A004166 Sum of digits of 3^n.

1, 3, 9, 9, 9, 9, 18, 18, 18, 27, 27, 27, 18, 27, 45, 36, 27, 27, 45, 36, 45, 27, 45, 54, 54, 63, 63, 81, 72, 72, 63, 81, 63, 72, 99, 81, 81, 90, 90, 81, 90, 99, 90, 108, 90, 99, 108, 126, 117, 108, 144, 117, 117, 135, 108, 90, 90, 108, 126, 117, 99

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• base

Comments

All terms a(n), n > 1, are divisible by 9. - M. F. Hasler, Sep 27 2017

Links

• Michel Marcus, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Vincenzo Librandi)
• Albert Frank, Solutions of International Contest Of Logical Sequences, 2002 - 2003. (The original Contest page without solutions was removed but remains available on web.archive.org.)

Crossrefs

Cf. A067500, A118872, A175435.

Cf. sum of digits of k^n: A001370 (k=2), this sequence (k=3), A065713 (k=4), A066001 (k=5), A066002 (k=6), A066003 (k=7), A066004 (k=8), A065999 (k=9), A066005 (k=11), A066006 (k=12), A175527 (k=13).

Programs

Mathematica

Total[IntegerDigits[#]]&/@(3^Range[0,60]) (* Harvey P. Dale, Mar 03 2013 *)
Table[Total[IntegerDigits[3^n]], {n, 0, 60}] (* Vincenzo Librandi, Oct 08 2013 *)

PARI

a(n)=sumdigits(3^n); \\ Michel Marcus, Nov 01 2013

Python

def a(n): return sum(map(int, str(3**n)))
print([a(n) for n in range(61)]) # Michael S. Branicky, Apr 25 2022

Formula

a(n) = A007953(A000244(n)). - Michel Marcus, Nov 01 2013

Extensions

Edited by M. F. Hasler, May 18 2017