

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.


2



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, 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, 3, 1, 4, 2, 1, 5, 6, 3, 2, 7, 1, 4, 8, 1, 9, 5, 3, 10, 2
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OFFSET

1,4


COMMENTS

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).


LINKS



FORMULA

Apparently:
 the greatest term in row n is A000045(n),
 a(n) = 1 iff n = 1 or n belongs to A060138,
 a(A048297(n+1)) = n (and this corresponds to the first occurrence of n),
 a(4^k) = A000045(2*k+1) for any k >= 0,
 a(2*4^k1) = A000045(2*k+2) for any k >= 0.


EXAMPLE

Triangle begins:
1: [1]
2: [1]
3: [1, 2]
4: [1, 2, 3, 1]
5: [1, 2, 3, 1, 4, 2, 1, 5]
6: [1, 2, 3, 1, 4, 2, 1, 5, 6, 3, 2, 7, 1, 4, 8, 1]


PROG

(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])))) }


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



