A384597 Integers k such that k + 1 has a divisor that is an anagram of k, which must have the same number of digits as k.

1, 41, 73, 631, 793, 6031, 6391, 6733, 7412, 7520, 7993, 8627, 9710, 25147, 37112, 43916, 49316, 51427, 60031, 60391, 60733, 62314, 63214, 63991, 66331, 67393, 67933, 70211, 71132, 72101, 74102, 74912, 75020, 75290, 78260, 79993, 81103, 85712, 86927, 89627

Offset

1,2

Comments

This sequence has infinitely many terms, since 60*10^m + 31 is a term for all positive integers m, as (60*10^m + 31) + 1 = 2*(30*10^m + 16).

A100412 is a subsequence of a(n), since if m is in A100412, then m + 1 = 2*reversal(m).

Links

Table of n, a(n) for n=1..40.

Example

73 is in this sequence since 73 + 1 = 37*2, where 37 is an anagram of 73.

Mathematica

{1}~Join~Select[Range[100000], ContainsAny[IntegerDigits/@Divisors[#+1], Complement[Permutations[IntegerDigits[#]], {IntegerDigits[#]}]]&] (* James C. McMahon, Jun 10 2025 *)

Prog

(Python)
def ok(k):
    return any((k+1)%d==0 and sorted(str(d))==sorted(str(k)) and len(str(d))==len(str(k)) for d in range(1, k+2))
print(", ".join(map(str, [k for k in range(1, 100000) if ok(k)])))
(PARI) isok(k) = my(s=vecsort(digits(k))); fordiv(k+1, d, if (vecsort(digits(d)) == s, return(1))); \\ Michel Marcus, Jun 04 2025

Crossrefs

Cf. A100412.

Sequence in context: A142038 A213047 A269426 * A224671 A044107 A044488

Adjacent sequences: A384594 A384595 A384596 * A384598 A384599 A384600

Keyword

nonn,base

Author

Gonzalo Martínez, Jun 04 2025

Status

approved