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#22 by Susanna Cuyler at Fri Mar 29 21:06:59 EDT 2019
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#21 by Michel Marcus at Fri Mar 29 13:58:05 EDT 2019
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#20 by Robert Price at Fri Mar 29 13:28:58 EDT 2019
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#19 by Robert Price at Fri Mar 29 13:28:55 EDT 2019
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| COMMENTS
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a(49) > 143800. - Robert Price, Mar 29 2019
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| MATHEMATICA
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A126659[n_] := Module[{k = 1}, If[n == 14, Return[0]]; While[! PrimeQ[((2 n - 1)^k + 1)/(2 n)], k++]; k]; Join[Table[A126659[n], {n, 2, 48}]] (* _}] (* _Robert Price_, Oct 29 2018 *)
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| STATUS
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approved
editing
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#18 by N. J. A. Sloane at Sat Dec 01 08:02:14 EST 2018
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#17 by Michel Marcus at Fri Nov 30 01:46:25 EST 2018
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#16 by Michel Marcus at Fri Nov 30 01:46:17 EST 2018
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| PROG
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(PARI) a(n) = {if ((p=ispower(2* (n-1)) && (p>2), ==14, return(0)); my(k=3); while (! isprime(((2*n-1)^k + 1)/(2*n)), k = nextprime(k+1)); k; } \\ Michel Marcus, Nov 23 2018
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| STATUS
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proposed
editing
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#15 by Michel Marcus at Fri Nov 23 04:00:23 EST 2018
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Discussion
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| Fri Nov 30
| 01:45
| Michel Marcus: Robert warned me that this is not ok for n=41
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#14 by Michel Marcus at Fri Nov 23 03:59:37 EST 2018
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| PROG
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(PARI) a(n) = {if ((p=ispower(2*n-1)) && (p>2), return(0)); my(k=3); while (! isprime(((2*n-1)^k + 1)/(2*n)), k = nextprime(k+1)); k; } \\ Michel Marcus, Nov 23 2018
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| STATUS
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approved
editing
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Discussion
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| Fri Nov 23
| 04:00
| Michel Marcus: I think rather (p=ispower(2*n-1)) && (p>2) than n=14 : See 1st comment of A084742.
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#13 by Susanna Cuyler at Tue Oct 30 20:19:30 EDT 2018
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