A063510 a(1) = 1, a(n) = a(floor(square root(n))) + 1 for n > 1.

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

Author

Reinhard Zumkeller, Jul 30 2001

Keywords

• easy
• nonn

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

Crossrefs

Cf. A010096, A000196.

Programs

Haskell

a063510 1 = 1
a063510 n = a063510 (a000196 n) + 1
-- Reinhard Zumkeller, Mar 16 2012

PARI

a(n)=if(n<2,1,floor(log(4*log(n)/log(2))/log(2)))

PARI

a(n)=if(n<2, 1, 2+logint(logint(n,2),2)) \\ Charles R Greathouse IV, Nov 28 2024

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