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#31 by N. J. A. Sloane at Sun Jun 28 10:34:22 EDT 2020
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#30 by N. J. A. Sloane at Sun Jun 28 10:34:20 EDT 2020
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#29 by N. J. A. Sloane at Wed Jun 24 22:31:37 EDT 2020
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#28 by Michel Marcus at Fri Jun 05 12:11:15 EDT 2020
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#27 by Michel Marcus at Fri Jun 05 12:10:59 EDT 2020
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upto(n) = res=List(); r=0; forstep(i=n, 1, -1, c= mindiff(2*i) / (2*i); if(c>r, r=c; listput(res, i))); Vecrev(Vec(res)) \\ using Michels mindiff \\ _)) \\ _David A. Corneth_, Jun 02 2020
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#26 by Michel Marcus at Thu Jun 04 01:52:37 EDT 2020
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#25 by Michel Marcus at Thu Jun 04 01:52:27 EDT 2020
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Integers k such that for all m>k, d(m)/m< < d(k)/k where d(j) = Min_{p & q odd primes, 2*j = p+q, p <= q} (q-p)/2.
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#24 by Michel Marcus at Tue Jun 02 04:05:28 EDT 2020
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Discussion
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| Tue Jun 02
| 04:07
| Michel Marcus: David, if you feel like adding code for A335298 ...
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| 04:15
| David A. Corneth: I had a brief look but I don't get that one. Can one of you adapt my code? Might be similar, it looks like an inequality sign switched somewhere. But then again I don't get it so I could be wrong.
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| 04:36
| David A. Corneth: Ya-Ping Lu, if you want to translate this to python, it works as follows. Suppose you like to search until U. Then start at i = U. Compute d(i). If it's a record so far, store i and the value of the record. Then check i = U-1. If it's a new record, store i and update the record. This all the way down to i = 1.
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| 04:49
| Ya-Ping Lu: Thanks, David. I'll give it a try. Please let me know or post the longer list of the numbers for A335298 here when you finish running the code for A335298.
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#23 by Michel Marcus at Tue Jun 02 04:04:48 EDT 2020
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lista(nn) = {my(v=vector(nn, k, k = 2*k; mindiff(k)/k)); for (n=1, #v-1, if (v[n] > vecmax(vector(#v-n, k, v[k+n])), print1(n, ", ")); ); } \\ Michel Marcus, Jun 01 2020
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Discussion
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| Tue Jun 02
| 04:05
| Michel Marcus: was really too slow
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#22 by David A. Corneth at Tue Jun 02 04:00:41 EDT 2020
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