A064672 a(0) = 0, a(1) = 1; for a(n), n >= 2, write n = x^2 + y with y >= 0 as small as possible, then a(n) = a(x) + a(y).

0, 1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 5, 6, 5, 6, 7, 2, 3, 4, 5, 4, 5, 6, 7, 6, 3, 4, 5, 6, 5, 6, 7, 8, 7, 6, 7, 4, 5, 6, 7, 6, 7, 8, 9, 8, 7, 8, 9, 10, 5, 6, 7, 8, 7, 8, 9, 10, 9, 8, 9, 10, 11, 10, 11, 4, 5, 6, 7, 6, 7, 8, 9, 8, 7, 8, 9, 10, 9, 10, 11, 6, 3, 4, 5, 6, 5, 6, 7, 8, 7, 6, 7, 8, 9, 8, 9, 10, 5, 6

Offset: 0

Author

Jonathan Ayres (jonathan.ayres(AT)btinternet.com), Oct 09 2001

Keywords

• nonn
• easy
• nice

Comments

Because of the definition of a(n), a(n^2) = a(n) and more generally a(n^(2m)) = a(n), so the sequence recursively contains itself.

a(A064689(n)) = n and a(m) < n for m < A064689(n).

Examples

			a(7) = 5 because 7 = 2^2 + 3, a(2) = 2 and a(3) = 3, giving 5
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A064689.

Cf. A048760.

Programs

Haskell

a064672 n = a064672_list !! n
a064672_list = 0 : 1 : f (drop 2 a000196_list) 1 1 (tail a064672_list)
   where f (r:rs) r' u (v:vs)
           | r == r' = (u + v) : f rs r u vs
           | r /= r' = u' : f rs r u' (tail a064672_list)
           where u' = a064672 $ fromInteger r
-- Reinhard Zumkeller, Apr 27 2012

Mathematica

a[0]=0; a[1]=1; a[n_] := a[n] = a[ Floor[ Sqrt[n] ] ] + a[ n - Floor[ Sqrt[n] ]^2 ]; Table[a[n], {n, 0, 98}] (* Jean-François Alcover, May 23 2012, after Reinhard Zumkeller *)

Formula

For n > 1: a(n) = a(A000196(n)) + a(A053186(n)), a(0) = 0, a(1) = 1. [Reinhard Zumkeller, Apr 27 2012]