A265332 - OEIS

%I #22 Dec 19 2021 04:17:37

%S 1,2,1,3,1,1,2,4,1,1,1,2,1,2,3,5,1,1,1,1,2,1,1,2,1,2,3,1,2,3,4,6,1,1,

%T 1,1,1,2,1,1,1,2,1,1,2,1,2,3,1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,7,1,1,1,1,

%U 1,1,2,1,1,1,1,2,1,1,1,2,1,1,2,1,2,3,1,1,1,2,1,1,2,1,2,3,1,1,2,1,2,3,1,2,3,4,1,1,2,1,2,3,1,2,3,4,1,2,3,4

%N a(n) is the index of the column in A265901 where n appears; also the index of the row in A265903 where n appears.

%C If all 1's are deleted, the remaining terms are the sequence incremented. - after _Franklin T. Adams-Watters_ Oct 05 2006 comment in A051135.

%C Ordinal transform of A162598.

%H Antti Karttunen, <a href="/A265332/b265332.txt">Table of n, a(n) for n = 1..8192</a>

%H T. Kubo and R. Vakil, <a href="http://dx.doi.org/10.1016/0012-365X(94)00303-Z">On Conway's recursive sequence</a>, Discr. Math. 152 (1996), 225-252.

%F a(1) = 1; for n > 1, a(n) = A051135(n).

%e Illustration how the sequence can be constructed by concatenating the frequency counts Q_n of each successive level n of A004001-tree:

%e --

%e              1                                      Q_0 = (1)

%e              |

%e             _2__                                    Q_1 = (2)

%e            /    \

%e          _3    __4_____                             Q_2 = (1,3)

%e         /     /  |     \

%e       _5    _6  _7    __8___________                Q_3 = (1,1,2,4)

%e      /     /   / |   /  |  \        \

%e    _9    10  11 12  13  14  15___    16_________    Q_4 = (1,1,1,2,1,2,3,5)

%e   /     /   /  / |  /  / |   |\  \   | \  \  \  \

%e 17    18  19 20 21 22 23 24 25 26 27 28 29 30 31 32

%e --

%e The above illustration copied from the page 229 of Kubo and Vakil paper (page 5 in PDF).

%t terms = 120;

%t h[1] = 1; h[2] = 1;

%t h[n_] := h[n] = h[h[n - 1]] + h[n - h[n - 1]];

%t seq[nmax_] := seq[nmax] = (Length /@ Split[Sort @ Array[h, nmax, 2]])[[;; terms]];

%t seq[nmax = 2 terms];

%t seq[nmax += terms];

%t While[seq[nmax] != seq[nmax - terms], nmax += terms];

%t seq[nmax] (* _Jean-François Alcover_, Dec 19 2021 *)

%o (Scheme) (define (A265332 n) (if (= 1 n) 1 (A051135 n)))

%Y Essentially same as A051135 apart from the initial term, which here is set as a(1)=1.

%Y Cf. A004001, A265901, A265903.

%Y Cf. A162598 (corresponding other index).

%Y Cf. A265754.

%Y Cf. also A267108, A267109, A267110.

%K nonn

%O 1,2

%A _Antti Karttunen_, Jan 09 2016