A358351 Number of values of m such that m + (sum of digits of m) + (product of digits of m) is n.

0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 2, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 3, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 2, 0, 0, 1, 1, 1, 1, 0, 2, 0, 0, 0, 2, 1, 0, 0, 1, 0, 2, 1, 0, 0, 0, 1, 2, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 2, 0

Offset: 1

Author

Bernard Schott, Nov 16 2022

Keywords

• nonn
• base

Comments

a(n) is the number of times n occurs in A161351.

a(n) > 0 iff n is in A358350.

Examples

			A161351(15) = 15 + (1+5) + (1*5) = 26 =  21 + (2+1) + (2*1) = A161351(21), so, a(26) = 2.
		

Crossrefs

Cf. A011540, A161351, A358350.

Similar: A230093 (m+digitsum), A230103 (m+digitprod).

Programs

Mathematica

f[n_] := n + Total[(d = IntegerDigits[n])] + Times @@ d; With[{m = 100}, BinCounts[Table[f[n], {n, 1, m}], {1, m, 1}]] (* Amiram Eldar, Nov 16 2022 *)

PARI

first(n) = {my(res = vector(n)); for(i = 1, n, c = i + sumdigits(i) + vecprod(digits(i)); if(c <= n, res[c]++ ) ); res } \\ David A. Corneth, Nov 16 2022

Python

from itertools import combinations_with_replacement
from math import prod
def A358351(n):
    c = set()
    for l in range(1,len(str(n))+1):
        l1, l2 = 10**l, 10**(l-1)
        for d in combinations_with_replacement(tuple(range(10)),l):
            s, p = sum(d), prod(d)
            if l1>(m:=n-s-p)>=l2 and sorted(int(d) for d in str(m)) == list(d):
                c.add(m)
    return len(c) # Chai Wah Wu, Nov 20 2022