A388137 a(n) = position of first even digit of 2^n, starting from left, or 0 if no even digit exists.

0, 1, 1, 1, 2, 2, 1, 2, 1, 3, 2, 1, 1, 1, 2, 2, 1, 4, 1, 2, 2, 1, 1, 1, 2, 5, 1, 3, 1, 3, 2, 1, 1, 1, 6, 2, 1, 4, 1, 2, 2, 1, 1, 1, 5, 4, 2, 2, 1, 2, 3, 1, 1, 2, 2, 2, 2, 2, 1, 3, 4, 1, 1, 2, 2, 2, 4, 2, 1, 3, 3, 1, 1, 2, 2, 5, 6, 8, 2, 1, 2, 1, 1, 2, 4, 2

Offset

0,5

Comments

The first digit of the powers of two are known to follow Benford's Law asymptotically (see Raimi reference), so it follows that the proportion of n with a(n) = 1 is log10(3*5*7*9) - log10(2*4*6*8) = 0.3911....

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

0 < a(n) <= A034887(n) for all n > 0, with equality iff 2^n in A272698. - Michael S. Branicky, Sep 20 2025

Example

a(5) = 2, because the first even digit of 2^5 = 32 is the 2nd one from the left.
a(11) = 1, because the first even digit of 2^11 = 2048 is the 1st one from the left.

Maple

f:= proc(n) local L, i;
  L:= convert(2^n, base, 10);
  for i from 1 do if L[-i]::even then return i fi od
end proc:
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Sep 21 2025

Mathematica

a[n_] := (FirstPosition[IntegerDigits[2^n], _?EvenQ] /. _Missing -> {0})[[1]]; Array[a, 100, 0] (* Amiram Eldar, Sep 20 2025 *)

Prog

(PARI) a(n) = my(v=select(x->!(x%2), digits(2^n), 1)); if (#v, v[1], 0); \\ Michel Marcus, Sep 15 2025
(Python)
def a(n): return next((i+1 for i, d in enumerate(str(1<<n)) if d in "02468"), 0)
print([a(n) for n in range(86)]) # Michael S. Branicky, Sep 15 2025

Crossrefs

Cf. A000079, A034887, A272698, A387907.

Sequence in context: A305298 A298824 A343384 * A237615 A343190 A256132

Adjacent sequences: A388134 A388135 A388136 * A388138 A388139 A388140

Keyword

nonn,base

Author

Harry Richman, Sep 15 2025

Status

approved