\(\displaystyle u_{n+1} = \left(\sum_{k=0}^{m} {m + 2\choose 2k} u_n^{m+2-2k} \cdot x^k\right) \cdot \left( \sum_{k=0}^{m-1}{m+2\choose 2k+1} u_n^{m+1-2k} \cdot x^{k} \right)^{-1}\) for even positive m.

What value does it get when n->infty? I have an idea, but can't explicitely prove it.

\(\displaystyle u_{n+1}=\left(\sum_{k=0}^{0.5m} {m \choose 2k}\cdot u_n^{m-2k}\cdot x^k\right) \cdot \left(\sum_{k=0}^{0.5m-1} {m \choose 2k+1}\cdot u_n^{m-2k-1}\cdot x^k\right)^{-1}\) for even positive m.

I have no time to explain* ( :cry: ) butagentredlum said:I ask the honorable gentleman from UTC-5 to post his solutions to his questions please.

http://oeis.org/A000028

and the Lambek-Moser paper linked therein should get you there.

* Personal matters: got in a terrible car wreck, now dealing with fallout.

I like the question but as usual am having trouble understanding the notation. It appears m is arbitrary even number > 0 and determines the number of terms in the summation , is that right?Hoempa said:No worries, agent, I'm happy to read CRG is fine. Hope you guys like the Q.

Would you be kind enough to give numerical values of \(\displaystyle \ \ \ u_{0 + 1} \ , \ u_{1 + 1} \ , \ and \ u_{2 + 1} \ \ \\) so that I may work backwards and make sure that I have not misinterpreted the notation ?

Yes, that's right.agentredlum said:I like the question but as usual am having trouble understanding the notation. It appears m is arbitrary even number > 0 and determines the number of terms in the summation , is that right?

Yes. I'll chose x = 100, u_0 = 10000, m = 2.agentredlum said:Would you be kind enough to give numerical values of \(\displaystyle \ \ \ u_{0 + 1} \ , \ u_{1 + 1} \ , \ and \ u_{2 + 1} \ \ \\) so that I may work backwards and make sure that I have not misinterpreted the notation ?

u_1 = 1000001/200 = 5000.005

u_2 = 1000006000001/400000400 ~= 2500.0125

u_3 = 1000028000070000028000001/800005600005600005600000800 ~= 1250.026.

This problem appears to require programming capability using Computer Algebra Systems in order to determine trends. Right now i don't have a computer ... unless you know how to program the free online version of Wolfram Alpha ? I don't think WIA can run program loops but i'm not sure ... i suppose i could input each value separately and write down each result but it's so tedious ...Hoempa said:Could you work backwards agent? I'll admit that my values are tricky, the next is about half the previous but that isn't the general case.

Do you know of any online CAS that does not require the user to download onto the hard drive? My PS3 hard drive won't accept any download unless it comes from Sony as an app and Sony hasn't made a CAS app as far as i know. The PS3 has very powerful computing power (in relation to the cost) but unless we hack it we can't use that power for anything other than gaming. I remember reading an article about a university that hacked together a bunch of PS3's and turned them into 1 supercomputer.

I hope another member with a computer can help you.

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