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

1,2

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.

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

Formula

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

Example

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)).

Mathematica

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

Prog

(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

Crossrefs

Cf. A055642, A082576, A373205.

Sequence in context: A129036 A290272 A291524 * A020391 A355318 A252807

Adjacent sequences: A373203 A373204 A373205 * A373207 A373208 A373209

Keyword

nonn,base

Author

Marco Ripà, May 28 2024

Status

approved