A373206 Numbers m such that m^m == m (mod 10^(len(m) + 2)), where len(m) is the number of digits of m (A055642).

1, 751, 1001, 2001, 2751, 3001, 4001, 5001, 5376, 6001, 6751, 7001, 8001, 9001, 10001, 18751, 20001, 30001, 40001, 50001, 58751, 60001, 69376, 70001, 80001, 90001, 98751, 100001, 110001, 120001, 130001, 138751, 140001, 150001, 160001, 170001, 180001, 190001

Offset: 1

Author

Marco Ripà, May 28 2024

Keywords

• nonn
• base

Comments

By definition, this sequence is a subsequence of A082576 and also a subsequence of A373205.

For each integer r >= 3 the sequence contains 10^r + 1.

All terms > 9001 end in 0001 (e.g., 10001), 0625 (e.g., 390625), 1249 (e.g., 781249), 8751 (e.g., 18751), 9376 (e.g., 69376), and possibly in 4193, 7057, or 9375.

Examples

			751 is a term since 751 is a 3-digit number and 751^751 == 500751 (mod 10^6) and thus 751^751 == 751 (mod 10^(3 + 2)).
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..351 (All terms up to 20 million)
• Index entries for sequences related to automorphic numbers

Crossrefs

Cf. A055642, A082576, A373205.

Programs

Mathematica

Select[Range[200000],PowerMod[#,#,10^(IntegerLength[#]+2)]==#&] (* Harvey P. Dale, Dec 05 2024 *)

PARI

for (len_m = 1, 5, for (m = 10^(len_m - 1), 10^len_m - 1, if (m == Mod(m, 10^(len_m + 1))^m, print1(m, ", "))));

Python

from itertools import count
def A373206_gen(): # generator of terms
    yield from (1, 751, 1001, 2001, 2751, 3001, 4001, 5001, 5376, 6001, 6751, 7001, 8001, 9001)
    for i in count(10000,10000):
        for j in (1,625,1249,4193,7057,8751,9375,9376):
            m = i+j
            if pow(m,m,100*10**(len(str(m)))) == m:
                yield m
A373206_list = list(islice(A373206_gen(),20)) # Chai Wah Wu, Jun 02 2024

Formula

For all terms m, m^m == m (mod 10^(floor(log_10(m)) + 3)).