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

Author

Gonzalo Martínez, Jun 04 2025

Keywords

• nonn
• base

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).

Examples

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

Crossrefs

Cf. A100412.

Programs

Mathematica

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

PARI

isok(k) = my(s=vecsort(digits(k))); fordiv(k+1, d, if (vecsort(digits(d)) == s, return(1))); \\ Michel Marcus, Jun 04 2025

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)])))