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

11, 14, 15, 18, 20, 27, 29, 31, 34, 38, 41, 43, 47, 48, 50, 53, 54, 58, 59, 63, 64, 65, 67, 69, 71, 72, 75, 77, 79, 83, 88, 90, 94, 98, 99, 102, 103, 107, 109, 112, 114, 118, 119, 123, 125, 131, 132, 134, 136, 139, 141, 142, 146, 150, 154, 159, 161, 164, 167

Offset: 1

Author

Oliver Bel, Mar 29 2014

Keywords

• nonn
• base

Examples

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

Links

• Robert Israel, Table of n, a(n) for n = 1..8009

Crossrefs

Cf. A004166, A007953.

Programs

Maple

N:= 1000: # to get the first N terms
threen:= 3:
digsum:= 3:
count:= 0:
for n from 1 while count < N do
  threen:= 3*threen;
  oldsum:= digsum;
  digsum:= convert(convert(threen,base,10),`+`);
  if oldsum > digsum then
     count:= count+1;
     A239935[count]:= n;
  fi
od: # Robert Israel, Apr 18 2014

Mathematica

lis = Table[Total[IntegerDigits[3^n, 10]], {n, 1, 100}];
Flatten[Position[Greater @@@ Partition[lis, 2, 1], True]]

PARI

isok(k) = sumdigits(3^k) > sumdigits(3^(k+1)); \\ Michel Marcus, Jul 03 2021

Extensions

More terms from Jon E. Schoenfield, Mar 29 2014