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A239935 Numbers k such that DigitSum(3^k) > DigitSum(3^(k+1)).

%I A239935 #17 Jul 03 2021 04:20:30
%S A239935 11,14,15,18,20,27,29,31,34,38,41,43,47,48,50,53,54,58,59,63,64,65,67,
%T A239935 69,71,72,75,77,79,83,88,90,94,98,99,102,103,107,109,112,114,118,119,
%U A239935 123,125,131,132,134,136,139,141,142,146,150,154,159,161,164,167
%N A239935 Numbers k such that DigitSum(3^k) > DigitSum(3^(k+1)).
%H A239935 Robert Israel, <a href="/A239935/b239935.txt">Table of n, a(n) for n = 1..8009</a>
%e A239935 For k=11, we have DigitSum(3^11) = 27 > 18 = DigitSum(3^12).
%p A239935 N:= 1000: # to get the first N terms
%p A239935 threen:= 3:
%p A239935 digsum:= 3:
%p A239935 count:= 0:
%p A239935 for n from 1 while count < N do
%p A239935   threen:= 3*threen;
%p A239935   oldsum:= digsum;
%p A239935   digsum:= convert(convert(threen,base,10),`+`);
%p A239935   if oldsum > digsum then
%p A239935      count:= count+1;
%p A239935      A239935[count]:= n;
%p A239935   fi
%p A239935 od: # _Robert Israel_, Apr 18 2014
%t A239935 lis = Table[Total[IntegerDigits[3^n, 10]], {n, 1, 100}];
%t A239935 Flatten[Position[Greater @@@ Partition[lis, 2, 1], True]]
%o A239935 (PARI) isok(k) = sumdigits(3^k) > sumdigits(3^(k+1)); \\ _Michel Marcus_, Jul 03 2021
%Y A239935 Cf. A004166, A007953.
%K A239935 nonn,base
%O A239935 1,1
%A A239935 _Oliver Bel_, Mar 29 2014
%E A239935 More terms from _Jon E. Schoenfield_, Mar 29 2014