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Topic: Integer Sequence? (Read 1543 times) |
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Barukh
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Integer Sequence?
« on: Feb 6th, 2006, 6:29am » |
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The sequence gn is defined iteratively as follows: g0 = 1 gn = (1 + g02 + .. + gn-12)/n, n = 1, 2, … Is it true that gn consists of integers only?
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thedkl
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Re: Integer Sequence?
« Reply #1 on: Feb 7th, 2006, 11:48am » |
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is it true that gn=1 for all n?
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towr
wu::riddles Moderator Uberpuzzler
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Re: Integer Sequence?
« Reply #2 on: Feb 7th, 2006, 12:46pm » |
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on Feb 7th, 2006, 11:48am, thedkl wrote:is it true that gn=1 for all n? |
| No, e.g. g1 = 2 in fact, the sequence explodes quite rapidly, over a googol in about a dozen step. 1 2 2 3 3 5 4 10 5 28 6 154 7 3520 8 1551880 9 267593772160 10 7160642690122633501504 11 4661345794146064133843098964919305264116096 12 1.8106787177169338e+84 13 2.521967245225415e+167
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« Last Edit: Feb 7th, 2006, 1:16pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Barukh
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Re: Integer Sequence?
« Reply #3 on: Feb 17th, 2006, 12:12am » |
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I would like to revive this thread, as there seem to be some subtleties I don't know how to deal with. Hint: Use modular arithmetic.
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Barukh
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Re: Integer Sequence?
« Reply #4 on: Mar 31st, 2006, 8:43am » |
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Somebody's still working on this?
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towr
wu::riddles Moderator Uberpuzzler
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Re: Integer Sequence?
« Reply #5 on: Mar 31st, 2006, 2:23pm » |
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Well, not me.. After finding the first few numbers from the sequence I just googled and found an answer.
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Wikipedia, Google, Mathworld, Integer sequence DB
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ecoist
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Re: Integer Sequence?
« Reply #6 on: Apr 2nd, 2006, 10:40am » |
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I ignored this problem until today, and expected to get nowhere with it until I remembered something about the Fibonacci sequence. I get that gn=f(n+2), where f(k) is the Fibonacci sequence. Hence gn is an integer, for all n[ge]0. This follows from [sum]_{i=1}^{i=n} f(i)^2 = f(n)f(n+1); easily proved by induction. The sum of squares of the gi in your iteration is the sum of the squares of the first n+1 Fibonacci numbers.
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Obob
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Re: Integer Sequence?
« Reply #7 on: Apr 2nd, 2006, 11:06am » |
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Ecoist, your solution doesn't coincide with towr's computed values for the sequence.
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ecoist
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Re: Integer Sequence?
« Reply #8 on: Apr 2nd, 2006, 11:51am » |
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Right! I'm an idiot!
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Eigenray
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Re: Integer Sequence?
« Reply #9 on: Mar 9th, 2007, 8:52am » |
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Another old problem: Let sn = 1 + g02 + ... + gn-12 = ngn, so that sn+1 = sn + gn2 = sn + (sn/n)2. Since sn grows too fast to compute directly, we compute it mod n, and make sure it's 0. If n is prime, this is easy: Code:For[p = 2, p < 100, p = NextPrime[p], S[1] = 2; For[k = 1, k < p, k++, S[k + 1] = Mod[S[k] + PowerMod[k, -2, p]S[k]^2, p]; ]; If[Mod[S[p], p] != 0, Print["S[", p, "]=", S[p]]]; ] |
| S[43]=24 S[61]=12 S[67]=18 S[83]=19 So this shows s43 isn't an integer. To show 43 is the first time this happens is a bit trickier. If pe | n, then we need to compute sn mod pe. However, if we know sk mod pa, and pb | k, then we only know sk+1 mod pa-2b, since we divided by k2. So we need to start by knowing s1 mod pa, where a = e + 2b = e + 2*# of times p divides (n-1)!. Thus: Code:For[n = 2, n <= 43, n++, fac = FactorInteger[n]; For[i = 1, i <= Length[fac], i++, (* fac[[i]] = {p, e} *) p = fac[[i, 1]]; a = fac[[i, 2]] + 2Sum[Floor[(n - 1)/p^k], {k, 1, Log[p, n - 1]}]; S[1] = 2; For[k = 1, k < n, k++, b = Log[p, GCD[p^a, k]]; a = a - 2b; (* S[k + 1] = S[k] + S[k]^2/k^2 *) S[k + 1] = Mod[S[k] + PowerMod[k/p^b, -2, p^a](S[k]/p^b)^2, p^a]; ]; If[S[n] != 0, Print["S[", n, "]=", S[n], " mod ", p^a]]; ] ] |
| gives S[43]=24 mod 43, so sn is an integer for n<43.
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