A372797
Smallest prime p such that the multiplicative order of 4 modulo p is 2*n, or 0 if no such prime exists.
Original entry on oeis.org
3, 17, 31, 73, 151, 433, 631, 337, 127, 241, 331, 601, 4421, 673, 3061, 257, 1429, 1657, 1103, 3121, 2143, 1321, 18539, 1777, 2351, 37441, 2971, 2857, 3191, 17401, 683, 15809, 17029, 9929, 38431, 1801, 11471, 63689, 49999, 13121, 17467, 21169, 83077, 25609, 5581, 5153, 26227
Offset: 1
In the following examples let () denote the reptend. The prime numbers themselves and the fractions are written out in decimal.
The base-4 expansion of 1/3 is 0.(1), so the reptend has length 1 = (3-1)/2. Also, the base-4 expansions of 1/3 = 0.(1) and 2/3 = 0.(2) have two cycles 1 and 2. 3 is the smallest such prime, so a(1) = 3.
The base-4 expansion of 1/17 is 0.(0033), so the reptend has length 4 = (17-1)/4. Also, the base-4 expansions of 1/17, 2/17, ..., 16/17 have four cycles 0033, 0132, 1023 and 1122. 17 is the smallest such prime, so a(2) = 17.
The base-4 expansion of 1/31 is 0.(00133), so the reptend has length 5 = (31-1)/6. Also, the base-4 expansions of 1/31, 2/31, ..., 30/31 have three cycles 00201, 01203, 02211, 03213, 11223 and 13233. 13 is the smallest such prime, so a(3) = 13.
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a[n_] := a[n] = For[p = 2, True, p = NextPrime[p], If[MultiplicativeOrder[4, p] == (p-1)/(2n), Return[p]]];
Table[Print[n, " ", a[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Nov 24 2024 *)
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a(n,{base=4}) = forprime(p=2, oo, if((base%p) && znorder(Mod(base,p)) == (p-1)/(n * if(issquare(base), 2, 1)), return(p)))
A372798
Smallest prime p such that the multiplicative order of 8 modulo p is n, or 0 if no such prime exists.
Original entry on oeis.org
3, 17, 13, 113, 251, 7, 1163, 89, 109, 431, 1013, 577, 4421, 953, 571, 257, 4523, 127, 15467, 3761, 3109, 7151, 18539, 73, 25301, 14327, 2971, 42953, 72269, 151, 683, 12641, 331, 2687, 42701, 5113, 18797, 1103, 8581, 13121, 172283, 631, 221021, 120737, 3061, 5153, 217517
Offset: 1
In the following examples let () denote the reptend. The prime numbers themselves and the fractions are written out in decimal.
The base-8 expansion of 1/3 is 0.(25), so the reptend has length 2 = (3-1)/1. Also, the base-8 expansions of 1/3 = 0.(25) and 2/3 = 0.(52) have only one cycle 25. 3 is the smallest such prime, so a(1) = 3.
The base-8 expansion of 1/17 is 0.(03607417), so the reptend has length 8 = (17-1)/2. Also, the base-8 expansions of 1/17, 2/17, ..., 16/17 have two cycles 03607417 and 13226455. 17 is the smallest such prime, so a(2) = 17.
The base-8 expansion of 1/13 is 0.(0473), so the reptend has length 4 = (13-1)/3. Also, the base-8 expansions of 1/13, 2/13, ..., 12/13 have three cycles 0473, 1166 and 2354. 13 is the smallest such prime, so a(3) = 13.
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a[n_] := a[n] = For[p = 2, True, p = NextPrime[p], If[MultiplicativeOrder[8, p] == (p-1)/n, Return[p]]];
Table[Print[n, " ", a[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Nov 24 2024 *)
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a(n,{base=8}) = forprime(p=2, oo, if((base%p) && znorder(Mod(base,p)) == (p-1)/(n * if(issquare(base), 2, 1)), return(p)))
A372800
Smallest prime p such that the multiplicative order of 16 modulo p is 2*n, or 0 if no such prime exists.
Original entry on oeis.org
3, 5, 31, 17, 151, 109, 631, 113, 127, 1181, 331, 433, 13963, 1709, 3331, 1217, 2687, 397, 1103, 241, 2143, 1013, 18539, 1777, 2351, 4421, 2971, 673, 3191, 3061, 683, 257, 58147, 1429, 38431, 1657, 11471, 22573, 49999, 3121, 17467, 33013, 252583, 1321, 23671, 51797, 26227, 4513
Offset: 1
In the following examples let () denote the reptend. The prime numbers themselves and the fractions are written out in decimal.
The base-16 expansion of 1/3 is 0.(5), so the reptend has length 1 = (3-1)/2. Also, the base-16 expansions of 1/3 = 0.(5) and 2/3 = 0.(A) have two cycles 5 and A. 3 is the smallest such prime, so a(1) = 3.
The base-16 expansion of 1/5 is 0.(3), so the reptend has length 1 = (5-1)/4. Also, the base-16 expansions of 1/5 = 0.(3), 2/5 = (0.6), 3/5 = 0.(9) and 4/5 = 0.(C) have four cycles 3, 6, 9 and A. 5 is the smallest such prime, so a(2) = 5.
The base-16 expansion of 1/31 is 0.(08421), so the reptend has length 5 = (31-1)/6. Also, the base-16 expansions of 1/31, 2/31, ..., 30/31 have six cycles 08421, 18C63, 294A5, 39CE7, 5AD6B and 7BDEF. 31 is the smallest such prime, so a(3) = 31.
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a[n_] := a[n] = For[p = 2, True, p = NextPrime[p], If[MultiplicativeOrder[16, p] == (p-1)/(2n), Return[p]]];
Table[Print[n, " ", a[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Nov 24 2024 *)
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a(n,{base=16}) = forprime(p=2, oo, if((base%p) && znorder(Mod(base,p)) == (p-1)/(n * if(issquare(base), 2, 1)), return(p)))
Showing 1-3 of 3 results.
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