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

1,1

Links

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

Example

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

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 *)

Prog

(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

Crossrefs

Cf. A034710, A061672.

Sequence in context: A195712 A296473 A213916 * A021920 A367960 A280703

Adjacent sequences: A297812 A297813 A297814 * A297816 A297817 A297818

Keyword

nonn,base

Author

Reiner Moewald, Jan 06 2018

Extensions

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

Status

approved