A290374 10-adic integer x = ...7057 satisfying x^5 = x.
7, 5, 0, 7, 7, 0, 2, 9, 0, 8, 1, 4, 3, 2, 5, 9, 5, 3, 6, 9, 1, 7, 1, 8, 7, 2, 0, 7, 3, 6, 9, 6, 1, 3, 3, 3, 7, 3, 3, 2, 8, 6, 5, 5, 4, 6, 7, 9, 1, 6, 8, 3, 2, 2, 4, 3, 3, 3, 1, 5, 0, 2, 4, 3, 0, 1, 9, 2, 0, 9, 6, 9, 5, 6, 1, 0, 0, 7, 2, 0, 4, 6, 6, 2, 9, 3, 5, 1
Offset: 0
Examples
7^5 - 7 == 0 mod 10, 57^5 - 57 == 0 mod 10^2, 57^5 - 57 == 0 mod 10^3, 7057^5 - 7057 == 0 mod 10^4. From _Seiichi Manyama_, Aug 01 2019: (Start) 2^(5^0) + 5^(2^0) == 7 mod 10, 2^(5^1) + 5^(2^1) == 57 mod 10^2, 2^(5^2) + 5^(2^2) == 57 mod 10^3, 2^(5^3) + 5^(2^3) == 7057 mod 10^4. (End)
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..9999
Crossrefs
Programs
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Ruby
def P(n) s1, s2 = 2, 8 n.times{|i| m = 10 ** (i + 1) (0..9).each{|j| k1, k2 = j * m + s1, (9 - j) * m + s2 if (k1 ** 5 - k1) % (m * 10) == 0 && (k2 ** 5 - k2) % (m * 10) == 0 s1, s2 = k1, k2 break end } } s1 end def Q(s, n) n.times{|i| m = 10 ** (i + 1) (0..9).each{|j| k = j * m + s if (k ** 2 - k) % (m * 10) == 0 s = k break end } } s end def A290374(n) str = (P(n) + Q(5, n)).to_s.reverse (0..n).map{|i| str[i].to_i} end p A290374(100)
Comments