January 2012 - mathfun@mailman.xmission.com

[math-fun] stupidity with decimals: a new(?) level
by Joerg Arndt 30 Jan '12

30 Jan '12

Let r = 100001.00000000009999800002999940001499964000... (the real root of x^3-100001*x^2-1). Now we print the powers of r with underscores for zeros to make the pattern apparent: [ 0, 1] [ 1, 1____1.__________99998____299994___1499964___8399796__512987___3311] [ 2, 1____2____1.____199998____199995___1399968___7199823__44998864_2882] [ 3, 1____3____3____3.999999999999997____899982___3999895__27299314_1746] [ 4, 1____4____6____8____4.9999999998____399994___159995___13199671__856] [ 5, 1____5___10___15___15____6.___________________49998____4999875__35_] [ 6, 1____6___15___26___33___24___10.______________199994___1199965__12_] [ 7, 1____7___21___42___63___63___42___14.9999999999999999999999993___35] [ 8, 1____8___28___64__110__136__120___72___20.99999999999999999998____8] [ 9, 1____9___36___93__180__261__282__225__117___31.____________________] [10, 1___10___45__130__280__462__585__570__405__190___46._______________] [11, 1___11___55__176__418__770_1111_1265_1111__715__308___67.__________] [12, 1___12___66__232__603_1224_1974_2556_2637_2108_1248__492___98._____] [13, 1___13___78__299__845_1872_3328_4810_5655_5330_3926_2145__780__144 E65] The diagonal seq. appears to be http://oeis.org/A001609 What could the triangle be (if the polynomial is replaced by x^3-(10^k+1)*x^2-1 for large k)? And: do I get a cookie?

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[math-fun] Fwd: Re: [EXTERNAL] Re: Fun with math: Dividing one by 998001 yields a surprising result
by David Wilson 30 Jan '12

30 Jan '12

I realized that this sequence would be a(n) = smallest prime >= 10^n with primitive root 10. This made it easier to extend 1 17 2 109 3 1019 4 10007 5 100019 6 1000171 7 10000019 8 100000007 9 1000000007 -------- Original Message -------- Subject: Re: [math-fun] [EXTERNAL] Re: Fun with math: Dividing one by 998001 yields a surprising result Date: Sat, 28 Jan 2012 10:41:33 -0500 From: David Wilson <davidwwilson(a)comcast.net> Reply-To: math-fun <math-fun(a)mailman.xmission.com> To: math-fun <math-fun(a)mailman.xmission.com> Which made me curious... What is the smallest value a(d) such that every d-digit sequence occurs in the decimal expansion of 1/d (after the decimal point). For small d>= 1 my program says 17, 109, 1019, 10007, 100019, ... but it seems stuck on 6 digits, so perhaps there is an overflow issue. All primes slightly above 10^d, as one might expect, but not necessarily the smallest such primes. _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] pFp-1[- 6^6/7^7]
by Bill Gosper 30 Jan '12

30 Jan '12

There are endlessly many identities of the form pFp-1[rational] = pile of Gammas(rationals), but, as Dick Askey pointed out to me decades ago, the rationals all seem to be made out of primes < 7. I've always known how to construct one with 7s, but the hardware (and my stomach) were never up to it. Just to get it over with, Sum[(((j^13 + ((946722645184512*j^12 + 3397234214391808*j^11 + 7324312847932416*j^10 + 10575837135607296*j^9 + 10789977229862400*j^8 + 7994097957085696*j^7 + 4349780625735744*j^6 + 1736115415231328*j^5 + 501325200097728*j^4 + 101649833417016*j^3 + 13682157776100*j^2 + 1093720984200*j + 39131717625)/ 119811560185856))*(-1)^j*((j - 1/2)!)^2*(6*j + 1/2)!*(7*j + 1/2)!)/((j + 1)!* (7*j + 6)!*(7*j + 7)!)), {j, 0, Infinity}] == Pi/14625434593 This is thirteen or so 16F15[-6^6/7^7]s. The XP Macsyma (compiled for a 286 in NT) that disgorged this was so effete that afterward, it claimed 2F1[1/2,1/2;2] = 0, and when asked for the pFp-1s for the above, said Error: FACTOR ran out of primes. --rwg

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Re: [math-fun] [EXTERNAL] Re: Fun with math: Dividing one by 998001 yields a surprising result
by Ray Tayek 30 Jan '12

30 Jan '12

At 03:18 PM 1/27/2012, you wrote: >998 starts 3rd row from bottom, 2 digits from righthand end ... that's 799 at the end of that row. thanks >On 1/27/12, Ray Tayek <rtayek(a)ca.rr.com> wrote: > > At 11:34 AM 1/27/2012, you wrote: > >>SetAccuracy[1/998001, 4000] gives the same problem (no 998) for me. > > > > i used java. > > > > import java.math.*; > > public class Main { > > public static void main(String[] args) { > > BigDecimal n = BigDecimal.valueOf(998001); > > MathContext mc = new MathContext(3010); > > BigDecimal x = BigDecimal.ONE.divide(n, mc); > > String s = x.toString(); > > for (int i = 0; i < 10; i++) > > System.out.println(s.substring(2 + 300 * i, 2 + 300 * (i + 1))); > > } > > } > > > > > 000001002003004005006007008009010011012013014015016017018019020021022023024025026027028029030031032033034035036037038039040041042043044045046047048049050051052053054055056057058059060061062063064065066067068069070071072073074075076077078079080081082083084085086087088089090091092093094095096097098099 > > > 100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199 > > > 200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299 > > > 300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399 > > > 400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499 > > > 500501502503504505506507508509510511512513514515516517518519520521522523524525526527528529530531532533534535536537538539540541542543544545546547548549550551552553554555556557558559560561562563564565566567568569570571572573574575576577578579580581582583584585586587588589590591592593594595596597598599 > > > 600601602603604605606607608609610611612613614615616617618619620621622623624625626627628629630631632633634635636637638639640641642643644645646647648649650651652653654655656657658659660661662663664665666667668669670671672673674675676677678679680681682683684685686687688689690691692693694695696697698699 > > > 700701702703704705706707708709710711712713714715716717718719720721722723724725726727728729730731732733734735736737738739740741742743744745746747748749750751752753754755756757758759760761762763764765766767768769770771772773774775776777778779780781782783784785786787788789790791792793794795796797798799 > > > 800801802803804805806807808809810811812813814815816817818819820821822823824825826827828829830831832833834835836837838839840841842843844845846847848849850851852853854855856857858859860861862863864865866867868869870871872873874875876877878879880881882883884885886887888889890891892893894895896897898899 > > > 900901902903904905906907908909910911912913914915916917918919920921922923924925926927928929930931932933934935936937938939940941942943944945946947948949950951952953954955956957958959960961962963964965966967968969970971972973974975976977978979980981982983984985986987988989990991992993994995996997999000 > > > > > > --- > > co-chair http://ocjug.org/ > > > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun --- co-chair http://ocjug.org/

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[math-fun] sums involving all integer powers
by Warren Smith 29 Jan '12

29 Jan '12

I think somebody was recently asking about something like this. Let S be the set of powers>1 of integers>1, S={4,8,9,16,25,27,32,36....} then SUM 1/(x-1) = 1 and SUM 1/(x+1) = pi^2 / 3 - 5/2 where the sums are over x in S.

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[math-fun] more fun with decimals
by Warren Smith 29 Jan '12

29 Jan '12

Jorg Arndt's example generating Pascal's triangle and a Fibonacci-like sequence was considerably cooler... another I invented quite a while back was 1000 / (500 + sqrt(249990)) = 1.00001000020000500014000420013200429014300... where the sequence 1,2,5,14,42,... is the Catalan numbers (which easily give you the central binomial coefficients). How did I do that? Well, again, one way is using generating functions. But the continued fraction of quadratic irrationals is ultimately periodic and once you know its period you can easily solve for the surd, so again, this is easy to construct from the desired sequence without any intelligence.

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Re: [math-fun] math-fun Digest, Vol 107, Issue 38
by Warren Smith 29 Jan '12

29 Jan '12

>Andy Latto: But without the "silly" generating functions, do we have any explanation of why if we take the continued fraction expansion of 1.004009016025036049064081100, and truncate before the first large entry, we end up with a decimal that continues with 121144169196225256289324361... Surely this is not coincidence! Continued fractions give us an efficient way to find the smallest-denominatored fraction that starts with the first ten three-digit squares, but unless I'm missing something, they give no insight as to why this expansion should continue with the next 10 3-digit squares, and the generating function approach does. --well, true, but this should be obvious :) Of course the only integer sequences one is going to be able to generate and keep going for a long time are those with rational functions of x as generating functions. But if you have any particular such sequence and want to know the P/Q, the easiest fastest way to find P and Q is using the continued fraction. So no: they*do* give an insight as to why this expansion should continue with the next 10 3-digit squares, which is: via the continued fraction, you have FOUND the generating function of x, just evaluated at x=1000 (in the present case).

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Re: [math-fun] Fun with math:
by Phil Carmody 28 Jan '12

28 Jan '12

From: David Wilson <davidwwilson(a)comcast.net> > What is the smallest value a(d) such that every d-digit sequence occurs > in the decimal expansion of 1/d (after the decimal point). > > For small d >= 1 my program says > > 17, 109, 1019, 10007, 100019, ... > > but it seems stuck on 6 digits, so perhaps there is an overflow issue. 1000171 ought to work. I cannot convince myself that the smallest /unique prime/ > 10^d is the minimal value that satisfies the condition, but it certainly matches the 5 values you've found so far. Phil

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[math-fun] Dirichlet's theorem on primes in arithmetic progressions
by Dan Asimov 28 Jan '12

28 Jan '12

Dirichlet's theorem on primes in arithmetic progressions says that for any sequence s_N := aN + b, with a,b in Z+ with GCD(a,b) = 1, there exists an infinite number of primes s_N. I'm reading detailed versions of his proof, circa 1837. I'm deeply impressed by the complexity of this proof, making use of Fourier analysis on finite abelian groups as well as the theory of (analytic) functions. The various modern versions I'm reading seem to agree almost exactly on the reasoning involved, so I presume that there isn't much to modernize. Question: Is it really true that there isn't an "elementary" proof that avoids analytic functions? It appears that at least such do exist for a number of specific arithmetic progressions, like 6N+1 or 6N-1. Somewhat related question: Given a finite abelian group G, its group G^ of "characters" is the group of all homomorphisms Hom(G,R/Z) from G to the circle group R/Z (the group operation is pointwise multiplication in R/Z). It's easy to see that |G^| = |G|, but to see that these two groups are isomorphic seems to be proved using the classification theorem for finite abelian groups. Is there a proof that G^ is isomorphic to G that doesn't rely on this classification theorem? --Dan ________________________________________________________________________________________ It goes without saying that .

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Re: [math-fun] Gosperian PaIn physics?
by Bill Gosper 28 Jan '12

28 Jan '12

I used to nonrigorously crank out conjectural identities (doubtless all now conquered by holonomics) by closing rectangular contours with non-integer sides, using the standard trick Prod(M,a,b) := Prod(M,a,oo).Prod(M,b+1,oo)^-1 . --rwg DWilson> I would wonder if a grid of matrices with path invariance could be generalized to a continuous surface of matrices with path invariance? This would require some notion of matrix product along a continuous path. If this were possible, perhaps we could formulate some smooth generalization for multinomial expressions and continued fractions. Better at imagining these things than doing them.

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