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.

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

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

Rémy Sigrist, Table of n, a(n) for n = 1..8192

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^k-1) = 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

Cf. A000045, A011782 (row lengths), A048297, A060138, A107946.

Sequence in context: A145726 A322984 A277822 * A181631 A348390 A133674

Adjacent sequences: A327613 A327614 A327615 * A327617 A327618 A327619

Keyword

nonn,look,tabf

Author

Rémy Sigrist, Jul 06 2020

Status

approved