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A393453
Integers k, not divisible by any of their digits, such that k = Sum_{j = digits of k} k/j.
2
263, 623, 3446, 4346, 4463, 4634, 4643, 6434, 6443, 26999, 29699, 29969, 34886, 38486, 38846, 43886, 44666, 46466, 46646, 46838, 46883, 48386, 48638, 48683, 48863, 62999, 64466, 64646, 64838, 64883, 66446, 68438, 68483, 68834, 68843, 69299, 69929, 83486, 83846
OFFSET
1,1
COMMENTS
From Michael S. Branicky, Feb 17 2026: (Start)
Also, integers k, not divisible by any of their digits, such that 1 = Sum_{j = digits of k} 1/j.
Terms cannot have any 0's or 1's and must have less than d digits d for d in 2..9. All terms must be < 999999999 since the reciprocals of the digits of any higher number will add to more than 1.
If the divisibility constraint is removed, then we recover A037268. (End)
LINKS
FORMULA
A037268 setminus A038770. - Michael S. Branicky, Feb 17 2026
EXAMPLE
a(1) = 263 -> 263/2 + 263/6 + 263/3 = 263.
a(9) = 6443 -> 6443/6 + 6443/4 + 6443/4 + 6443/3 = 6443.
MATHEMATICA
Select[Range[10^5], Function[{k, d}, If[MemberQ[d, 0], False, And[NoneTrue[Union[d], Divisible[k, #] &], k == Total[k/d] ] ] ] @@ {#, IntegerDigits[#]} &] (* Michael De Vlieger, Feb 16 2026 *)
PROG
(Python)
from collections import Counter
from fractions import Fraction
def ok(k):
digs = Counter(map(int, str(k)))
if 0 in digs or 1 in digs or any(k%d == 0 for d in digs): return False
return 1 == sum(digs[d]*Fraction(1, d) for d in digs)
print([k for k in range(10**5) if ok(k)]) # Michael S. Branicky, Feb 16 2026
(PARI) isok(k) = my(d=digits(k)); if (vecmin(d), my(v=vector(#d, i, k/d[i])); (#select(x->(type(x)!="t_INT"), v) == #d) && (vecsum(v) == k)); \\ Michel Marcus, Feb 17 2026
CROSSREFS
Subsequence of A037268.
Sequence in context: A320710 A142754 A245877 * A142379 A128654 A236245
KEYWORD
nonn,base,fini,full
AUTHOR
Paolo P. Lava, Feb 15 2026
STATUS
approved