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A327616 Irregular table read by rows; the first row contains a single 1; for any n > 1, row n+1 corresponds to the ordinal transform of the terms in rows 1..n.

%I A327616 #48 Jul 09 2020 02:59:15
%S A327616 1,1,1,2,1,2,3,1,1,2,3,1,4,2,1,5,1,2,3,1,4,2,1,5,6,3,2,7,1,4,8,1,1,2,
%T A327616 3,1,4,2,1,5,6,3,2,7,1,4,8,1,9,5,3,10,2,6,11,2,1,4,7,1,12,3,1,13,1,2,
%U A327616 3,1,4,2,1,5,6,3,2,7,1,4,8,1,9,5,3,10,2
%N A327616 Irregular table read by rows; the first row contains a single 1; for any n > 1, row n+1 corresponds to the ordinal transform of the terms in rows 1..n.
%C A327616 The ordinal transform of a sequence b(n) is the sequence t(n) = number of values in b(1), ..., b(n) which are equal to b(n).
%H A327616 Rémy Sigrist, <a href="/A327616/b327616.txt">Table of n, a(n) for n = 1..8192</a>
%F A327616 Apparently:
%F A327616 - the greatest term in row n is A000045(n),
%F A327616 - a(n) = 1 iff n = 1 or n belongs to A060138,
%F A327616 - a(A048297(n+1)) = n (and this corresponds to the first occurrence of n),
%F A327616 - a(4^k) = A000045(2*k+1) for any k >= 0,
%F A327616 - a(2*4^k-1) = A000045(2*k+2) for any k >= 0.
%e A327616 Triangle begins:
%e A327616   1:  [1]
%e A327616   2:  [1]
%e A327616   3:  [1, 2]
%e A327616   4:  [1, 2, 3, 1]
%e A327616   5:  [1, 2, 3, 1, 4, 2, 1, 5]
%e A327616   6:  [1, 2, 3, 1, 4, 2, 1, 5, 6, 3, 2, 7, 1, 4, 8, 1]
%o A327616 (PARI) { for (n=1, #a=vector(85), print1 (a[n]=if (n==1, a[n]=1, a[n]=o[a[k++]]++)", "); if (hammingweight(n)==1, k=0; o=vector(vecmax(a[1..n])))) }
%Y A327616 Cf. A000045, A011782 (row lengths), A048297, A060138, A107946.
%K A327616 nonn,look,tabf
%O A327616 1,4
%A A327616 _Rémy Sigrist_, Jul 06 2020