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Good day all. I am not a mathematician by a long shot. Please bear with me...

I am playing with "descending-consecutive-prime composite numbers" (I don't think that's the term). These are numbers that contain all the prime factors from 2 until $p_n$, such that their prime factorization, say $2^{a_1} \times 3^{a_2} \times ... \times p_n^{a_n}$ have the property $1 \leq a_n \leq a_{n-1} \leq ... \leq a_1$

They tend to have really interesting patterns, especially as they become larger: $120120, 180180, 240240, 360360, 720720, 1081080, 1441440, 1801800$, etc.

I was wondering, is it at all possible to get the next "descending-consecutive-prime composite number" given the number, $n$, and its prime factorization, i.e. $n=77636318760$ with $n = 2^3 \times 3^2 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29$ without painstakingly checking each number?

I fail to see the pattern... :-(

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    $\begingroup$ Note that the count of all such numbers where $\sum_{i=1}^n a_i=k$ is the value $p(k)$ of the partition function at $k$. $\endgroup$
    – Paul Tanenbaum
    Commented Jun 7 at 11:42
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    $\begingroup$ The mentioned pattern is simply a consquence of $7\cdot 11\cdot 13=1001$ and all of the numbers you are interested with $p_n\ge 13$ have the prime factors $7,11,13$ and are hence divisible by $1001$. For small multiples , this gives a clear pattern , but for larger values this pattern soon breaks down. $\endgroup$
    – Peter
    Commented Jun 7 at 12:25
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    $\begingroup$ Is this oeis.org/A126098 ? $\endgroup$
    – Gerry Myerson
    Commented Jun 7 at 12:58
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    $\begingroup$ @GerryMyerson Indeed it is! That is helpful, thank you. $\endgroup$
    – Jaco Van Niekerk
    Commented Jun 7 at 13:20
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    $\begingroup$ @GerryMyerson I think A126098 is not it... take for example $2^3×3^2×5^2=1800$. It fits the definition but its not in the sequence... Powers of 2 should all be there as well if I understood correctly.... $\endgroup$
    – François Huppé
    Commented Jun 11 at 22:54

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I don't have a set pattern but after the $n = 2^3 \times 3^2 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29$

The next one with a new prime added would be $2\times 3 \times 5 \times.... \times 29\times 31$ which will be $\frac {31}{2^2\cdot 3}\approx 2.6$ times larger.

Before that we have one that is only two times larger but $2^4\times 3\times 5 .... \times 29$.

And before that we have one that is one and a half time larger by $2^2 \times 3^3 \times 5 \times.... \times 29$.

Basically we can get larger ones but not very much larger ones by decreasing the powers of $2,3$ and increasing a power for $p_n$.

$\frac 53 = 1\frac 23$ so we can get one that is $1\frac 23$ larger by $2^3\times 3 \times 5^2 \times .... \times 29$.

And $\frac 76$ or $\frac {11}{10}$ and so on... play with it.

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  • $\begingroup$ As I understand it, OP insists on exponents decreasing as primes increase, so anything starting $2^3\times3\times5^2$ won't fly. $\endgroup$
    – Gerry Myerson
    Commented Jun 7 at 22:02
  • $\begingroup$ Oh, I misunderstood. Then.... well I actually don't undestand the problems. .... okay. Well, then it's easier. The only things less then $2\cdot....\cdot 31$ (which is $2.6$ times as big as what we've got is to increase $2$ then we have then 2 to 31. But then we want 3 times what we had with $2^3\times 2^3 \times.... 29$. $\endgroup$
    – fleablood
    Commented Jun 7 at 22:58
  • $\begingroup$ If the link in my comment gives the right sequence, then the next one is $107084577600=2^63^25^2\times7\times11\times13\times17\times19\times23$ $\endgroup$
    – Gerry Myerson
    Commented Jun 8 at 0:25

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