A003634 Smallest positive integer that is n times its digit sum, or 0 if no such number exists.

1, 18, 27, 12, 45, 54, 21, 72, 81, 10, 198, 108, 117, 126, 135, 144, 153, 162, 114, 180, 378, 132, 207, 216, 150, 234, 243, 112, 261, 270, 372, 576, 594, 102, 315, 324, 111, 342, 351, 120, 738, 756, 516, 792, 405, 230, 423, 432, 441, 450, 918, 312, 954, 972

Offset: 1

Author

N. J. A. Sloane, Mira Bernstein

Keywords

• nonn
• base
• easy
• look
• nice

Comments

a(n) = 0 for n = 62, 63, 65, ... (A003635). - Robert G. Wilson v, Aug 15 2000

Examples

			a(3) = 27 because no number less than 27 has a digit sum equal to 3 times the number.
		

References

• J. H. Conway, personal communication.
• Anthony Gardiner, Mathematical Puzzling, Dover Publications, Inc., Mineola, NY, 1987, Page 11.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Donovan Johnson, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A005349, A003635.

Programs

Mathematica

Do[k = n; While[Apply[Plus, RealDigits[k][[1]]]*n != k, k += n]; Print[k], {n, 1, 61}]
With[{ll=Select[Table[{n,n/Total[IntegerDigits[n]]},{n,1000}],IntegerQ[ #[[2]]]&]},Table[Select[ll,#[[2]]==i&,1][[1,1]],{i,60}]] (* Harvey P. Dale, Mar 09 2012 *)

Python

def sd(n): return sum(map(int, str(n)))
def a(n):
  m = 1
  while m != n*sd(m): m += 1
  return m
print([a(n) for n in range(1,62)]) # Michael S. Branicky, Jan 18 2021

Python

from itertools import count, combinations_with_replacement
def A003634(n):
    for l in count(1):
        if 9*l*n < 10**(l-1): return 0
        c = 10**l
        for d in combinations_with_replacement(range(10),l):
            if sorted(str(a:=sum(d)*n)) == [str(e) for e in d] and a>0:
                c = min(c,a)
        if c < 10**l:
            return c # Chai Wah Wu, May 09 2023