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A063510
a(1) = 1, a(n) = a(floor(square root(n))) + 1 for n > 1.
2
1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
OFFSET
1,2
COMMENTS
a(n) = A010096(n) until n = 255, but 5 = a(256) <> A010096(256) = 4.
The least k such that a(k)=n for n >= 2 is given by k = 2^(2^(n-2)) so the closed form for a(n) follows. - Benoit Cloitre, Apr 28 2005
FORMULA
a(1)=1; for n >= 2, a(n) = floor(log(4*log(n)/log(2))/log(2)). - Benoit Cloitre, Apr 28 2005
Equivalently, a(n) = 2 + floor(log_2(log_2(n))) for n > 1. - Charles R Greathouse IV, Dec 19 2011
PROG
(PARI) a(n)=if(n<2, 1, floor(log(4*log(n)/log(2))/log(2)))
(Haskell)
a063510 1 = 1
a063510 n = a063510 (a000196 n) + 1
-- Reinhard Zumkeller, Mar 16 2012
CROSSREFS
Cf. A010096.
Cf. A000196.
Sequence in context: A157639 A010096 A230864 * A156878 A214454 A140474
KEYWORD
easy,nonn
AUTHOR
Reinhard Zumkeller, Jul 30 2001
STATUS
approved