A027870 Number of zero digits in 2^n.
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 3, 1, 1, 1, 1, 1, 0, 1, 0, 2, 3, 2, 2, 2, 1, 1, 2, 2, 3, 2, 2, 2, 1, 1, 0, 1, 3, 3, 1, 0, 1, 1, 1, 0, 0, 2, 4, 2, 0, 2, 3, 1, 1, 0, 3, 5, 3, 3, 4, 2, 3, 4, 1, 1, 4, 5, 5, 6, 6, 7, 5, 5
Offset: 0
Examples
2^31 = 2147483648 so a(31) = 0 and 2^42 = 4398046511104 so a(42) = 2.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..20000 (first 1001 terms from _Harry J. Smith_, Oct 27 2009)
Crossrefs
Similar for other digits: A065712 (1's), A065710 (2's), A065714 (3's), A065715 (4's), A065716 (5's), A065717 (6's), A065718 (7's), A065719 (8's), A065744 (9's).
Cf. A031146 (index of first appearance of n in this sequence), A094776 (index of last occurrence of digit n in powers of 2).
Cf. A305932 (table with n in row a(n)).
Programs
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Haskell
a027870 = a055641 . a000079 -- Reinhard Zumkeller, Apr 30 2013
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Mathematica
Table[ Count[ IntegerDigits[2^n], 0], {n, 0, 100} ] DigitCount[2^Range[0,110],10,0] (* Harvey P. Dale, Nov 20 2011 *)
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PARI
A027870(n)=#select(d->!d,digits(2^n)) \\ M. F. Hasler, Jun 14 2018
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Python
def A027870(n): return str(2**n).count('0') # Chai Wah Wu, Feb 14 2020
Formula
Extensions
Edited by M. F. Hasler, Jul 09 2025
Comments