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A088157
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Value of (n+1)-th digit in sexagesimal representation of n^n.
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9
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1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 7, 21, 2, 1, 59, 5, 49, 2, 19, 57, 20, 45, 35, 30, 0, 5, 28, 50, 4, 19, 50, 23, 32, 10, 23, 38, 16, 45, 29, 6
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OFFSET
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0,62
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COMMENTS
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a(n) = d(n) with n^n = Sum(d(k)*60^k: 0 <= d(k) < 60, k >= 0).
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LINKS
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FORMULA
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a(n) = floor(n^n / 60^n) mod 60.
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EXAMPLE
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a(0) = 1, a(k) = 0 for 0 < k < 60 and a(60) = 1.
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MATHEMATICA
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f[n_] := IntegerDigits[n^n, 60, n + 1][[1]]; f[0] = 1; Array[f, 92, 0] (* Robert G. Wilson v, Dec 27 2012 *)
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PROG
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(PARI) a(n)=lift(chinese(chinese(Mod(n, 3^(n+1))^n, Mod(n, 4^(n+1))^n), Mod(n, 5^(n+1))^n))\60^n \\ Charles R Greathouse IV, Dec 27 2012
(Haskell)
a088157 n = mod (div (n ^ n) (60 ^ n)) 60
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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