A306737 -id:A306737 - cp's OEIS frontend

A307322 Irregular triangle where row n is a list of indices in A002110 with multiplicity whose product is A004394(n).

0, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 3, 1, 1, 3, 2, 3, 1, 1, 1, 3, 1, 2, 3, 1, 1, 2, 3, 1, 1, 4, 2, 4, 1, 1, 1, 4, 1, 2, 4, 1, 1, 2, 4, 1, 1, 1, 2, 4, 1, 2, 2, 4, 1, 1, 3, 4, 1, 2, 5, 1, 1, 2, 5, 1, 1, 1, 2, 5, 1, 2, 2, 5, 1, 1, 3, 5, 1, 1, 2, 2

Offset: 1

Michael De Vlieger, Apr 02 2019

Analogous to A306737.
The first 52 terms of a(n) and A306737 are identical, since the first 19 terms of A002182 and A004394 are the same, and the first two terms of row 20 are the same. a(20) = 4,2,1,1,1, while A306737(20) = 4,2,2.
Each superabundant number A004394(n) can be expressed as a product of primorials in A002110.
Row 1 = {0} by convention.
Maximum value in row n = A001221(A004394(n)).
Row n in reverse order is the conjugate of the list of the multiplicities of the prime divisors of A004394(n).

			Terms in the first rows n of this sequence, followed by the corresponding primorials whose product = A004394(n):
   n    T(n,k)       A002110(T(n,k))   A004394(n)
  -----------------------------------------------
   1:   0;              1                =     1
   2:   1;              2                =     2
   3:   1, 1;           2 * 2            =     4
   4:   2;              6                =     6
   5:   1, 2;           2 * 6            =    12
   6:   1, 1, 2;        2 * 2 * 6        =    24
   7:   2, 2;           6 * 6            =    36
   8:   1, 1, 1, 2;     2 * 2 * 2 * 6    =    48
   9:   1, 3;           2 * 30           =    60
  10:   1, 1, 3;        2 * 2 * 30       =   120
  11:   2, 3;           6 * 30           =   180
  12:   1, 1, 1, 3;     2 * 2 * 2 * 30   =   240
  13:   1, 2, 3;        2 * 6 * 30       =   360
  14:   1, 1, 2, 3;     2 * 2 * 6 * 30   =   720
  15:   1, 1, 4;        2 * 2 * 210      =   840
  ...
Row 6 = {1,1,2} since A002110(1)*A002110(1)*A002110(2) = 2*2*6 = 24 and A004394(6) = 24. The conjugate of {1,1,2} = {3,1} and 24 = 2^3 * 3^1.
Row 10 = {1,1,3} since A002110(1)*A002110(1)*A002110(3) = 2*2*30 = 120 and A004394(10) = 120. The conjugate of {1,1,3} = {3,1,1} and 120 = 2^3 * 3^1 * 5^1.

Michael De Vlieger, Table of n, a(n) for n = 1..10664 (rows 1 <= n <= 1200, flattened).
Michael De Vlieger, Relation between A307322 and A306737.

Cf. A001221, A002110, A002182, A004394, A306737.

Block[{s = Array[DivisorSigma[1, #]/# &, 10^6]}, Map[Table[LengthWhile[#, # >= i &], {i, Max@ #}] &@ If[# == 1, {0}, Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> Last@ # &, f]]@ FactorInteger@ #] &@ FirstPosition[s, #][[1]] &, Union@ FoldList[Max, s]] /. {} -> {0}] // Flatten

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A307322 Irregular triangle where row n is a list of indices in A002110 with multiplicity whose product is A004394(n).

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