

A353322


A variant of Van Eck's sequence where we only consider powers of 2: for n >= 0, if a(n) = a(n2^e) for some e, take the least such e and set a(n+1) = 2^e; otherwise a(n+1) = 0. Start with a(1) = 0.


1



0, 0, 1, 0, 2, 0, 2, 2, 1, 0, 4, 0, 2, 8, 0, 0, 1, 8, 4, 8, 2, 8, 2, 2, 1, 8, 4, 8, 2, 8, 2, 2, 1, 8, 4, 8, 2, 8, 2, 2, 1, 8, 4, 8, 2, 8, 2, 2, 1, 8, 4, 8, 2, 8, 2, 2, 1, 8, 4, 8, 2, 8, 2, 2, 1, 8, 4, 8, 2, 8, 2, 2, 1, 8, 4, 8, 2, 8, 2, 2, 1, 8, 4, 8, 2, 8, 2
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OFFSET

1,5


COMMENTS

The sequence is eventually 8periodic.
The variant with powers of 4 is 3periodic: (0 0 1)*.


LINKS



FORMULA

a(n) = a(n8) for n >= 25.


EXAMPLE

a(1) = 0 by definition.
a(2) = 0 as there is only one occurrence of a(1) = 0 so far.
a(3) = 2^0 = 1 as a(2) = a(22^0).
a(4) = 0 as there is only one occurrence of a(3) = 1 so far.
a(5) = 2^1 = 2 as a(4) = a(42^1).
a(6) = 0 as there is only one occurrence of a(5) = 2 so far.
a(7) = 2^1 = 2 as a(6) = a(62^1).
a(8) = 2^1 = 2 as a(7) = a(72^1).
a(9) = 2^0 = 1 as a(8) = a(82^0).
a(10) = 0 as a(9) <> a(92^e) for any admissible e.


PROG

(PARI) { for (n=1, #a=vector(87), for (e=0, oo, m = n1d=2^e; if (m<1, break, a[n1]==a[m], a[n]=d; break)); print1 (a[n]", ")) }


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



