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A308175 Let EM denote the Ehrenfeucht-Mycielski sequence A038219, and let P(n) = [EM(1),...,EM(n)]. To compute EM(n+1) for n>=3, we find the longest suffix S (say) of P(n) which has previously appeared in P(n). Suppose the most recent appearance of S began at index n-t(n). Then a(n) = t(n), while the length of S is given in A308174. 3

%I #17 May 22 2019 13:17:54

%S 2,1,4,1,5,4,8,4,7,2,8,12,2,13,10,17,7,3,8,19,14,3,15,21,19,24,18,28,

%T 17,25,27,19,34,9,23,7,38,21,32,20,38,14,30,34,29,45,24,39,35,4,36,41,

%U 27,49,33,54,36,52,41,4,42,54,39,31,65,24,44,9,36,53

%N Let EM denote the Ehrenfeucht-Mycielski sequence A038219, and let P(n) = [EM(1),...,EM(n)]. To compute EM(n+1) for n>=3, we find the longest suffix S (say) of P(n) which has previously appeared in P(n). Suppose the most recent appearance of S began at index n-t(n). Then a(n) = t(n), while the length of S is given in A308174.

%C Then EM(n+1) is the complement of the bit following the most recent appearance of S.

%H Rémy Sigrist, <a href="/A308175/b308175.txt">Table of n, a(n) for n = 3..50000</a>

%H Rémy Sigrist, <a href="/A308175/a308175.pl.txt">Perl program for A308175</a>

%e Tableau showing calculation of terms 3 through 13

%e 1 2 3 4 5 6 7 8 9 10 11 12 13 n

%e 0 1 0 0 1 1 0 1 0 1 1 1 0 A038219(n)

%e - - 0 0 01 1 10 01 010 101 011 11 110 S

%e - - 1 1 2 1 2 2 3 3 3 2 3 s = A308174(n)

%e - - 1 3 1 5 2 4 1 6 4 10 5 previous

%e - - 2 1 4 1 5 4 8 4 7 2 8 t = A308175(n)

%e "Previous" = index of start of most recent previous occurrence of S; s = |S|; t = n - "previous" = A308175(n)

%o (Perl) See Links section.

%Y Cf. A038219, A308174.

%K nonn,look

%O 3,1

%A _N. J. A. Sloane_, May 21 2019, corrected and extended May 21 2019

%E More terms from _Rémy Sigrist_, May 21 2019

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Last modified March 29 01:36 EDT 2024. Contains 371264 sequences. (Running on oeis4.)