Factorial problem - Page 2

Annunaki · 2003-07-20, 19:52

however assumption that 4th element in this row could only be such n^2 that is not divisible by two is correct as any factorial except 1! will produce pair number and pair number +1 isn`t divisible by two..

Annunaki · 2003-07-20, 20:22

just thought about that a little bit.. and what i figured out is here..
for every unpair natural n^2-1 or just (2n+1)^2 - 1 the result if you will divide it by 8 is sum of arithmethic progression..
for example..
1^2-1=0 {contains 0 8ghts}
3^2-1=0 {contains 1 8ghts}
5^2-1=0 {contains 3 8ghts}
7^2-1=0 {contains 6 8ghts} and so on..
so you can just calculate the result if you know as it increases as arithmethic progression..
((2n+1)^2-1)/8=(n^2-n)/2

now lets take a lok at factorials..
4!=1*2*3*4=3*8 which corresponds to 5^2-1
5!=5*3*8 or 15 eights wich corresponds to 11^2-1
6!={6*5*3}*8 90 eights
7!={7*90}*8 which is 630 eights 630*2= 1260 and 1260=36^2-36 and (36*2)-1=71..
actualy if we need to prove that 4th such equation insn`t posible.. then we need to prove that
(n^2-n)/2 =/ x!/8 (=/ means not equal or not posible.. if x>7)..

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Multi-Factorial Search	rogue	And now for something completely different	1	2015-06-02 23:51
Factorial puzzle	henryzz	Puzzles	5	2015-04-02 12:58
Factorial primes	Unregistered	Information & Answers	2	2011-09-11 21:32
Factorial	mfgoode	Puzzles	6	2007-07-24 14:24
Factorial in C programming	Unregistered	Programming	7	2005-04-09 20:13

2003-07-20, 19:52	#12
Annunaki Jul 2003 37₈ Posts	however assumption that 4th element in this row could only be such n^2 that is not divisible by two is correct as any factorial except 1! will produce pair number and pair number +1 isn`t divisible by two..

2003-07-20, 20:22	#13
Annunaki Jul 2003 31 Posts	just thought about that a little bit.. and what i figured out is here.. for every unpair natural n^2-1 or just (2n+1)^2 - 1 the result if you will divide it by 8 is sum of arithmethic progression.. for example.. 1^2-1=0 {contains 0 8ghts} 3^2-1=0 {contains 1 8ghts} 5^2-1=0 {contains 3 8ghts} 7^2-1=0 {contains 6 8ghts} and so on.. so you can just calculate the result if you know as it increases as arithmethic progression.. ((2n+1)^2-1)/8=(n^2-n)/2 now lets take a lok at factorials.. 4!=1234=38 which corresponds to 5^2-1 5!=538 or 15 eights wich corresponds to 11^2-1 6!={653}8 90 eights 7!={790}8 which is 630 eights 6302= 1260 and 1260=36^2-36 and (36*2)-1=71.. actualy if we need to prove that 4th such equation insn`t posible.. then we need to prove that (n^2-n)/2 =/ x!/8 (=/ means not equal or not posible.. if x>7)..