A246878 a(0) = 1, then a(n) = sum(a(k), k = floor(log_2(n)) .. n - 1).

1, 1, 1, 2, 3, 6, 12, 24, 47, 94, 188, 376, 752, 1504, 3008, 6016, 12030, 24060, 48120, 96240, 192480, 384960, 769920, 1539840, 3079680, 6159360, 12318720, 24637440, 49274880, 98549760, 197099520, 394199040, 788398077

Offset: 0

Author

Benoit Jubin, Sep 06 2014

Keywords

• nonn
• easy

Comments

a(n) = Theta(2^n), and more precisely, for n >= 4, (13/16)*(3/16)2^n <= a(n) <= (3/16)*2^n.

Indeed, from the formula, one gets a(n) <= (3/16)*2^n, and injecting this in the formula, one gets a(n) >= 2*a(n - 1) - (3/32)*n. Then by induction, and using the formula sum(k*2^k, k = 1 .. n) = (n - 1)*2^(n + 1) + 2, one obtains a(n) >= (13/16)*(3/16)2^n + (3/32)*n + (3/8).

Examples

			a(2) = a(1) = a(0) = 1.
a(3) = a(2) + a(1) = 2.
a(4) = a(3) + a(2) = 3.
a(5) = a(4) + a(3) + a(2) = 6.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• W. Kuszmaul, Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations, arXiv preprint arXiv:1509.08216, 2015

Crossrefs

Cf. A000523.

Programs

Haskell

import Data.List (genericDrop)
a246878 n = a246878_list !! n
a246878_list = 1 : f [1] a000523_list where
   f xs (k:ks) = y : f (xs ++ [y]) ks where y = sum $ genericDrop k xs
-- Reinhard Zumkeller, Sep 16 2014

Formula

If n >= 1 is not a power of 2, then a(n) = 2*a(n - 1), and if k >= 1, then a(2^k) = 2*a(2^k - 1) - a(k - 1).