A348907 -id:A348907 - cp's OEIS frontend

A349191 a(n) = A000720(A348907(n+1)).

1, 2, 1, 3, 2, 4, 1, 3, 2, 5, 4, 6, 1, 3, 2, 7, 5, 8, 4, 6, 1, 9, 3, 2, 7, 5, 8, 10, 4, 11, 6, 1, 9, 3, 2, 12, 7, 5, 8, 13, 10, 14, 4, 11, 6, 15, 1, 9, 3, 2, 12, 16, 7, 5, 8, 13, 10, 17, 14, 18, 4, 11, 6, 15, 1, 19, 9, 3, 2, 20, 12, 21, 16, 7, 5, 8, 13, 22, 10

Offset: 1

Michael De Vlieger, Nov 09 2021

Regarding this sequence as an irregular triangle T(m,j) where the rows m terminate with 1 exhibits row length A338237(m). In such rows m, we have a permutation of the range of natural numbers 1..A338237(m).

Records are the natural numbers.

			Table showing a(n) for the first rows m of this sequence seen as an irregular triangle T(m,j). "New" numbers introduced for prime (n+1) are shown in parentheses:
  m\j   1   2   3   4   5   6   7   8   9  10  11   A338237(m)
  ------------------------------------------------------------
  1:   (1)                                                1
  2:   (2)  1                                             2
  3:   (3)  2  (4)  1                                     4
  4:    3   2  (5)  4  (6)  1                             6
  5:    3   2  (7)  5  (8)  4   6   1                     8
  6:   (9)  3   2   7   5   8 (10)  4 (11)  6   1        11
  ...

Michael De Vlieger, Table of n, a(n) for n = 1..10237 (as an irregular triangle, rows 1 <= n <= 36, flattened)
Michael De Vlieger, Log-log scatterplot of a(n) for 1 <= n <= 11636, showing 36 rows if read as an irregular table.

Cf. A000027, A000040, A000720, A338237, A348907, A349192.

c = 0; 1 + Reap[Do[Set[a[i], If[PrimeQ[i], i; c++, a[i - c]] ]; Sow[a[i]], {i, 2, 2^24}] ][[-1, -1]]

A349192 Irregular triangle T(m,k) = inverse permutation of S(m,k) = A349191 read as an irregular triangle.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 3, 6, 2, 1, 4, 3, 5, 8, 2, 1, 6, 4, 7, 3, 5, 11, 3, 2, 8, 5, 10, 4, 6, 1, 7, 9, 15, 3, 2, 11, 6, 13, 5, 7, 1, 9, 12, 4, 8, 10, 14, 18, 3, 2, 14, 7, 16, 6, 8, 1, 10, 15, 4, 9, 12, 17, 5, 11, 13, 24, 4, 3, 19, 10, 21, 9, 11, 2, 14, 20, 6, 12, 16

Offset: 1

Michael De Vlieger, Nov 09 2021

We find k at S(m,k) where S is A349191 read as an irregular triangle. Alternatively, we find prime(k) at U(m,k) where U is A348907 read as an irregular triangle.

			First rows of T(m,k):
  m\k   1   2   3   4   5   6   7   8   9  10  11
  -----------------------------------------------
  1:    1
  2:    2   1
  3:    4   2   1   3
  4:    6   2   1   4   3   5
  5:    8   2   1   6   4   7   3   5
  6:   11   3   2   8   5  10   4   6   1   7   9
  ...

Michael De Vlieger, Table of n, a(n) for n = 1..10237 (rows 1 <= n <= 35, flattened)
Michael De Vlieger, Log-log scatterplot of T(m,k) 1 <= m <= 36.

Cf. A000027, A000040, A000720, A338237, A348907, A349191.

c = 0; Flatten@ Map[Table[If[k == 1, Length[#] + 1, FirstPosition[#, k - 1][[1]]], {k, If[IntegerQ@ #, # + 1, 1] &@ Max[#]}] &, {{}}~Join~Most@ SplitBy[Reap[Do[Set[a[i], If[PrimeQ[i], i; c++, a[i - c]]]; Sow[a[i]], {i, 2, 100}]][[-1, -1]], # == 0 &][[2 ;; -1 ;; 2]]]

Row lengths are in A338237.

cp's OEIS Frontend

A349191 a(n) = A000720(A348907(n+1)).

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