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

1,26

Comments

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

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

Links

Table of n, a(n) for n=1..87.

Example

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

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 *)

Prog

(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

Crossrefs

Cf. A011540, A161351, A358350.

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

Sequence in context: A337939 A319020 A099200 * A093578 A172398 A070107

Adjacent sequences: A358348 A358349 A358350 * A358352 A358353 A358354

Keyword

nonn,base

Author

Bernard Schott, Nov 16 2022

Status

approved