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:There may, of course, be more than one way to decompose a J-P number into a product of factorials but the idea is to choose the way which uses the largest factorials and present these in highest to lowest order. I've added a sentence to the task description to try and clarify this. --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 09:57, 9 June 2023 (UTC)
:I believe (bicbw) there is no possible way to constructmake 439,378,587,648,000 = 14! * 7! with only prime factorials. --[[User:Petelomax|Petelomax]] ([[User talk:Petelomax|talk]]) 02:53, 10 June 2023 (UTC)
:Yes, your right.14! = 13!*2*7 2= 2! but no way to get a lonely 7. Of course there 16!= 2!^4 * 15! . To much work for to less advantage --[[User:Horst|Horst]] ([[User talk:Horst|talk]]) 16:28, 11 June 2023 (UTC)
== Factoring the 1050th number ==
The 1050th jp number is 139,345,920,000. OneIt andfactors Ias suspect7! the* only5!^3 possible* factorisation2!^4 isor 78! * 5!^3 * 2!^4, but the <del>Phix</del>/Julia/Wren entries are not getting that. IUpdate: havenew thoughtalgorithm ofposted, ahowever bettermy strategy,faffing basedabout on thewith prime powers, butis it isprobably not straightforward and would naturally produce the ''lowest'' factorialspoint, one-at- I ''think'' I might have just thought of a-time/less waygreedy toprobably convertis. that--[[User:Petelomax|Petelomax]] to([[User thetalk:Petelomax|talk]]) highest05:31, factorials...10 InJune the2023 meantime, some examples:(UTC)
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:Well spotted! Rather than translate yours, I've modified the decomposition algorithm I was using for both Wren versions using recursion and it now seems to be working OK. --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 11:31, 10 June 2023 (UTC)
92,160 = 6! * 2!^7, or
92,160 = 5! * 3! * 2!^7
18,345,885,696 = 4!^7 * 2!^2, or
18,345,885,696 = 3!^7 * 2!^16
139,345,920,000 = 7! * 5!^3 * 2!^4 (only one?)
18,345,885,696 = 4!^7 * 2!^2, or
18,345,885,696 = 3!^7 * 2!^16
724,775,731,200 = 6! * 5! * 2!^23, or
724,775,731,200 = 5!^2 * 3! * 2!^23
9,784,472,371,200 = 6!^2 * 4!^2 * 2!^15, or
9,784,472,371,200 = 5!^2 * 3!^4 * 2!^19
439,378,587,648,000 = 14! * 4!^2 * 2!^15 (only one?)
7,213,895,789,838,336 = 4!^8 * 2!^16, or
7,213,895,789,838,336 = 3!^8 * 2!^32
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There are probably only a few ways to do each, maybe show them all? --[[User:Petelomax|Petelomax]] ([[User talk:Petelomax|talk]]) 23:52, 9 June 2023 (UTC)
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