%I #17 Jul 03 2021 04:20:30

%S 11,14,15,18,20,27,29,31,34,38,41,43,47,48,50,53,54,58,59,63,64,65,67,

%T 69,71,72,75,77,79,83,88,90,94,98,99,102,103,107,109,112,114,118,119,

%U 123,125,131,132,134,136,139,141,142,146,150,154,159,161,164,167

%N Numbers k such that DigitSum(3^k) > DigitSum(3^(k+1)).

%H Robert Israel, <a href="/A239935/b239935.txt">Table of n, a(n) for n = 1..8009</a>

%e For k=11, we have DigitSum(3^11) = 27 > 18 = DigitSum(3^12).

%p N:= 1000: # to get the first N terms

%p threen:= 3:

%p digsum:= 3:

%p count:= 0:

%p for n from 1 while count < N do

%p threen:= 3*threen;

%p oldsum:= digsum;

%p digsum:= convert(convert(threen,base,10),`+`);

%p if oldsum > digsum then

%p count:= count+1;

%p A239935[count]:= n;

%p fi

%p od: # _Robert Israel_, Apr 18 2014

%t lis = Table[Total[IntegerDigits[3^n, 10]], {n, 1, 100}];

%t Flatten[Position[Greater @@@ Partition[lis, 2, 1], True]]

%o (PARI) isok(k) = sumdigits(3^k) > sumdigits(3^(k+1)); \\ _Michel Marcus_, Jul 03 2021

%Y Cf. A004166, A007953.

%K nonn,base

%O 1,1

%A _Oliver Bel_, Mar 29 2014

%E More terms from _Jon E. Schoenfield_, Mar 29 2014