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.

A116365 Sum of the sizes of the tails below the Durfee squares of all partitions of n.

Original entry on oeis.org

0, 1, 3, 6, 11, 20, 33, 56, 86, 136, 200, 301, 429, 621, 868, 1219, 1669, 2297, 3091, 4171, 5542, 7357, 9648, 12652, 16402, 21250, 27298, 35003, 44556, 56637, 71515, 90160, 113046, 141464, 176189, 219053, 271149, 335044, 412447, 506787, 620597
Offset: 1

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Author

Emeric Deutsch, Feb 12 2006

Keywords

Examples

			a(4) = 6 because the bottom tails of the five partitions of 4, namely [4], [3,1], [2,2], [2,1,1] and [1,1,1,1], are { }, [1], { }, [1,1] and [1,1,1], respectively, having total size 0+1+0+2+3=6.
		

References

  • G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976 (pp. 27-28).
  • G. E. Andrews and K. Eriksson, Integer Partitions, Cambridge Univ. Press, 2004 (pp. 75-78).

Crossrefs

Programs

  • Maple
    g:=sum(z^(k^2)/product((1-z^j)*(1-(t*z)^j),j=1..k),k=1..10): dgdt1:=simplify(subs(t=1,diff(g,t))): dgdt1ser:=series(dgdt1,z=0,55): seq(coeff(dgdt1ser,z,n),n=1..48);
    # second Maple program:
    b:= proc(n, i) option remember;
          `if`(n=0, 1, `if`(i<1, 0, b(n, i-1)+`if`(i>n, 0, b(n-i, i))))
        end:
    a:= n-> add(k*add(b(k, d) *b(n-d^2-k, d),
                d=0..floor(sqrt(n))), k=0..n-1):
    seq(a(n), n=1..40);  # Alois P. Heinz, Apr 2012
  • Mathematica
    b[n_, i_] := b[n, i] = If[n==0, 1, If[i<1, 0, b[n, i-1] + If[i>n, 0, b[n-i, i]]]]; a[n_] := Sum[k*Sum[b[k, d]*b[n-d^2-k, d], {d, 0, Floor[Sqrt[n]]}], {k, 0, n-1}]; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Mar 31 2015, after Alois P. Heinz *)

Formula

a(n) = Sum_{k=0..n-1} k*A114087(n,k).
G.f.: [(d/dt){sum(q^(k^2)/product((1-q^j)(1-(tq)^j), j=1..k), k=1..oo)}]_{t=1}.
a(n) = (n*A000041(n)-A116503(n))/2. - Vladeta Jovovic, Feb 18 2006
a(n) ~ (1/(8*sqrt(3)) - sqrt(3) * (log(2))^2 / (4*Pi^2)) * exp(Pi*sqrt(2*n/3)). - Vaclav Kotesovec, Jan 03 2019