A088157 Value of (n+1)-th digit in sexagesimal representation of n^n.

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

Offset: 0

Author

Reinhard Zumkeller, Sep 20 2003

Keywords

• nonn
• base
• look

Comments

a(n) = d(n) with n^n = Sum(d(k)*60^k: 0 <= d(k) < 60, k >= 0).

Examples

			a(0) = 1, a(k) = 0 for 0 < k < 60 and a(60) = 1.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
• Eric Weisstein's World of Mathematics, Sexagesimal

Crossrefs

Cf. A000312, A088150, A088151, A088152, A088153, A088154, A088155, A088156, A159991.

Programs

Haskell

a088157 n = mod (div (n ^ n) (60 ^ n)) 60
-- Reinhard Zumkeller, Mar 14 2014

Mathematica

f[n_] := IntegerDigits[n^n, 60, n + 1][[1]]; f[0] = 1; Array[f, 92, 0] (* Robert G. Wilson v, Dec 27 2012 *)

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

Formula

a(n) = floor(n^n / 60^n) mod 60.