A297815 Number of positive integers with n digits whose digit sum is equal to its digit product.

9, 1, 6, 12, 40, 30, 84, 224, 144, 45, 605, 495, 1170, 1092, 210, 240, 2448, 4896, 15846, 3420, 1750, 462, 15939, 0, 8100, 67925, 80730, 19656, 11774, 164430, 930, 29760, 197472, 0, 0, 1260, 23976, 50616, 54834, 395200, 1248860, 4253340, 75852, 0, 42570

Offset: 1

Author

Reiner Moewald, Jan 06 2018

Keywords

• nonn
• base

Examples

			The only term with two digits is 22: 2 * 2 = 2 + 2.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A034710, A061672.

Programs

Mathematica

cperm[w_] := Length[w]!/Times @@ ((Last /@ Tally[w])!); ric[s_, p_, w_, tg_] := Block[{d}, If[tg == 0, If[s == p, tot += cperm@ w], Do[ If[p*d > s + d + (tg-1)*9, Break[]]; ric[s+d, p*d, Append[w,d], tg-1], {d, Last@ w, 9}]]]; a[n_] := (tot=0; ric[#, #, {#}, n-1] & /@ Range[9]; tot); Array[a, 45] (* Giovanni Resta, Feb 05 2018 *)

Python

import math
def digitProd(natNumber):
    digitProd = 1
    for letter in str(natNumber):
        digitProd *= int(letter)
    return digitProd
def digitSum(natNumber):
    digitSum = 0
    for letter in str(natNumber):
        digitSum += int(letter)
    return digitSum
for n in range(24):
    count = 0
    for a in range(int(math.pow(10,n)), int(math.pow(10, n+1))):
        if digitProd(a) == digitSum(a):
            count += 1
    print(n+1, count)

Python

from sympy.utilities.iterables import combinations_with_replacement
from sympy import prod, factorial
def A297815(n):
    f = factorial(n)
    return sum(f//prod(factorial(d.count(a)) for a in set(d)) for d in combinations_with_replacement(range(1,10),n) if prod(d) == sum(d)) # Chai Wah Wu, Feb 06 2018

Extensions

a(10) and a(23) corrected by and a(25)-a(45) from Giovanni Resta, Feb 05 2018