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A100403
Digital root of 6^n.
2
1, 6, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9
OFFSET
0,2
COMMENTS
Also the digital root of k^n for any k == 6 (mod 9). - Timothy L. Tiffin, Dec 02 2023
FORMULA
From Timothy L. Tiffin, Dec 01 2023: (Start)
a(n) = 9 for n >= 2.
G.f.: (1+5x+3x^2)/(1-x).
a(n) = A100401(n) for n <> 1.
a(n) = A010888(A000400(n)) = A010888(A001024(n)) = A010888(A009968(n)) = A010888(A009977(n)) = A010888(A009986(n)) = A010888(A159991(n)). (End)
E.g.f.: 9*exp(x) - 3*x - 8. - Elmo R. Oliveira, Aug 09 2024
EXAMPLE
For n=8, the digits of 6^8 = 1679616 sum to 36, whose digits sum to 9. So, a(8) = 9. - Timothy L. Tiffin, Dec 01 2023
MATHEMATICA
PadRight[{1, 6}, 100, 9] (* Timothy L. Tiffin, Dec 03 2023 *)
PROG
(PARI) a(n) = if( n<2, [1, 6][n+1], 9); \\ Joerg Arndt, Dec 03 2023
KEYWORD
easy,nonn,base
AUTHOR
Cino Hilliard, Dec 31 2004
STATUS
approved