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[math-fun] Fractions in primorial base
by Keith F. Lynch 02 Feb '21

02 Feb '21

I calculated fractions in the primorial base. Consecutive places after the point represent the product of the first n primes, i.e. 1/2, 1/6, 1/30, 1/210, 1/2310, etc. I only go through the 15th prime, 47, so that every number will fit in 64 bits. Fractions with a square-free denominator all terminate. All other fractions do not. The expansion of 1/4 appears to be A102781. Other expansions (e.g. 1/8, 1/9, 1/25) aren't in OEIS. As usual, A through Z represent 10 through 35. When I ran off the end of the alphabet, I started over with the lowercase alphabet; a through z represent 36 through 61. Here are the first few integer reciprocals. For more fractions, and the code that generated them, see http://KeithLynch.net/primorial.txt 1/2: 0.1 1/3: 0.02 1/4: 0.01235689BEFIKLN... 1/5: 0.011 1/6: 0.01 1/7: 0.0042 1/8: 0.003529CE5LN9AAZ... 1/9: 0.003238B67JACDSV... 1/10: 0.003 1/11: 0.00251 1/12: 0.00235689BEFIKLN... 1/13: 0.002219 1/14: 0.0021 1/15: 0.002 1/16: 0.001614EGEAR4PQf... 1/17: 0.00153B8 1/18: 0.0014745CF9KOREF... 1/19: 0.0014078I 1/20: 0.00135689BEFIKLN...

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[math-fun] Question about compact convex polyhedra in 3-space
by Dan Asimov 02 Feb '21

02 Feb '21

Suppose P is a compact convex polyhedron in 3-space, topologically equivalent to the sphere, with finitely many convex planar faces each bounded by a convex polygon with finitely many straight sides. Then every point x of P that is not a vertex has a neighborhood that is isometric to a neighborhood in the plane. And each vertex is locally isometric to a neighborhood of the vertex of a cone, with total angle around the vertex* equal to some theta in the open interval (0, pi). For the jth vertex, call this number theta_j. If this cone is approximated by a smooth surface then its unit normals translated to the origin will cover an area of the unit sphere equal to 2pi - theta. So if we assign the number 2pi - theta_j to the jth vertex, the Gauss-Bonnet theorem tells us that Sum (2pi - theta_j) = 4pi or equivalently Sum theta_j = 2pi(V-2) if V is the number of vertices.** Now suppose we have a finite number of regions P_j in the plane, each bounded by a polygon with straight sides, and that we are given a pairing of all the edges with edges of equal length so that, if the resulting abstract surface has V vertices, E edges, and F faces, the following two conditions are satisfied: 1) V - E + F = 2 (The Euler characteristic of the resulting surface is 2, from which we know that it must be a sphere.) and 2) The total angle about vertex j is theta_j in the interval (0, pi), such that Sum theta_j = 2pi(V-2). Question: --------- Does there then necessarily exist a convex polyhedron in 3-space that is isometric to the resulting abstract surface? —Dan ————— * The total angle of a cone can be seen if you slit the cone from the base to the vertex and flatten it out, it will fill a certain fraction f of a disk. Then its total angle is 2pi*f. ** E.g., for a cube V = 8 and each theta_j = 3pi/2, and indeed 8(3pi/2) = 2pi(8-2).

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[math-fun] The only explicit formula for the Heighway Dragon function,
by Bill Gosper 01 Feb '21

01 Feb '21

as far as I know, is my old Fourier approximation <http://gosper.org/explicittwindragon.pdf>, The formula approaches exactitude as the summation ranges over integers ever farther from 0 . I wrote this as a piece of evidence supporting my pitch to Wolfram Inc to at long last support (optionally infinite) MatrixProducts. (ContinuedFactionK would become a special case.) Product[m_?MatrixQ,...] unfortunately does elementwise rather than matrix multiplication. As I incessantly remind, Path Invariance <http://gosper.org/pathi.pdf> is a nice introduction to (2⨉2) iterated matrix products. —rwg

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[math-fun] Reciprocals of integers in binary
by Keith F. Lynch 01 Feb '21

01 Feb '21

I wondered what reciprocals of integers looked like in binary, so I computed them. I noticed that lots of them had first and second halves of the repeating section that were complements of each other. For instance 1/97 is 0.000000101010001110100000111111010101110001011111.... That's 0.000000101010001110100000 111111010101110001011111.... For which numbers is this the case? It turns out to be precisely those in http://oeis.org/A014657 (except the first two) times any power of two (including zero). And I see why.

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[math-fun] Reciprocals of integers in binary
by Keith F. Lynch 31 Jan '21

31 Jan '21

Dan Asimov <asimov(a)msri.org> wrote: > Very interesting. Thank you. > Which leads me to wonder what reciprocal integers look like in > factorial base, where x ? (0, 1) is represented as > x = ? a_n / n! > where the sum is over n ? 2 with a_n ? Z+, 0 ? a_n < n, using the > greedy algorithm. I didn't understand most of that, due to -- as I've mentioned several times before -- non-ASCII characters being converted by this list's digest option into question marks, which makes lots of posts almost entirely incomprehensible. Here's my best guess as to what you were asking for. The first place after the point is 1/2!, the next place is 1/3!, the next place is 1/4!, etc. I only go to 20! since that's as far as 64-bit integers go. Also, my digits would soon run off the end of the alphabet. Note that in this base all rationals terminate. 1/2: 0.1 1/3: 0.02 1/4: 0.012 1/5: 0.0104 1/6: 0.01 1/7: 0.003206 1/8: 0.003 1/9: 0.00232 1/10: 0.0022 1/11: 0.002053140A 1/12: 0.002 1/13: 0.00141254850C 1/14: 0.001333 1/15: 0.0013 1/16: 0.00123 1/17: 0.001202368909270G 1/18: 0.00114 1/19: 0.001116209526BBD80I 1/20: 0.0011 Here's my code. It works right to left, placing all of 1/N in the 1/N! bin, then distributing as much of it as possible as far left as possible. #include <stdio.h> int main() { long long i,j,k,a[21],f[21]; char c; f[0] = 1; for (i=1;i<21;i++) f[i] = f[i-1]*i; for (i=2;i<21;i++) { for (j=0;j<21;j++) a[j] = 0; a[i] = f[i-1]; for (j=i-1;j;j--) { a[j] = a[j+1]/(j+1); a[j+1] %= (j+1); } for (j=0;j<21;j++) if (a[j]) k=j; printf("1/%lld: 0.",i); for (j=2;j<k+1;j++) { c = a[j]+48; if (c>57) c=a[j]+55; printf("%c",c); } printf("\n"); } return(0); } For my next trick I will calculate reciprocals of integers in negabinary (base -2). Remember, there are 110 kinds of people: Those who understand negabinary, and those who don't.

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Re: [math-fun] Reciprocals of integers in binary
by Keith F. Lynch 31 Jan '21

31 Jan '21

Since there was so much interest in reciprocals in factorial base, and since 19 wasn't very many, I decided to calculate *all* fractions between 0 and 1 exclusive whose numerators and denominators did not exceed 20. (Who am I kidding? I'm just using this as an excuse to procrastinate on debugging my negabinary code.) Note the countdown pattern in the last digit, and the countup pattern in the next to last digit, whenever the denominator is a prime. 1/2: 0.1 1/3: 0.02 2/3: 0.11 1/4: 0.012 3/4: 0.112 1/5: 0.0104 2/5: 0.0213 3/5: 0.1022 4/5: 0.1131 1/6: 0.01 5/6: 0.12 1/7: 0.003206 2/7: 0.012415 3/7: 0.022124 4/7: 0.101333 5/7: 0.111042 6/7: 0.120251 1/8: 0.003 3/8: 0.021 5/8: 0.103 7/8: 0.121 1/9: 0.00232 2/9: 0.01114 4/9: 0.02232 5/9: 0.10114 7/9: 0.11232 8/9: 0.12114 1/10: 0.0022 3/10: 0.0131 7/10: 0.1104 9/10: 0.1213 1/11: 0.002053140A 2/11: 0.0101462819 3/11: 0.0122424328 4/11: 0.0203355737 5/11: 0.0224317246 6/11: 0.1010250655 7/11: 0.1031212164 8/11: 0.1112143573 9/11: 0.1133105082 10/11: 0.1214036491 1/12: 0.002 5/12: 0.022 7/12: 0.102 11/12: 0.122 1/13: 0.00141254850C 2/13: 0.003325306A1B 3/13: 0.01124105542A 4/13: 0.013153613939 5/13: 0.021106362348 6/13: 0.023022120857 7/13: 0.100434669266 8/13: 0.102350427775 9/13: 0.110303176184 10/13: 0.112215734693 11/13: 0.1201314830A2 12/13: 0.1220442415B1 1/14: 0.001333 3/14: 0.011042 5/14: 0.020251 9/14: 0.103206 11/14: 0.112415 13/14: 0.122124 1/15: 0.0013 2/15: 0.0031 4/15: 0.0122 7/15: 0.0231 8/15: 0.1004 11/15: 0.1113 13/15: 0.1204 14/15: 0.1222 1/16: 0.00123 3/16: 0.01023 5/16: 0.01323 7/16: 0.02223 9/16: 0.10123 11/16: 0.11023 13/16: 0.11323 15/16: 0.12223 1/17: 0.001202368909270G 2/17: 0.0024047477154E1F 3/17: 0.010110326521762E 4/17: 0.01131270532A9D3D 5/17: 0.013015274136C54C 6/17: 0.020220652A430C5B 7/17: 0.02142323184C346A 8/17: 0.0231256106585B79 9/17: 0.1003311794648388 10/17: 0.102033558270AA97 11/17: 0.103236137079D2A6 12/17: 0.11044151598619B5 13/17: 0.11214408479241C4 14/17: 0.11334646359B68D3 15/17: 0.1210520423A790E2 16/17: 0.1222544211B3B7F1 1/18: 0.00114 5/18: 0.01232 7/18: 0.02114 11/18: 0.10232 13/18: 0.11114 17/18: 0.12232 1/19: 0.001116209526BBD80I 2/19: 0.002235418A5098AG1H 3/19: 0.00335462847775872G 4/19: 0.0110140379A1525F3F 5/19: 0.0121332474082E364E 6/19: 0.0132524569320B0E5D 7/19: 0.020411666358C7E56C 8/19: 0.022031075882A4BD7B 9/19: 0.0231502852A981948A 10/19: 0.1003065048135D6C99 11/19: 0.10142571423A3A43A8 12/19: 0.103045123764171BB7 13/19: 0.11020433318AD3F2C6 14/19: 0.1113235426B4B0CAD5 15/19: 0.11244275211B8CA1E4 16/19: 0.1201021616456979F3 17/19: 0.12122137106C4650G2 18/19: 0.1223405805962328H1 1/20: 0.0011 3/20: 0.0033 7/20: 0.0202 9/20: 0.0224 11/20: 0.1011 13/20: 0.1033 17/20: 0.1202 19/20: 0.1224 Here's my code. It's all mine, except for the small part I borrowed from Euclid to avoid duplicate fractions. #include <stdio.h> int main() { long long i,i2,j,k,n,n2,a[21],f[21]; char c; f[0] = 1; for (i=1;i<21;i++) f[i] = f[i-1]*i; for (i=2;i<21;i++) { for (n=1;n<i;n++) { i2 = i; n2 = n; while (i2 != n2) { if (i2 > n2) i2 -= n2; if (n2 > i2) n2 -= i2; } if (i2 == 1) { for (j=0;j<21;j++) a[j] = 0; a[i] = n*f[i-1]; for (j=i-1;j;j--) { a[j] = a[j+1]/(j+1); a[j+1] %= (j+1); } for (j=0;j<21;j++) if (a[j]) k=j; printf("%lld/%lld: 0.",n,i); for (j=2;j<k+1;j++) { c = a[j]+48; if (c>57) c=a[j]+55; printf("%c",c); } printf("\n"); } } printf("\n"); } return(0); }

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[math-fun] Fwd: Coeffts of modular forms of exceptional groups
by James Propp 29 Jan '21

29 Jan '21

Noam Elkies writes: ---------- Forwarded message --------- From: Elkies, Noam <elkies(a)math.harvard.edu> Date: Fri, Jan 29, 2021 at 1:48 PM Subject: Re: [math-fun] Coeffts of modular forms of exceptional groups To: James Propp <jamespropp(a)gmail.com> CC: Henry Cohn <Henry.Cohn(a)microsoft.com> Hi Jim, You write: > Anything to contribute to the discussion? > ---------- Forwarded message --------- > From: Neil Sloane <njasloane(a)gmail.com> > Date: Fri, Jan 29, 2021 at 1:16 PM > Subject: [math-fun] Coeffts of modular forms of exceptional groups > To: fun <math-fun(a)mailman.xmission.com> > In the Feb 2021 Notices of AMS, there is an article by Aaron Pollack about > modular forms associated with exceptional groups. The coeffts of the > modular form associated with E_8 are famously connected with the densest > sphere packing in 8-D (and the Theory Of Everything), so naturally I was > interested in this. The first reference in the article is to > Walter L. Baily, Jr., An Exceptional Arithmetic Group and its Eisenstein > Series, Annals of Mathematics , May, 1970, Vol. 91, No. 3 (May, 1970), pp. > 512-549 : https://www.jstor.org/stable/1970636 > and Baily's last theorem there shows that the Fourier coefficients of the > various Eisenstein series associated with this group (which is connected > with E_7) are rational numbers. But no numbers are given (this is the > Annals, after all). And that is just the first of 20 references. So what > are the values of these Fourier coefficients? > [ . . . ] That paper seems to concern not the E_7 *lattice* (whose theta series coefficients, though not as nice as those for E_8, are well known, and appear in Conway-Sloane = SPLAG), but the exceptional Lie *group* E_7 and its associated "exceptional tube domain". I actually wrote a joint paper with B.H.Gross on this: Noam D. Elkies, Benedict H. Gross: The exceptional cone and the Leech lattice,, International Math. Research Notices 1996 #14, 665-698. But this was almost half a lifetime ago, and anyhow the "tube domain" part was all Gross's, so you should probably ask him directly if you need more information in that direction. Our paper did give some actual numbers. NDE

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[math-fun] Coeffts of modular forms of exceptional groups
by Neil Sloane 29 Jan '21

29 Jan '21

In the Feb 2021 Notices of AMS, there is an article by Aaron Pollack about modular forms associated with exceptional groups. The coeffts of the modular form associated with E_8 are famously connected with the densest sphere packing in 8-D (and the Theory Of Everything), so naturally I was interested in this. The first reference in the article is to Walter L. Baily, Jr., An Exceptional Arithmetic Group and its Eisenstein Series, Annals of Mathematics , May, 1970, Vol. 91, No. 3 (May, 1970), pp. 512-549 : https://www.jstor.org/stable/1970636 and Baily's last theorem there shows that the Fourier coefficients of the various Eisenstein series associated with this group (which is connected with E_7) are rational numbers. But no numbers are given (this is the Annals, after all). And that is just the first of 20 references. So what are the values of these Fourier coefficients? I also looked at Baily's book "Intro. Lectures on Automorphic Forms" which has a chapter on Fourier coeffts, but again there were no numbers given. Do we have any experts on automorphic forms on the list? Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com

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[math-fun] Question about "central" binomial coefficients binomial(2n+1, n+1)
by Dan Asimov 27 Jan '21

27 Jan '21

The OEIS sequence for binomial(2n+1, n+1) = (2n+1)! / (n! (n+1)!) (<https://oeis.org/A001700>) begins with 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052, (0 ≤ n ≤ 25), for which binomial(2n+1, n+1) = is odd if and only if n is of the form 2^K - 1. Is this true for all n ≥ 0 ??? —Dan

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[math-fun] Chaos Continues... Another Missing OEIS Sequence?
by Brad Klee 27 Jan '21

27 Jan '21

Section 4 of this student paper has a nice tabulation of periodic orbits on the Rössler attractor: http://chaosbook.org/projects/Carroll/Carr12.pdf We probably would want to define a sequence: > a(n) counts the number of distinct cycles of winding > number n on the Rössler attractor with (a,b,c)=(.2,.2,5.7). For the numbers a(n), from n=1, I calculated that: a(n) = 1, 1, 2, 1, 2, 3, 4, 6, 8, 10, 17 . . . These were determined using a numerical algorithm with randomized searching across a linear Poincare section. If there are hidden cycles, which are difficult to reach by iterating Newton's method, I'm worried these numbers could be wrong. Does anyone else get the same or better than? --Brad

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