A346217 Integers m, with k digits, such that Sum_{i=1..k} (m without its i-th digit)/(its i-th digit) is an integer.
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 111, 122, 124, 126, 142, 144, 155, 162, 168, 186, 221, 222, 244, 248, 261, 263, 284, 288, 324, 326, 333, 342, 346, 362, 364, 366, 442, 444, 488, 555, 621, 623, 648, 663, 666, 684, 728, 742, 777, 812
Offset: 1
Examples
124 gives 12/4 + 14/2 + 24/1 = 34, an integer, so 124 is a term. 221 gives 21/2 + 21/2 + 22/1 = 43, an integer, so 221 is a term.
Links
- Michel Marcus, Table of n, a(n) for n = 1..5000
- Carlos Rivera, Puzzle 1045. One nice puzzle from Paolo Lava, The Prime Puzzles and Problems Connection.
Programs
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PARI
subs(d, j) = {my(x=""); for (k=1, #d, if (j != k, x = concat(x, d[k]));); eval(x);} isok(m) = {my(d=digits(m), res); if (vecmin(d), res = sum(j=1, #d, subs(d, j)/d[j]); (denominator(res)==1););} -
Python
from fractions import Fraction def ok(n): s = str(n) if '0' in s: return False if len(s) == 1: return True return sum(Fraction(int(s[:i]+s[i+1:]), int(s[i])) for i in range(len(s))).denominator == 1 print(list(filter(ok, range(813)))) # Michael S. Branicky, Jul 11 2021