A297815 - OEIS

%I #26 Sep 25 2023 14:59:34

%S 9,1,6,12,40,30,84,224,144,45,605,495,1170,1092,210,240,2448,4896,

%T 15846,3420,1750,462,15939,0,8100,67925,80730,19656,11774,164430,930,

%U 29760,197472,0,0,1260,23976,50616,54834,395200,1248860,4253340,75852,0,42570

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

%H Chai Wah Wu, <a href="/A297815/b297815.txt">Table of n, a(n) for n = 1..10000</a>

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

%t 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 *)

%o (Python)

%o import math

%o def digitProd(natNumber):

%o     digitProd = 1

%o     for letter in str(natNumber):

%o         digitProd *= int(letter)

%o     return digitProd

%o def digitSum(natNumber):

%o     digitSum = 0

%o     for letter in str(natNumber):

%o         digitSum += int(letter)

%o     return digitSum

%o for n in range(24):

%o     count = 0

%o     for a in range(int(math.pow(10,n)), int(math.pow(10, n+1))):

%o         if digitProd(a) == digitSum(a):

%o             count += 1

%o     print(n+1, count)

%o (Python)

%o from sympy.utilities.iterables import combinations_with_replacement

%o from sympy import prod, factorial

%o def A297815(n):

%o     f = factorial(n)

%o     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

%Y Cf. A034710, A061672.

%K nonn,base

%O 1,1

%A _Reiner Moewald_, Jan 06 2018

%E a(10) and a(23) corrected by and a(25)-a(45) from _Giovanni Resta_, Feb 05 2018