A387907 - OEIS

A387907

a(n) = position of first odd digit of 2^n, starting from left, or 0 if no odd digit exists.

1, 0, 0, 0, 1, 1, 0, 1, 2, 1, 1, 0, 3, 2, 1, 1, 2, 1, 4, 1, 1, 3, 2, 2, 1, 1, 2, 1, 5, 1, 1, 2, 3, 2, 1, 1, 3, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 1, 3, 1, 1, 3, 2, 1, 1, 1, 1, 1, 5, 1, 1, 2, 3, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 7, 1, 3, 3, 1, 1, 1

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OFFSET

0,9

COMMENTS

It is conjectured that a(n) is positive for n >= 12. This is sometimes known as the "2048 problem".

The first digit of 2^n follows Benford's law asymptotically (see Raimi reference), so the proportion of n with a(n) = 1 is log10(2*4*6*8*10) - log10(3*5*7*9) ~= 0.6088... .

LINKS

Michael S. Branicky, Table of n, a(n) for n = 0..10000

Ralph A. Raimi, The first digit problem, Amer. Math. Monthly, 83 (1976), 521-538.

Wikipedia, Benford's law

FORMULA

a(n) = 0 iff 2^n in A068994. - Michael S. Branicky, Sep 20 2025

EXAMPLE

a(6) = 0 because 2^6 = 64 has no odd digits.

a(10) = 1 because the first digit of 2^10 = 1024, from the left, is odd.

a(18) = 4 because the first odd digit of 2^18 = 262144, from the left, is the 4th digit.

MATHEMATICA

a[n_]:=Module[{so=Boole[OddQ/@IntegerDigits[2^n]]}, If[Total[so]==0, 0, Position[so, 1][[1]]]]; Array[a, 86, 0]//Flatten (* James C. McMahon, Sep 20 2025 *)

PROG

(PARI) a(n) = my(x=select(x->(x%2), digits(2^n), 1)); if (#x, x[1], 0); \\ Michel Marcus, Sep 14 2025

(Python)

def a(n): return next((i+1 for i, d in enumerate(str(1<<n)) if d in "13579"), 0)

print([a(n) for n in range(86)]) # Michael S. Branicky, Sep 15 2025

(Haskell)

import Data.Char

a387907 n = firstOddDigitPos $ 2^n

firstOddDigitPos n =

let digits = map digitToInt (show n)

indexedDigits = zip [1..] digits

oddPositions = filter (\(pos, digit) -> odd digit) indexedDigits