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

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

0,3

Comments

A 3/4-power Fibonacci sequence.

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. This sequence is related to: A112961 "a cubic Fibonacci sequence" a(n) = a(n-1)^3 + a(n-2)^3, A112969 "a quartic Fibonacci sequence" a(n) = a(n-1)^4 + a(n-2)^4, just as A000283 is the quadratic analog of the Fibonacci sequence.

Links

Table of n, a(n) for n=0..74.

Example

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.

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 *)

Crossrefs

Cf. A000283, A112961, A112969, A114793.

Sequence in context: A342760 A244912 A249745 * A039256 A039197 A286057

Adjacent sequences: A114953 A114954 A114955 * A114957 A114958 A114959

Keyword

easy,nonn

Author

Jonathan Vos Post, Feb 21 2006

Extensions

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

Status

approved