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wu :: forums    riddles    medium (Moderators: towr, ThudnBlunder, Eigenray, SMQ, william wu, Icarus, Grimbal)    find « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: find  (Read 555 times) tony123 Junior Member     Posts: 61 find   « on: Jul 7th, 2007, 3:43pm » Quote Modify http://www.sosmath.com/CBB/latexrender/pictures/aa1c1da929c9347fddda5d32 bb20d5e5.gif IP Logged denis Uberpuzzler     Gender: Posts: 1222 Re: find   « Reply #1 on: Jul 7th, 2007, 5:59pm » Quote Modify n=1: 1/2 n=2: 3/8 n=3: 21/64 n=4: 315/256 315/1024 . . .   So we have basically two sequences to figure out   The denominators 2,8,64, 1024 ...  which is 22n  22n-1 2n(n+1)/2   and the sequence of nominators     1,3,21,315,... which is the 2 factorial sequence:   (1, 1*3, 1*3*7, 1*3*7*15, ...   =[n]2!  as shown in:   http://www.emis.de/journals/JIS/VOL9/Morrison/morrison37.pdf   So we get for general n:   [n]2!/22n-1 [n]2!/2n(n+1)/2     EDIT: Thanks to Fibster for finding my erroneous calculations. « Last Edit: Jul 7th, 2007, 7:31pm by denis » IP Logged FiBsTeR Senior Riddler     Gender: Posts: 581 Re: find   « Reply #2 on: Jul 7th, 2007, 6:56pm » Quote Modify on Jul 7th, 2007, 5:59pm, denis wrote: The denominators 2,8,64,256 etc  which is 22n   Ummm, maybe I'm doing something wrong, but 22n = 4n which is {4,16,64,256,...}. « Last Edit: Jul 7th, 2007, 6:57pm by FiBsTeR » IP Logged denis Uberpuzzler     Gender: Posts: 1222 Re: find   « Reply #3 on: Jul 7th, 2007, 7:01pm » Quote Modify Ooops. Went too fast.     2,8,64,256 = 22n/2=22n-1   Thanks for pointing out my error ...   Corrected in my post above... « Last Edit: Jul 7th, 2007, 7:03pm by denis » IP Logged FiBsTeR Senior Riddler     Gender: Posts: 581 Re: find   « Reply #4 on: Jul 7th, 2007, 7:06pm » Quote Modify on Jul 7th, 2007, 7:01pm, denis wrote: 2,8,64,256 = 22n/2=22n-1   I think you're still going too fast!     22n/2 = 4n/2, which is {2,8,32,128,...} IP Logged denis Uberpuzzler     Gender: Posts: 1222 Re: find   « Reply #5 on: Jul 7th, 2007, 7:11pm » Quote Modify Darn... Your right   the correct seqence for the denominators is 2,8,64,1024, ...   how about   2, 8, 64, 1024 ... = 2n(n+1)/2   Corrected again in post above...     « Last Edit: Jul 7th, 2007, 7:39pm by denis » IP Logged FiBsTeR Senior Riddler     Gender: Posts: 581 Re: find   « Reply #6 on: Jul 7th, 2007, 7:20pm » Quote Modify Now you're thinking too much and you've completely overstepped the problem! Remember, you're trying to find a sequence for {2,8,64,256,...} and not {2,8,64,1024,...}   And BTW, if someone comes along thinking they'll use that encyclopedia for integer sequences, don't bother.   EDIT: Nevermind, I misread the post. « Last Edit: Jul 7th, 2007, 7:22pm by FiBsTeR » IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy => medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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