%I #15 May 25 2023 21:40:04
%S 1,2,4,12,8,24,16,48,144,720,32,96,288,1440,864,4320,21600,151200,64,
%T 192,576,2880,1728,8640,43200,302400,128,384,1152,5760,3456,17280,
%U 86400,604800,10368,51840,259200,1814400,256,768,2304,11520,6912,34560,172800,1209600,20736,103680,518400,3628800,62208
%N Numbers in A363063 arranged in lexicographic order according to multiplicities of prime power factors p^k, written in order of p.
%C Numbers m in A363063 are products of prime powers p(j)^S(j), j = 1..N, where p(j) is the j-th prime, such that p(j+1)^S(j+1) < p(j)^S(j). As consequence of definition of A363063, S(j) > S(j+1), hence multiplicities S(j) are distinct. Consequently, A363063 is a subset of A025487; m is a product of primorials. A025487 in turn is a subset of A055932.
%C These qualities enable us to write an algorithm that increments S(j) or drops the last term in S until we can increment S(j) to attain a solution. This algorithm generates terms in lexical order as described in the Name. The same qualities enable expression of m = Product p(j)^S(j) instead as Sum 2^(S(j)-1), a strictly increasing sequence.
%H Michael De Vlieger, <a href="/A363250/b363250.txt">Table of n, a(n) for n = 0..14158</a> (rows i = 0..30, flattened)
%H Michael De Vlieger, <a href="/A363250/a363250.png">Plot p^e | a(n) at (x,y) = (n,e)</a>, n = 1..3526, 12X vertical exaggeration
%F Seen as an irregular triangle, the first term in row i is 2^i, and the last term in row i is A347284(i).
%e Table of n, a(n), and multiplicities S(j) written such that Product p(j)^S(j) = a(n). a(n) = A000079(i) is shown in the penultimate column, while a(n) = A347284(k) appears in the last column.
%e n a(n) multiplicities i k
%e -----------------------------------
%e 0: 1 0 0
%e 1: 2 1 1 1
%e 2: 4 2 2
%e 3: 12 2 1 2
%e 4: 8 3 3
%e 5: 24 3 1 3
%e 6: 16 4 4
%e 7: 48 4 1
%e 8: 144 4 2
%e 9: 720 4 2 1 4
%e 10: 32 5 5
%e 11: 96 5 1
%e 12: 288 5 2
%e 13: 1440 5 2 1
%e 14: 864 5 3
%e 15: 4320 5 3 1
%e 16: 21600 5 3 2
%e 17: 151200 5 3 2 1 5
%e ...
%e Sequence read as an irregular triangle T(n, k):
%e n\k 1 2 3 4 5 6 7 8
%e ---------------------------------------------------
%e 0: 1
%e 1: 2
%e 2: 4 12
%e 3: 8 24
%e 4: 16 48 144 720
%e 5: 32 96 288 1440 864 4320 21600 151200
%e 6: 64 192 576 2880 1728 8640 43200 302400
%e ...
%t nn = 12;
%t f[x_] := Times @@ MapIndexed[Prime[First[#2]]^#1 &, x];
%t {1}~Join~Reap[Do[s = {i}; Sow[2^i]; Set[k, 1];
%t Do[
%t If[Prime[k]^s[[-1]] > Prime[k + 1],
%t AppendTo[s, 1]; k++; Sow[f[s]],
%t If[Length[s] == 1, Break[],
%t If[Prime[k - 1]^(s[[-2]]) > Prime[k]^(s[[-1]] + 1),
%t s[[-1]]++; Sow[f[s]],
%t While[And[k > 1,
%t Prime[k - 1]^(s[[-2]]) < Prime[k]^(s[[-1]] + 1)], k--;
%t s = s[[1 ;; k]]]; If[k == 1, Break[], s[[-1]]++; Sow[f[s]] ]
%t ] ] ], {j, Infinity}], {i, nn}]][[-1, -1]] ]
%o (Python)
%o from sympy import nextprime,oo
%o from itertools import islice
%o primes = [2] # global list of first primes
%o def f(pi, ppmax):
%o # Generate numbers with nonincreasing prime-powers <= ppmax, starting at the (pi+1)-st prime.
%o if len(primes) <= pi: primes.append(nextprime(primes[-1]))
%o p0 = primes[pi]
%o if ppmax < p0:
%o yield 1
%o return
%o pp = 1
%o while pp <= ppmax:
%o for x in f(pi+1, pp):
%o yield pp*x
%o pp *= p0
%o def A363250_list(nterms):
%o return list(islice(f(0,oo),nterms)) # _Pontus von Brömssen_, May 25 2023
%Y Cf. A000079, A025487, A055932, A067255, A347284, A363063.
%K nonn
%O 0,2
%A _Michael De Vlieger_, May 23 2023