A332711 Sum of all numbers in bijective base-n numeration with digit sum n.

0, 1, 5, 28, 203, 1936, 23517, 349504, 6149495, 124999936, 2881935953, 74300836864, 2118007738035, 66142897770496, 2245609694259557, 82351536043343872, 3244079458377786863, 136619472483668525056, 6125138252818308310041, 291271111111111111081984

Offset: 0

Author

Alois P. Heinz, Feb 20 2020

Keywords

• nonn

Comments

The number of numbers in bijective base-n numeration with digit sum n equals the number of compositions of n: A000079(n).

Examples

			a(0) =  0.
a(1) =  1 = 1_bij1.
a(2) =  5 = 3 + 2 = 11_bij2 + 2_bij2.
a(3) = 28 = 13 + 7 + 5 + 3 = 111_bij3 + 21_bij3 + 12_bij3 + 3_bij3.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..387
• Wikipedia, Bijective numeration
• Wikipedia, Digit sum

Crossrefs

Cf. A000079, A007953, A214676.

Programs

Maple

b:= proc(n, k) option remember; `if`(n=0, [1, 0], add((p->
      [p[1], p[2]*k+p[1]*d])(b(n-d, k)), d=1..min(n, k)))
    end:
a:= n-> b(n$2)[2]:
seq(a(n), n=0..23);

Mathematica

b[n_, k_] := b[n, k] = If[n == 0, {1, 0}, Sum[Function[p,{p[[1]], p[[2]]*k + p[[1]]*d}][b[n - d, k]], {d, 1, Min[n, k]}]];
a[n_] := b[n, n][[2]];
a /@ Range[0, 23] (* Jean-François Alcover, Apr 23 2021, after Alois P. Heinz *)

Formula

a(n) = ((n+1)^n - 2^n) / (n - 1) for n >= 2. - Peter Bala, Sep 28 2023