January 2021 - mathfun@mailman.xmission.com

[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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[math-fun] Topology of the Lorenz System
by Brad Klee 22 Jan '21

22 Jan '21

A question was asked whether or not the "template" of the Lorenz strange attractor could have the same topology of the YGB surface, and I said, for typical parameters, the answer is No. See Figure 3 of: https://arxiv.org/pdf/1610.07079.pdf Cut along the midline and glue together the two loose ends, then you have a mobius strip. Repeat: When self-intersecting a mobius strip on an edge, there is a choice between two alternatives. The Lorenz template features an intersection with a point-like join between two cycles. The YGB surface involves a line-like join between two cycles (Try drawing graphs with two T-nodes and two cycles). Is there a chaotic system of ODE's whose attractor template is topologically equivalent to the YGB surface? The other issue is almost-periodic analysis of the Lorenz system, when considered as two joined oscillation disks. Has anything been written on Period Functions? Does this question sound interesting to anyone else? --Brad

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Re: [math-fun] September 10000, 1993
by Keith F. Lynch 22 Jan '21

22 Jan '21

M F Hasler <mhasler(a)dsi972.fr> wrote: > Except for the fact that there was certainly no month more that > 10^20 (or so) years ago, because the universe didn't exist yet > and/or more specifically, time as we define it doesn't reach that > far into the past. Yes, as I mentioned. The Big Bang was a mere 10^10.14 years ago. Whether time existed before that, nobody knows. But this is math, not physics or history. Numbers go on forever. Both Earth and the known universe were around 10^12 days ago, but not 10^13. 33.0% of all days in a Gregorian period (400 years or 146097 days) are the 10^Nth day of some month in the past 10^12 days. That's just under a third. The Gregorian calendar was in use 10^5 days ago (albeit only in Catholic countries), but not 10^6. 18.8% of all days in a Gregorian period are the 10^Nth day of some month in the past 10^5 days. > But yes, the answer to the original question relies only on residues > (mod 7) of powers of 10 , so it is essentially the same for any > calender that has weeks of seven days and no exceptions to that. It > is quite obvious that one particular value will never appear as such > a residue, Right, since 10 and 7 are relatively prime. > slightly less obvious that all others will (and do so for N < 7 ? > incidentally corresponding to roughly the age of mankind). Only because the decimal expansion of 1/7 repeats after 6 digits. Had I tried a different base, the expansion could repeat after any of 1, 2, 3, or 6 digits. If not 6, then several days of the week would never appear. And it's not obvious to me that if I were to re-run my second question for other bases that I'd get a solution for every day, e.g. that every day is the 11^Nth day of some prior month (possibly in the absurdly distant past) for some N. Here's how many of the 146097 days in a Gregorian period are the 10^Nth day of some month for each N, not counting days for which a smaller N applies. There are 4800 for N=0 since every month has a first day, and there's obviously never any previous month with a smaller N. There are 4800 for N=1 since every month has a tenth day, which is of course never also its first day. But there are only 4097 for N=2 since some 100th days of a month land on the 10th day of subsequent month. The smallest N that doesn't appear is 75. The largest N that does appear is 306, for just 9 out of every 146097 days. The largest N with 4800 is 9; the billionth day of each month somehow manages to never land on any 1st, 10th, 100th, 1000th, ... or hundred millionth day of any month. 4800 0 4800 1 4097 2 4800 3 4800 4 4198 5 3234 6 3247 7 4466 8 4800 9 1316 10 2322 11 1354 12 4264 13 4632 14 2834 15 2906 16 3565 17 3246 18 2807 19 2169 20 1768 21 1881 22 2949 23 3752 24 799 25 551 26 259 27 4466 28 989 29 1896 30 753 31 1861 32 1144 33 1827 34 2505 35 1505 36 1579 37 1482 38 2345 39 982 40 721 41 1882 42 856 43 1440 44 1622 45 1546 46 560 47 1641 48 843 49 405 50 1169 51 975 52 619 53 714 54 1099 55 1003 56 460 57 137 58 777 59 257 60 1149 61 905 62 748 63 287 64 124 65 39 66 632 67 74 68 397 69 1222 70 58 71 141 72 341 73 620 74 181 76 110 77 153 78 10 79 446 80 346 81 647 82 371 83 63 84 258 85 173 86 175 87 243 88 243 89 44 90 44 91 17 92 12 93 195 94 22 95 101 96 539 97 219 98 87 99 93 100 134 101 94 102 266 103 304 104 173 105 107 106 376 107 404 108 22 109 102 110 77 111 280 112 37 114 58 115 121 116 12 118 187 119 66 120 17 121 44 122 35 123 240 124 106 125 80 126 121 127 16 128 114 130 114 131 99 132 12 133 6 134 21 135 18 136 140 137 14 138 411 140 87 141 9 142 9 143 54 145 33 146 48 147 20 148 28 149 15 150 5 151 10 152 7 154 2 155 6 156 307 159 2 161 189 162 7 163 12 164 15 165 48 166 10 169 9 170 9 172 81 176 82 177 23 178 25 181 2 182 28 183 26 184 9 186 5 187 16 188 33 190 17 197 16 199 4 202 2 204 35 206 23 213 12 216 17 219 3 222 2 223 13 225 33 227 3 229 1 238 1 261 2 280 9 306 A hundred days ago was last October. A thousand days ago was in April of 2018. Ten thousand days ago was in September 1993. A hundred thousand days ago was in April 1747, less than six years before Britain and its colonies adopted the Gregorian calendar. A million days ago was in February of 718 BCE, 35 years after Rome's founding. Ten million days ago was in the middle of the last glacial period, long before recorded history. A hundred million days ago was early in the third from last glacial period. A billion days ago was during the Pliocene epoch, before our species existed. Ten billion days ago was during the Oligocene epoch, early in the age of mammals. A hundred billion days ago was during the Permian period, the last part of the Paleozoic era, before mammals, or even dinosaurs, existed, but after the great coal forests. A trillion days ago was during the late Archean or early Proterozoic eon, when the only life was single cells, and there was little or no oxygen in the atmosphere. Ten trillion days ago was twice as long ago as the Big Bang. But there are isotopes that will last much longer than that, stars that will still be shining longer than that, and computer programs that would take much longer than that to finish.

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[math-fun] Alternatives to Gram-Schmidt
by Henry Baker 21 Jan '21

21 Jan '21

Gram-Schmidt takes a set of vectors, puts them into some arbitrary order, and then "orthogonalizes" them by subtracting off non-orthogonal components. The problem with G-S is that there seems to be little rhyme or reason for the initial sort order. I was curious about trying to *minimize* the *length* of the *description* of the final set of orthogonal vectors. By *description*, I mean the total length in bits of all of the numerators and denominators. My first attempt is a "greedy" algorithm that find the 2 vectors which are at the least orthogonal (greatest |angle-90|) from one another, replace them with the sum and difference of their normalizations, and then rinse and repeat. Note that the sum and difference of same-length vectors are orthogonal. Thus, if one starts with N independent vectors, then it appears that this "greedy" algorithm converges on a mutually orthogonal set. While the convergence isn't fast, it does appear to keep the total description lengths from exploding. The convergence rate appears to be similar to Jacobi/Givens. I would imagine that this algorithm would have interesting behavior when trying to mutually orthogonalize N+1 vectors in an N-dimensional space. It would attempt to find a distribute the vectors over the surface of the N-dimensional unit hypersphere with the arrow heads as far from one another as possible. Ditto for N+2, N+3, etc., vectors.

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