# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a116520 Showing 1-1 of 1 %I A116520 #80 Mar 06 2023 17:20:29 %S A116520 0,1,5,9,25,29,45,61,125,129,145,161,225,241,305,369,625,629,645,661, %T A116520 725,741,805,869,1125,1141,1205,1269,1525,1589,1845,2101,3125,3129, %U A116520 3145,3161,3225,3241,3305,3369,3625,3641,3705,3769,4025,4089,4345,4601,5625 %N A116520 a(0) = 0, a(1) = 1; a(n) = max { 4*a(k) + a(n-k) | 1 <= k <= n/2 }, for n > 1. %C A116520 Equivalently, a(n) = r*a(ceiling(n/2)) + s*a(floor(n/2)), a(0)=0, a(1)=1, for (r,s) = (1,4). - _N. J. A. Sloane_, Feb 16 2016 %C A116520 A 5-divide version of A084230. %C A116520 Zero together with the partial sums of A102376. - _Omar E. Pol_, May 05 2010 %C A116520 Also, total number of cubic ON cells after n generations in a three-dimensional cellular automaton in which A102376(n-1) gives the number of cubic ON cells in the n-th level of the structure starting from the top. An ON cell remains ON forever. The structure looks like an irregular stepped pyramid, with n >= 1. - _Omar E. Pol_, Feb 13 2015 %C A116520 From _Gary W. Adamson_, Aug 27 2016: (Start) %C A116520 The formula of Mar 26 2010 is equivalent to lim_{k->infinity} M^k of the following production matrix M: %C A116520 1, 0, 0, 0, 0, 0, ... %C A116520 5, 0, 0, 0, 0, 0, ... %C A116520 4, 1, 0, 0, 0, 0, ... %C A116520 0, 5, 0, 0, 0, 0, ... %C A116520 0, 4, 1, 0, 0, 0, ... %C A116520 0, 0, 5, 0, 0, 0, ... %C A116520 0, 0, 4, 1, 0, 0, ... %C A116520 0, 0, 0, 5, 0, 0, ... %C A116520 ... %C A116520 The sequence with offset 1 divided by its aerated variant is (1, 5, 4, 0, 0, 0, ...). (End) %H A116520 Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 %H A116520 K.-N. Chang and S.-C. Tsai, Exact solution of a minimal recurrence, Inform. Process. Lett. 75 (2000), 61-64. %H A116520 H. Harborth, Number of Odd Binomial Coefficients, Proc. Amer. Math. Soc. 62, 19-22, 1977. %H A116520 Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, pp. 27, 32. %H A116520 D. E. Knuth, Problem 11320, The American Mathematical Monthly, Vol. 114, No. 9 (Nov., 2007), p. 835. %H A116520 N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS %H A116520 Eric Weisstein's World of Mathematics, Stolarsky-Harborth Constant %H A116520 Index entries for sequences related to cellular automata %F A116520 a(0) = 1, a(1) = 1; thereafter a(2n) = 5a(n) and a(2n+1) = 4a(n) + a(n+1). %F A116520 Let r(x) = (1 + 5x + 4x^2). Then (1 + 5x + 9x^2 + 25x^3 + ...) = r(x) * r(x^2) * r(x^4) * r(x^8) * ... . - _Gary W. Adamson_, Mar 26 2010 %F A116520 a(n) = Sum_{k=0..n-1} 4^wt(k), where wt = A000120. - _Mike Warburton_, Mar 14 2019 %F A116520 a(n) = Sum_{k=0..floor(log_2(n))} 4^k*A360189(n-1,k). - _Alois P. Heinz_, Mar 06 2023 %p A116520 a:=proc(n) if n=0 then 0 elif n=1 then 1 elif n mod 2 = 0 then 5*a(n/2) else 4*a((n-1)/2)+a((n+1)/2) fi end: seq(a(n),n=0..52); %t A116520 b[0] := 0 b[1] := 1 b[n_?EvenQ] := b[n] = 5*b[n/2] b[n_?OddQ] := b[n] = 4*b[(n - 1)/2] + b[(n + 1)/2] a = Table[b[n], {n, 1, 25}] %o A116520 (Haskell) %o A116520 import Data.List (transpose) %o A116520 a116520 n = a116520_list !! n %o A116520 a116520_list = 0 : zs where %o A116520 zs = 1 : (concat $ transpose %o A116520 [zipWith (+) vs zs, zipWith (+) vs $ tail zs]) %o A116520 where vs = map (* 4) zs %o A116520 -- _Reinhard Zumkeller_, Apr 18 2012 %Y A116520 Cf. A000120, A006046, A077465, A130665, A130667, A102376, A360189. %Y A116520 Sequences of the form a(n) = r*a(ceiling(n/2)) + s*a(floor(n/2)), a(1)=1, for (r,s) = (1,1), (1,2), (2,1), (1,3), (2,2), (3,1), (1,4), (2,3), (3,2), (4,1): A000027, A006046, A064194, A130665, A073121, A268524, A116520, A268525, A268526, A268527. %K A116520 nonn,look %O A116520 0,3 %A A116520 _Roger L. Bagula_, Mar 15 2006 %E A116520 Edited by _N. J. A. Sloane_, Apr 16 2006, Jul 02 2008 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE