A290373 10-adic integer x = ...2943 satisfying x^5 = x.
3, 4, 9, 2, 2, 9, 7, 0, 9, 1, 8, 5, 6, 7, 4, 0, 4, 6, 3, 0, 8, 2, 8, 1, 2, 7, 9, 2, 6, 3, 0, 3, 8, 6, 6, 6, 2, 6, 6, 7, 1, 3, 4, 4, 5, 3, 2, 0, 8, 3, 1, 6, 7, 7, 5, 6, 6, 6, 8, 4, 9, 7, 5, 6, 9, 8, 0, 7, 9, 0, 3, 0, 4, 3, 8, 9, 9, 2, 7, 9, 5, 3, 3, 7, 0, 6, 4, 8
Offset: 0
Examples
3^5 - 3 == 0 mod 10, 43^5 - 43 == 0 mod 10^2, 943^5 - 943 == 0 mod 10^3, 2943^5 - 2943 == 0 mod 10^4. From _Seiichi Manyama_, Aug 01 2019: (Start) 8^(5^0) - 5^(2^0) == 3 mod 10, 8^(5^1) - 5^(2^1) == 43 mod 10^2, 8^(5^2) - 5^(2^2) == 943 mod 10^3, 8^(5^3) - 5^(2^3) == 2943 mod 10^4. (End)
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..9999
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 } } s2 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 A290373(n) str = (10 ** (n + 1) + P(n) - Q(5, n)).to_s.reverse (0..n).map{|i| str[i].to_i} end p A290373(100)
Comments