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A373205
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Numbers m such that m^m == m (mod 10^(len(m) + 1)), where len(m) is the number of digits of m (A055642).
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2
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1, 51, 57, 101, 151, 176, 201, 301, 351, 401, 501, 551, 576, 601, 625, 701, 751, 801, 901, 951, 976, 1001, 1376, 2001, 2057, 2751, 3001, 4001, 4193, 4751, 5001, 5376, 6001, 6249, 6751, 7001, 8001, 8751, 9001, 9375, 9376, 10001, 10751, 11001, 12001, 13001
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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By definition, the present sequence is a subsequence of A082576.
For each integer r >= 2 this sequence contains 10^r + 1.
All terms > 1 end in 01, 25, 49, 51, 57, 75, 76, or 93.
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LINKS
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EXAMPLE
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51 is a term since 51 is a 2-digit number and 51^51 == 5051 (mod 10^4) and thus 51^51 == 51 (mod 10^(2 + 1)).
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PROG
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(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 A373205_gen(): # generator of terms
for i in count(0, 100):
for j in (1, 25, 49, 51, 57, 75, 76, 93):
m = i+j
if pow(m, m, 10*10**(len(str(m)))) == m:
yield m
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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