A114956 a(n) = ceiling(a(n-1)^(3/4) + a(n-2)^(3/4)), with a(0) = a(1) = 1.

1, 1, 2, 3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16

Offset: 0

Author

Jonathan Vos Post, Feb 21 2006

Keywords

• easy
• nonn

Comments

a(17) = 16 is exactly 16^(3/4) + 16^(3/4) = 16. This is a fixed point, so a(n) = 16 for all n>14.

Examples

			a(2) = ceiling(a(0)^(3/4) + a(1)^(3/4)) = ceiling(1^(3/4) + 1^(3/4)) = 2.
a(3) = ceiling(a(1)^(3/4) + a(2)^(3/4)) = ceiling(1^(3/4) + 2^(3/4)) = ceiling(2.68179283) = 3.
a(4) = ceiling(2^(3/4) + 3^(3/4)) = ceiling(3.96129989) = 4.
a(5) = ceiling(3^(3/4) + 4^(3/4)) = ceiling(5.10793418) = 6.
a(6) = ceiling(4^(3/4) + 6^(3/4)) = ceiling(6.66208575) = 7.
		

Crossrefs

Cf. A000283, A112961, A112969, A114793.

Programs

Mathematica

RecurrenceTable[{a[0]==1,a[1]==1,a[n]==Ceiling[a[n-1]^(3/4)+ a[n-2]^(3/4)]}, a[n],{n,80}] (* Harvey P. Dale, Jul 22 2011 *)

Extensions

Edited by N. J. A. Sloane, May 20 2006