%I #17 Mar 08 2019 20:23:42
%S 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,
%T 1,1,4,2,4,1,1,1,4,1,2,4,1,1,2,4,2,2,4,1,1,1,2,4,1,2,2,4,1,1,1,1,2,4,
%U 1,1,3,4,1,2,5,2,2,2,4,1,1,1,3,4,1,1,2,5,2,2,5,1,1,1,2,5,1,2,2,5,1,1,1,1,2,5
%N Irregular triangle where row n is a list of indices in A002110 with multiplicity whose product is A002182(n).
%C Each highly composite number A002182(n) can be expressed as a product of primorials in A002110.
%C Row 1 = {0} by convention.
%C Maximum value in row n is given by A001221(A002182(n)).
%C Row n in reverse order is the conjugate of A067255(A002182(n)), a list of the multiplicities of the prime divisors of A002182(n).
%H Michael De Vlieger, <a href="/A306737/b306737.txt">Table of n, a(n) for n = 1..10198</a>, rows 1 <= n <= 1200, flattened.
%H Michael De Vlieger, <a href="/A306737/a306737.png">Charts showing terms in A002182 as a product of terms in A002110</a>.
%H Michael De Vlieger, <a href="/A306737/a306737.txt">Condensed text table showing terms in rows 1 <= n <= 10000</a>.
%H Benny Lim, <a href="https://www.parabola.unsw.edu.au/2010-2019/volume-54-2018/issue-3/article/prime-numbers-generated-highly-composite-numbers">Prime Numbers Generated From Highly Composite Numbers</a>, Parabola (2018) Vol. 54, Issue 3.
%e Terms in the first rows n of this sequence, followed by the corresponding primorials whose product = A002182(n):
%e n T(n,k) A002110(T(n,k)) A002182(n)
%e -----------------------------------------------
%e 1: 0; 1 = 1
%e 2: 1; 2 = 2
%e 3: 1, 1; 2 * 2 = 4
%e 4: 2; 6 = 6
%e 5: 1, 2; 2 * 6 = 12
%e 6: 1, 1, 2; 2 * 2 * 6 = 24
%e 7: 2, 2; 6 * 6 = 36
%e 8: 1, 1, 1, 2; 2 * 2 * 2 * 6 = 48
%e 9: 1, 3; 2 * 30 = 60
%e 10: 1, 1, 3; 2 * 2 * 30 = 120
%e 11: 2, 3; 6 * 30 = 180
%e 12: 1, 1, 1, 3; 2 * 2 * 2 * 30 = 240
%e 13: 1, 2, 3; 2 * 6 * 30 = 360
%e 14: 1, 1, 2, 3; 2 * 2 * 6 * 30 = 720
%e 15: 1, 1, 4; 2 * 2 * 210 = 840
%e ...
%e Row 6 = {1,1,2} since A002110(1)*A002110(1)*A002110(2) = 2*2*6 = 24 and A002182(6) = 24. The conjugate of {2,1,1} = {3,1} and 24 = 2^3 * 3^1.
%e Row 10 = {1,1,3} since A002110(1)*A002110(1)*A002110(3) = 2*2*30 = 120 and A002182(10) = 120. The conjugate of {3,1,1} = {3,1,1} and 120 = 2^3 * 3^1 * 5^1.
%t With[{s = DivisorSigma[0, Range[250000]]}, Map[Reverse@ 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
%Y Cf. A001221, A002110, A002182, A067255, A304886.
%K nonn,tabf
%O 1,5
%A _Michael De Vlieger_, Mar 06 2019