A186775 Numbers k such that digitsum(2^k) > digitsum(2^(k+1)).

3, 4, 8, 9, 15, 16, 20, 21, 23, 24, 26, 28, 29, 33, 34, 36, 39, 40, 41, 46, 48, 51, 52, 55, 56, 57, 60, 63, 64, 67, 68, 69, 74, 75, 76, 77, 80, 82, 83, 85, 86, 88, 91, 92, 94, 95, 97, 98, 100, 102, 106, 108, 112, 113, 116, 118, 121, 124, 126

Offset: 1

Author

Thomas Nordhaus, Feb 26 2011

Keywords

• base
• easy
• nonn

Comments

If 2^k and 2^(k+1) acted like random numbers of their size, the probability that k would be in the sequence would be 1/2 + O(1/k). So very possibly a(n) ~ 2n. - Charles R Greathouse IV, Aug 08 2022

Examples

			3 is in the sequence because digitsum(2^3) = 8 > 7 = digitsum(2^4).
		

Links

• J.W.L. (Jan) Eerland, Table of n, a(n) for n = 1..49688

Crossrefs

Cf. A001370.

Programs

Mathematica

DeleteCases[Table[If[Total[Total[IntegerDigits[2^n]]]>Total[IntegerDigits[2^(n+1)]],n,k],{n,0, 10^5}],k] (* J.W.L. (Jan) Eerland, Aug 08 2022 *)

Python

from itertools import count, islice, pairwise
def ds2(n): return sum(map(int, str(1< t[1])
print(list(islice(agen(), 60))) # Michael S. Branicky, Aug 08 2022

Sage

def is_A186775(n): return sum((2^n).digits()) > sum((2^(n+1)).digits()) # D. S. McNeil, Feb 27 2011