A114955 A 2/3-power Fibonacci sequence.

1, 1, 2, 3, 4, 5, 6, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 0

Author

Jonathan Vos Post, Feb 21 2006

Keywords

• easy
• nonn

Comments

a(n) is also the minimum number of distinct palindromes (not counting the empty string) occurring as substrings of an n-bit binary string. For example, the string 11001 contains the five distinct palindromes 0, 00, 1, 11, and 1001. In fact, every 5-bit binary string contains five distinct palindromes, so a(5) = 5. - Austin Shapiro, Feb 15 2023

Examples

			a(2) = ceiling(a(0)^(2/3) + a(1)^(2/3)) = ceiling(1^(2/3) + 1^(2/3)) = 2.
a(3) = ceiling(a(1)^(2/3) + a(2)^(2/3)) = ceiling(1^(2/3) + 2^(2/3)) = ceiling(2.58740105) = 3.
a(4) = ceiling(2^(2/3) + 3^(2/3)) = ceiling(3.66748488) = 4.
a(5) = ceiling(3^(2/3) + 4^(2/3)) = ceiling(4.59992592) = 5.
a(6) = ceiling(4^(2/3) + 5^(2/3)) = ceiling(5.44385984) = 6.
a(7) = ceiling(5^(2/3) + 6^(2/3)) = ceiling(6.22594499) = 7.
a(8) = ceiling(6^(2/3) + 7^(2/3)) = ceiling(6.96123296) = 7.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1).

Crossrefs

Cf. A000283, A112961, A112969, A114793.

Programs

Mathematica

nxt[{a_,b_}]:={b,Ceiling[b^(2/3)+a^(2/3)]}; Transpose[NestList[nxt,{1,1},80]][[1]] (* Harvey P. Dale, Jan 03 2013 *)

PARI

{a(n)=if(n<1, n==0, if(n>8, 8, n-(n>7)))} /* Michael Somos, Aug 31 2006 */

Formula

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

a(n) = 8 for all n>8.

Euler transform of length 8 sequence [ 1, 1, 1, 0, 0, -1, 0, -1]. - Michael Somos, Aug 31 2006

G.f.: (1-x^6)(1-x^8)/((1-x)(1-x^2)(1-x^3)). - Michael Somos, Aug 31 2006