A347354 a(n) = sum of T(n,k) - T(n-1,k) for row n of A347285.

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

Offset: 1

Author

Michael De Vlieger, Sep 16 2021

Keywords

• nonn

Comments

If for row n, k > A089576(n-1), we interpret T(n-1,k) = 0.

Also the length of d = T(n,k) - T(n-1,k) in row n of A347285 such that d > 1, with 0 <= d <= 1.

Compactification of A347284 via indices k of primorials A002110(k). This is the most efficient compactification of A347284, superior to binary compactification via A347287. It employs the fact that A347284 concerns products of primorials, i.e., is a subset of A025487.

We can construct row n of A347285 by summing a constant array of a(k) 1's for 1 <= k <= n-1.

Examples

			Relation of a(n) and irregular triangle A347285, placing "." after the term in the current row where T(n,k) no longer exceeds T(n-1,k). Since the rows of A347285 reach a fixed point of 0, we interpret T(n,k) for vacant T(n-1,k) as exceeding same.
n    Row n of A347285   a(n)
-----------------------------
0:    0
1:    1.                   1
2:    2  1.                2
3:    3. 1                 1
4:    4  2  1.             3
5:    5  3  2  1.          4
6:    6. 3  2  1           1
7:    7  4. 2  1           2
8:    8  5  3  2  1.       5
9:    9. 5  3  2  1        1
10:  10  6  4  3  2  1.    6
...
a(3) = 1 since row 3 of A347285 has {3,1} while row 2 has {2,1}; only the first term of the former exceeds the analogous term in the latter.
a(4) = 3 since row 4 = {4,2,1} and row 3 = {3,1}; all 3 terms of the former are larger than the analogous term in the latter, etc.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002110, A025487, A089576, A347284, A347285, A347287.

Programs

Mathematica

Block[{nn = 84, a = {0}, c, e, m}, e[1] = 0; Do[c = 1; e[1]++; Do[Set[m, j]; Which[e[j - 1] == 1, Break[], IntegerQ@ e[j], If[e[j] < #, e[j]++; c++] &@ Floor@ Log[Prime[j], Prime[j - 1]^e[j - 1]], True, Set[e[j], 1]], {j, 2, k}]; AppendTo[a, c + Boole[c == m - 2]], {k, 2, nn}]; MapAt[# - 1 &, a, 4]]

Formula

A347284(n) = Product_{k=1..n} A002110(a(k)).