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A255577
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Numbers n not coprime to 10 such that there exists an integer k > 1 where n^k contains n as its last digits in base 10.
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1
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2, 4, 5, 6, 8, 12, 16, 24, 25, 28, 32, 36, 44, 48, 52, 56, 64, 68, 72, 75, 76, 84, 88, 92, 96, 104, 112, 125, 128, 136, 144, 152, 168, 176, 184, 192, 208, 216, 224, 232, 248, 256, 264, 272, 288, 296, 304, 312, 328
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OFFSET
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1,1
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COMMENTS
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Intersection of A065502 (numbers not coprime with 10) and A072495 (k-morphic numbers).
Also defined as all n not coprime with 10 where there exists k > 1 such that n^k mod 10^floor(log_10(n)) = n.
For n > 1, a(n) <= a(n-1) + 2^(ceiling(log_10(a(n))) + 1) (conjectured).
For a(n) >= 10^k where k >= 1, there exists a(m) = a(n) mod 10^j where m < n and j < k.
n with d digits is in the sequence if and only if n is either divisible by 2^d but not by 5, or divisible by 5^d but not by 2.
For d >= 2 the number of terms with d digits is 4*5^(d-1) + 2^(d-1) - 4*floor(5^d/50) - floor(2^d/20) - x(d) where x(d) = 3 if d == 2 or 3 mod 4, 2 otherwise.
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LINKS
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FORMULA
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For n > 6, a(n) > 2n since no term is divisible by 10 (but all are divisible by either 2 or 5). - Charles R Greathouse IV, May 13 2015
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EXAMPLE
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For n = 2, we have n^5 = 2^5 = 32, whose last digit is 2 = n, so 2 is in the sequence.
For n = 3, we have n^5 = 3^5 = 243, so 3 is in the sequence.
For n = 4, we have n^3 = 4^3 = 64, so 4 is in the sequence.
...
As a counterexample, n = 41 is not in the sequence because it is coprime with 10, even though we have 41^6 = 4750104241, whose last 2 digits are 41.
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MAPLE
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F:= d -> (seq(seq(2^d*(5*j+i), i=1..4), j=0..5^(d-1)-1), seq(5^d*(2*j+1), j=0..2^(d-1)-1)):
sort(convert({seq(F(d), d=1..4)}, list)); # Robert Israel, May 14 2015
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PROG
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See a255577.tcl in LINKS
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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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