%I #51 Feb 17 2024 12:38:57
%S 1,1,1,1,1,1,1,1,1,1,12,7,5,4,4,3,3,3,3,1,11,12,4,7,3,5,2,4,2,1,11,6,
%T 12,3,3,7,2,2,5,1,11,11,4,12,3,4,2,7,2,1,11,6,4,3,12,2,2,2,2,1,11,11,
%U 11,6,3,12,2,3,4,1,11,6,4,3,3,2,12,2,2,1,11,11
%N Let d_r d_{r-1} ... d_1 d_0 be the decimal expansion of n; a(n) is the number of nonnegative integer solutions x_r ... x_0 to the Diophantine equation d_r*x_r + ... + d_0*x_0 = n.
%C For a formula for a(n), please see the Samsonadze article in Links section. a(n) = P(b) if n = b AND the nonzero digits of b are the coefficients a_i (in the article).
%C a(n) is the number of partitions of n into parts that are nonzero digits of n. - _Stefano Spezia_, Feb 17 2024
%H Robert P. P. McKone, <a href="/A364135/b364135.txt">Table of n, a(n) for n = 1..636</a>
%H Eteri Samsonadze, <a href="https://arxiv.org/abs/2108.04756">On the number of integer non-negative solutions of a linear Diophantine equation</a>, arXiv:2108.04756 [math.NT], 2021.
%F a(n) = 1 for n = (d 0 ... 0), the digit d >= 1, the number of zeros >= 0.
%F a(n) = (1 ... 1) + 1 for n = (d ... d), the digit d >= 1, n >= 10.
%e For n = 10: 1*x_1 + 0*x_0 = 10, the solution is x_1 = 10, thus a(10) = 1.
%e For n = 22: 2*x_1 + 2*x_0 = 22, the solutions are (0,11), (2,10), ..., (11,0), thus a(22) = 12.
%t a[n_Integer] := Module[{ds = IntegerDigits[n], p, t, v}, p = Table[If[d == 0, {0}, Range[0, Quotient[n, d]]], {d, ds}]; t = Tuples[p]; v = Select[t, ds . # == n &]; Length[v]]; Table[a[n], {n, 1, 82}] (* _Robert P. P. McKone_, Aug 25 2023 *)
%t a[n_]:=Length[IntegerPartitions[n, All,DeleteCases[ IntegerDigits[n],0]]]; Array[a,82] (* _Stefano Spezia_, Feb 17 2024 *)
%Y Cf. A007954, A034838, A052423, A055642.
%K base,look,nonn
%O 1,11
%A _Ctibor O. Zizka_, Jul 10 2023