A146211 -id:A146211 - cp's OEIS frontend

A180511 Fermat quotients for base 4: (4^(p - 1) - 1)/p, where p = prime(n).

5, 51, 585, 95325, 1290555, 252645135, 3616814565, 764877654105, 2484744621997515, 37191016277640225, 127631526564044465235, 29485995600356809139175, 449832863112420158030205

Offset: 2

Jani Melik, Jan 20 2011

nonn

Cf. A007663, A146211.

A180511:= n-> map (p-> (4^(p-1)-1)/p, ithprime(n)):
seq (A180511(n), n=2..20);  # Peter Luschny, Jan 21 2010

Table[(4^(Prime[n] - 1) - 1)/Prime[n], {n, 2, 20}] (* Alonso del Arte, Jan 20 2011 *)

A238490 Odd primes p that divide a Lucas quotient studied by H. C. Williams: A001353(p - (3/p))/p, where (3/p) is a Jacobi symbol.

Original entry on oeis.org

103, 2297860813

Offset: 1

John Blythe Dobson, Mar 28 2014

The condition for an odd prime p to be a member of this sequence is that p^2 divides A001353(p - (3/p)).
Neither this quotient, nor the Lucas sequence U(4, 1) on which it is based, has a common name; but its fundamental discriminant of 3 places it between the quotient based on the Pell sequence U(2, -1) with discriminant 2 (A000129), and that based on the Fibonacci sequence U(1, -1) with discriminant 5 (A000045). Values of p dividing the Pell quotient will be found under A238736, while for the Fibonacci quotient it is known that there is no such p < 9.7*10^14.
The interest in this family of number-theoretic quotients derives from H. C. Williams, "Some formulas concerning the fundamental unit of a real quadratic field," p. 440, which proves a formula connecting the present quotient with the Fermat quotient base 2 (A007663), the Fermat quotient base 3 (A146211), and the harmonic number H(floor(p/12)) (see the Formula section below). As is well known, the vanishing of each of these Fermat quotients is a necessary condition for the failure of the first case of Fermat's Last Theorem (see discussions under A001220 and A014127); and a corresponding result concerning this type of harmonic number was proved by Dilcher and Skula. Thus, the vanishing mod p of the quotient based on U(4, 1) is also a necessary condition for the failure of the first case of Fermat's Last Theorem.
The pioneering computation for this quotient appears to be that of Elsenhans and Jahnel, "The Fibonacci sequence modulo p^2," p. 5, who report 103 as the only value of a(n) < 10^9. Extending the search to p < 2.5*10^10 has found only one further solution, 2297860813.
Let LucasQuotient(p) = A001353(p - (3/p))/p, q_2 = (2^(p-1) - 1)/p = A007663(p) be the corresponding Fermat quotient of base 2, q_3 = (3^(p-1) - 1)/p = A146211(p) be the corresponding Fermat quotient of base 3, H(floor(p/12)) be a harmonic number. Then Williams (1991) shows that 6*(3/p)*LucasQuotient(p) == -6*q_2 - 3*q_3 - 2*H(floor(p/12)) (mod p).
Also with an initial 2, primes p such that p^2 divides A001353(p - Kronecker(12,p)) (note that 12 is the discriminant of the characteristic polynomial of A001353, x^2 - 4x + 1). - Jianing Song, Jul 28 2018

			LucasQuotient(103) = 103*851367555454046677501642274766916900879231854719584128208.

John Blythe Dobson, Table of n, a(n) for n = 1..2
Karl Dilcher and Ladislav Skula, A new criterion for the first case of Fermat's Last Theorem, Mathematics of Computation, 64 (1995) 363-392.
Andreas-Stephan Elsenhans and Jörg Jahnel, The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 1014, arxiv 1006.0824 [math.NT], 2010.
H. C. Williams, Some formulas concerning the fundamental unit of a real quadratic field, Discrete Mathematics, 92 (1991), 431-440.

Cf. A001353, A238736.

The following criteria are equivalent:
PrimeQ[p] &&
  Mod[(MatrixPower[{{1,2},{1,3}}, p-JacobiSymbol[3,p]-1].{{1},{1}})[[2,1]], p^2]==0
PrimeQ[p] && Mod[Last[LinearRecurrence[{4,-1},{0,1}, p-JacobiSymbol[3,p]+1]], p^2]==0

isprime(p) && (Mod([2, 2; 1, 0], p^2)^(p-kronecker(3, p)))[2, 1]==0 \\ This test, which was used to find the second member of this sequence, is based on the test for A238736 devised by Charles R Greathouse IV

A213327 Analog of Fermat quotients: a(n) = (round((phi_2)^p)-2)/p, where phi_2 is silver ratio 1+sqrt(2) and p = prime(n).

Original entry on oeis.org

2, 4, 16, 68, 1476, 7280, 189120, 986244, 27676612, 4346071600, 23696518916, 3930960079760, 120508933265760, 669708812842692, 20814182249890948, 3654563002853231440, 650000099752136709444, 3664265812073801505200, 660535426260570501876228

Offset: 1

Vladimir Shevelev and Peter J. C. Moses, Mar 03 2013

nonn

For similar sequence for base 2, see A007663, for a similar sequence for golden ratio, see A064723.

Cf. A007663, A096060, A146211, A180511, A164740, A222207, A064723, A213328.

A213328 Analog of Fermat quotients: a(n) = (round((phi_3)^p)-3)/p, where phi_3 = (3+sqrt(13))/2 and p = prime(n).

Original entry on oeis.org

4, 11, 78, 612, 46374, 428040, 38948910, 380144556, 37367223558, 38467601033550, 392545092308724, 426897839167539480, 45841425452161683630, 476794964068892779068, 51906117696097060014342, 59746844088106673671809870, 69664778857791165966384195366

Offset: 1

Vladimir Shevelev and Peter J. C. Moses, Mar 03 2013

nonn

For similar sequence for base 2, see A007663. For similar sequences for golden ratio and silver ratio, see A064723 and A213327. Note that phi_3 is called "bronze ratio".

Cf. A007663, A096060, A146211, A180511, A164740, A222207, A064723, A213327.

A364883 Consider the Fermat quotient for base n: Fq(n,k) = (n^(p - 1) - 1)/p, where p = prime(k), for k >= 1. a(n) is the least k >= 1 such that Fq(n,j) is divisible by n^2 - 1 for all j >= k.

Original entry on oeis.org

3, 3, 4, 4, 5, 5, 5, 4, 6, 6, 7, 7, 7, 5, 8, 8, 9, 9, 9, 6, 10, 10, 10, 7, 7, 7, 11, 11, 12, 12, 12, 8, 8, 8, 13, 13, 13, 9, 14, 14, 15, 15, 15, 10, 16, 16, 16, 5, 8, 8, 17, 17, 17, 6, 9, 11, 18, 18, 19, 19, 19, 12, 7, 7, 20, 20, 20, 10, 21, 21, 22, 22, 22, 13, 9, 9, 23, 23, 23

Offset: 2

Robert G. Wilson v, Aug 17 2023

nonn

Conjecture: numbers appear in the sequence only a finite number of times. Terms appear in runs of length 1, 2, or 3, never more. The first time a term k appears is when the index is even. The terms appear for the first time in their natural order.

			For a(2), examine A007663 and notice that beginning with the second term, offset is 2, all terms are divisible by 3;
For a(3), examine A146211 and notice that beginning with the first term, offset is 3, all terms are divisible by 8;
For a(4), examine A180511 and notice that beginning with the third term, offset is 2, all terms are divisible by 15; etc.

Jean Bourgain, Kevin Ford, Sergei V. Konyagin, and Igor E. Shparlinski, On the Divisibility of Fermat Quotients, Michigan Mathematical Journal, Vol. 59 (Aug 2010), pp. 313-328.
Chris Caldwell, PrimePages, Fermat quotient.
nLab, Fermat quotient.
H. S. Vandiver, Fermat's Quotient and related arithmetic functions, Proceedings of the National Academy of Sciences of the United States of America, Vol. 31, 1945.
Eric Weisstein's World of Mathematics, Fermat Quotient.
Wikipedia, Fermat quotient.

Cf. A007663, A096060, A146211, A180511.

a[n_] := Block[{k = Floor[(1/2.3) n^(87/100) + 100]}, While[p = Prime@ k; PowerMod[n, p - 1, (n^2 - 1)*p] == 1, k--]; ++k]; Array[a, 79, 2]

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A180511 Fermat quotients for base 4: (4^(p - 1) - 1)/p, where p = prime(n).

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A238490 Odd primes p that divide a Lucas quotient studied by H. C. Williams: A001353(p - (3/p))/p, where (3/p) is a Jacobi symbol.

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A213327 Analog of Fermat quotients: a(n) = (round((phi_2)^p)-2)/p, where phi_2 is silver ratio 1+sqrt(2) and p = prime(n).

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A213328 Analog of Fermat quotients: a(n) = (round((phi_3)^p)-3)/p, where phi_3 = (3+sqrt(13))/2 and p = prime(n).

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A364883 Consider the Fermat quotient for base n: Fq(n,k) = (n^(p - 1) - 1)/p, where p = prime(k), for k >= 1. a(n) is the least k >= 1 such that Fq(n,j) is divisible by n^2 - 1 for all j >= k.

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