login
A373206
Numbers m such that m^m == m (mod 10^(len(m) + 2)), where len(m) is the number of digits of m (A055642).
2
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.
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)).
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
KEYWORD
nonn,base
AUTHOR
Marco Ripà, May 28 2024
STATUS
approved