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A114955 A 2/3-power Fibonacci sequence.

%I A114955 #21 Jul 14 2023 12:15:48
%S A114955 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,
%T A114955 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,
%U A114955 8,8,8,8,8,8,8,8,8,8
%N A114955 A 2/3-power Fibonacci sequence.
%C A114955 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
%H A114955 <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (1).
%F A114955 a(0) = a(1) = 1, for n>1 a(n) = ceiling(a(n-1)^(2/3) + a(n-2)^(2/3)).
%F A114955 a(n) = 8 for all n>8.
%F A114955 Euler transform of length 8 sequence [ 1, 1, 1, 0, 0, -1, 0, -1]. - _Michael Somos_, Aug 31 2006
%F A114955 G.f.: (1-x^6)(1-x^8)/((1-x)(1-x^2)(1-x^3)). - _Michael Somos_, Aug 31 2006
%e A114955 a(2) = ceiling(a(0)^(2/3) + a(1)^(2/3)) = ceiling(1^(2/3) + 1^(2/3)) = 2.
%e A114955 a(3) = ceiling(a(1)^(2/3) + a(2)^(2/3)) = ceiling(1^(2/3) + 2^(2/3)) = ceiling(2.58740105) = 3.
%e A114955 a(4) = ceiling(2^(2/3) + 3^(2/3)) = ceiling(3.66748488) = 4.
%e A114955 a(5) = ceiling(3^(2/3) + 4^(2/3)) = ceiling(4.59992592) = 5.
%e A114955 a(6) = ceiling(4^(2/3) + 5^(2/3)) = ceiling(5.44385984) = 6.
%e A114955 a(7) = ceiling(5^(2/3) + 6^(2/3)) = ceiling(6.22594499) = 7.
%e A114955 a(8) = ceiling(6^(2/3) + 7^(2/3)) = ceiling(6.96123296) = 7.
%t A114955 nxt[{a_,b_}]:={b,Ceiling[b^(2/3)+a^(2/3)]}; Transpose[NestList[nxt,{1,1},80]][[1]] (* _Harvey P. Dale_, Jan 03 2013 *)
%o A114955 (PARI) {a(n)=if(n<1, n==0, if(n>8, 8, n-(n>7)))} /* _Michael Somos_, Aug 31 2006 */
%Y A114955 Cf. A000283, A112961, A112969, A114793.
%K A114955 easy,nonn
%O A114955 0,3
%A A114955 _Jonathan Vos Post_, Feb 21 2006