0..40); # Emeric Deutsch, Feb 27 2007
    # second Maple program:
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i^2n, 0, b(n-2*i+1, i-1))))
        end:
    a:= n-> b(n*(n+1)/2, ceil(n*(n+1)/4)*2-1):
    seq(a(n), n=0..40);  # Alois P. Heinz, Jan 31 2018
  • Mathematica
    a[n_] := SeriesCoefficient[QPochhammer[-x, x^2], {x, 0, n*(n+1)/2}];
    Table[a[n], {n, 0, 40}] (* Jean-François Alcover, May 25 2018 *)

Extensions

More terms from Emeric Deutsch, Feb 27 2007
a(0)=1 prepended by Alois P. Heinz, Jan 31 2018

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 1-5 of 5 results.