

A293722


Number of distinct nonempty subsequences of the binary expansion of n.


2



1, 1, 3, 2, 5, 6, 5, 3, 7, 10, 11, 9, 8, 9, 7, 4, 9, 14, 17, 15, 16, 19, 17, 12, 11, 15, 16, 13, 11, 12, 9, 5, 11, 18, 23, 21, 24, 29, 27, 20, 21, 29, 32, 27, 25, 28, 23, 15, 14, 21, 25, 22, 23, 27, 24, 17, 15, 20, 21, 17, 14, 15, 11, 6, 13, 22, 29, 27, 32, 39, 37
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OFFSET

0,3


COMMENTS

The subsequence does not need to consist of adjacent terms.


LINKS

Andrew Howroyd, Table of n, a(n) for n = 0..4096
Geeks for Geeks, Count Distinct Subsequences


FORMULA

a(2^n) = 2n + 1.
a(2^n1) = n if n>0.
a(n) = A293170(n)  1.  Andrew Howroyd, Apr 27 2020


EXAMPLE

a(4) = 5 because 4 = 100_2, and the distinct subsequences of 100 are 0, 1, 00, 10, 100.
Similarly a(7) = 3, because 7 = 111_2, and 111 has only three distinct subsequences: 1, 11, 111.
a(9) = 10: 9 = 1001_2, and we get 0, 1, 00, 01, 10, 11, 001, 100, 101, 1001.


PROG

(Python)
def a(n):
if n == 0: return 1
r, l = 1, [0, 0]
while n:
r, l[n%2] = 2*r  l[n%2], r
n >>= 1
return r  1


CROSSREFS

Cf. A141297.
If the empty subsequence is also counted, we get A293170.
Sequence in context: A057028 A195112 A276618 * A153152 A153153 A175058
Adjacent sequences: A293719 A293720 A293721 * A293723 A293724 A293725


KEYWORD

nonn,base,easy


AUTHOR

Orson R. L. Peters, Oct 15 2017


EXTENSIONS

Terms a(50) and beyond from Andrew Howroyd, Apr 27 2020


STATUS

approved



