A238638 Position of n-th row of Pascal's triangle in Mathematica-ordered list of partitions of 2^n.

1, 2, 4, 14, 109, 3366, 380480, 592178710, 12245355432908, 42590813279958575804, 35428820136077436448479258280, 643572551892460566707053818908283349242945, 1088540944742787295982636155758383327725184898133092177544054

Offset: 0

Author

Clark Kimberling, Mar 04 2014

Keywords

• nonn

Examples

			The partitions of 4 in Mathematica order are 4, 31, 22, 211, 111.  a(2) = 4 is the position of 211, which as a partition is equivalent to row 2 of Pascal's triangle:  1 2 1 (where the top row is counted as row 0).
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..14
• Manfred Scheucher, C-Code

Crossrefs

Cf. A007318, A080577 (Mathematica ordering), A238639, A238640.

Programs

Maple

p:= (n, k)-> binomial(n, iquo(2*n-k+1, 2)):
g:= (n, k, i)-> `if`(n=0, 1, g(n-p(k, i-1), k, i-1)
    +add(b(n-j, j), j=p(k, i-1)+1..min(n, p(k, i)))):
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-> (m-> add(b(m-j, min(j, m-j)), j=p(n$2)+1..m)
            +g(m-p(n$2), n$2))(2^n):
seq(a(n), n=0..10);  # Alois P. Heinz, Jun 03 2015

Mathematica

r[n_] := Reverse[Sort[Table[Binomial[n, k], {k, 0, n}]]]; Flatten[Table[Position[IntegerPartitions[2^n], r[n]], {n, 0, 6}]]
(* second program: *)
$RecursionLimit = 2000;
p[n_, k_] := Binomial[n, Quotient[2*n - k + 1, 2]];
g[n_, k_, i_] := If[n == 0, 1, g[n - p[k, i - 1], k, i - 1] + Sum[b[n - j, j], {j, p[k, i - 1] + 1, Min[n, p[k, i]]}]];
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_] := Function[m, Sum[b[m - j, Min[j, m - j]], {j, p[n, n] + 1, m}] + g[m - p[n, n], n, n]][2^n];
Table[a[n], {n, 0, 10}] (* Jean-François Alcover, Aug 30 2016, after Alois P. Heinz *)

Extensions

a(7) from Manfred Scheucher, May 29 2015

a(8)-a(12) from Alois P. Heinz, Jun 03 2015