A238490 -id:A238490 - cp's OEIS frontend

A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3(ny)^2 = 1.

1, 2, 3, 2, 3, 6, 4, 4, 9, 6, 5, 6, 6, 4, 3, 8, 9, 18, 5, 6, 12, 10, 11, 12, 15, 6, 27, 4, 15, 6, 16, 16, 15, 18, 12, 18, 18, 10, 6, 12, 7, 12, 11, 10, 9, 22, 23, 24, 28, 30, 9, 6, 9, 54, 15, 4, 15, 30, 29, 6, 30, 16, 36, 32, 6, 30, 17, 18, 33, 12, 7, 36, 18, 18, 15

Offset: 1

A.H.M. Smeets, Jan 15 2018

nonn

The fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1 is the smallest solution of x^2 - 3*y^2 = 1 satisfying y == 0 (mod n).

For primes p > 2, 2^p-1 is a Mersenne prime if and only if a(2^p-1) = 2^(p-1). For example, a(7) = 4, a(31) = 16, a(127) = 64, but a(2047) = 495 < 1024. - Jianing Song, Jun 02 2022

Michael J. Jacobson, Jr. and Hugh C. Williams, Solving the Pell Equation, Springer, 2009, pages 1-17.

A.H.M. Smeets, Table of n, a(n) for n = 1..20000
H. W. Lenstra Jr., Solving the Pell Equation, Notices of the AMS, Vol.49, No.2, Feb. 2002, p.182-192.

Cf. A091338, A298210, A298212.

With[{s = Array[ChebyshevU[-1 + #, 2] &, 75]}, Table[FirstPosition[s, k_ /; Divisible[k, n]][[1]], {n, Length@ s}]] (* Michael De Vlieger, Jan 15 2018, after Eric W. Weisstein at A001353 *)

xf, yf = 2, 1
x, n = 2*xf, 0
while n < 20000:
    n = n+1
    y1, y0, i = 0, yf, 1
    while y0%n != 0:
        y1, y0, i = y0, x*y0-y1, i+1
    print(n, i)

a(n) <= n.
a(A038754(n)) = A038754(n).
A001075(a(n)) = A002350(3*n^2).
A001353(a(n)) = A002349(3*n^2).
if n | m then a(n) | a(m).
a(3^m) = 3^m  and a(2*3^m) = 2*3^m for m>=0.
In general: if p is prime and p == 3 (mod 4) then: a(n) = n iff n = p^m or n = 2*p^m, for m>=0.
a(k*A005385(n)) = a(k)*A005384(n) for n>2 and k > 0 (conjectured).
a(p) | (p-A091338(p)) for p is an odd prime. - A.H.M. Smeets, Aug 02 2018
From Jianing Song, Jun 02 2022: (Start)
a(p) | (p-A091338(p))/2 for p is an odd prime > 3.
a(p^e) = a(p)*p^(e-r) for e >= r, where r is the largest number such that a(p^r) = a(p). r can be greater than 1, for p = 2, 103, 2297860813 (Cf. A238490).
If gcd(m,n) = 1, then a(m*n) = lcm(a(m),a(n)). (End)

A306825 Primitive part of A001353(n).

Original entry on oeis.org

1, 4, 15, 14, 209, 13, 2911, 194, 2703, 181, 564719, 193, 7865521, 2521, 34945, 37634, 1525870529, 2701, 21252634831, 37441, 6779137, 489061, 4122901604639, 37633, 274758906449, 6811741, 19726764303, 7263361, 11140078609864049, 40321, 155161278879431551

Offset: 1

Jianing Song, Mar 16 2019

nonn

A prime p is called a unique-period prime in base b if there is no other prime q such that the period length of 1/q is equal to that of 1/p. If q = a(2p) = A001353(2*p)/(4*A001353(p)) = ((2 + sqrt(3))^p + (2 - sqrt(3))^p)/4 is prime (this happens for p = 3, 5, 7, 11, 13, 17, 19, 79, 151, 199, 233, 251, 317, ...), where p is an odd prime, then q is a unique-period prime in base b = (sqrt(12*q^2 - 3) - 1)/2 (1/q has period length 3) as well as in base b' = (sqrt(12*q^2 - 3) + 1)/2 (1/q has period length 6). For example, a(6) = 13 is prime, so 13 is the only prime whose reciprocal has period length 3 in base 22 and the only prime whose reciprocal has period length 6 in base 23. Compare: If q = A000129(p) = A008555(p), then q is a unique-period prime in base b = sqrt(2*q^2 - 1) (1/q has period length 4).
By Lucas-Lehmer test, p is a Mersenne prime > 3 if and only if the smallest k such that p divides a(k) is k = (p - 1)/2.
For primes p, p^2 divides a(k) for some k if and only if p = 2 or p is in A238490. If p > 2, the only possible values for k are the divisors of (p - Legendre(3,p))/2 (e.g., 103^2 divides a(52) = 53028360515521 = 103^2 * 4998431569).
Conjecturally there must be infinitely many primes p such that a(p) is prime, but no such p is known. [By the formula below, there is no such p. - Jianing Song, Oct 31 2024]

			For n = 8 we have: a(1) = A001353(1), a(1)*a(2) = A001353(2), a(1)*a(2)*a(4) = A001353(4), a(1)*a(2)*a(4)*a(8) = A001353(8). The solution is a(1) = 1, a(2) = 4, a(4) = 14 and a(8) = 194.

Eric Weisstein's World of Mathematics, Sylvester Cyclotomic Number
Wikipedia, Lucas Lehmer Primality Test

Cf. A001353, A238490, A309040.

Similar sequences: A061446, A008555.

b(n) = if(n==1, [1], my(v=vector(n)); v[1]=1; v[2]=4; for(i=3, n, v[i]=4*v[i-1]-v[i-2]); v)
a(n) = my(d=divisors(n)); prod(i=1, #d, (b(n)[d[i]])^moebius(n/d[i]))

Product_{d|n} a(d) = A001353(n), that is, a(n) = A001353(n)/(Product_{dA001353(d)^mu(n/d), where mu = A008683.

a(n) = A309040(2*n) for even n; A309040(n)*A309040(2*n) for odd n > 1. - Jianing Song, Oct 31 2024

A320161 Irregular triangle read by rows: row n lists 0 <= k < p^2 such that p^2 divides A316269(k, p-Kronecker(k^2-4, p)), p = prime(n).

Original entry on oeis.org

0, 1, 3, 0, 1, 8, 0, 1, 24, 0, 1, 10, 39, 48, 0, 1, 27, 36, 37, 84, 85, 94, 120, 0, 1, 6, 29, 34, 61, 108, 135, 140, 163, 168, 0, 1, 25, 45, 56, 75, 82, 132, 157, 207, 214, 233, 244, 264, 288, 0, 1, 42, 43, 73, 88, 106, 120, 161, 200, 241, 255, 273, 288, 318, 319, 360

Offset: 1

Jianing Song, Oct 06 2018

p always divides A316269(k, p-Kronecker(k^2-4, p)), so it's interesting to see when p^2 also divides A316269(k, p-Kronecker(k^2-4, p)).
In the following comments, let p = prime(n). Note that A316269(0, m) and A316269(1, m) is not defined, so here k must be understood as a remainder modulo p^2. because A316269(k+s*p^2, m) == A316269(k, m) (mod p^2).
Let p = prime(n). Every row contains 0, 1 and p^2 - 1. For n >= 3, the n-th row contains p - 2 numbers, whose remainders modulo p form a permutation of {0, 1, 3, 4, ..., p - 3, p - 1}.
Every row is antisymmetric, that is, k is a member iff p^2 - k is, k > 0. As a result, the sum of the n-th row is prime(n)^2*(prime(n) - 3)/2.
Equivalently, for n >= 2, row n lists 0 <= k < p^2 such that p^2 divides A316269(k, (p-Kronecker(k^2-4, p))/2), p = prime(n).

			Table starts
p = 2: 0, 1, 3,
p = 3: 0, 1, 8,
p = 5: 0, 1, 24,
p = 7: 0, 1, 10, 39, 48,
p = 11: 0, 1, 27, 36, 37, 84, 85, 94, 120,
p = 13: 0, 1, 6, 29, 34, 61, 108, 135, 140, 163, 168,
p = 17: 0, 1, 25, 45, 56, 75, 82, 132, 157, 207, 214, 233, 244, 264, 288,
p = 19: 0, 1, 42, 43, 73, 88, 106, 120, 161, 200, 241, 255, 273, 288, 318, 319, 360,
p = 23: 0, 1, 12, 15, 60, 86, 105, 141, 142, 156, 223, 306, 373, 387, 388, 424, 443, 469, 514, 517, 528,
p = 29: 0, 1, 42, 46, 80, 101, 107, 120, 226, 227, 327, 330, 358, 409, 432, 483, 511, 514, 614, 615, 721, 734, 740, 761, 795, 799, 840,
...

Jianing Song, Table of n, a(n) for n = 1..29083 (primes below 600)
Jianing Song, Table of n, a(n) for n = 1..75796 (primes below 1000)
Wikipedia, Wall-Sun-Sun prime

Cf. A143548, A316269, A320162 (discriminant k^2+4, a more studied case).

Cf. A238490 (primes p such that 4 occurs in the corresponding row), A238736 (primes p such that 6 occurs in the corresponding row).

B(k, p) = (([k, -1; 1, 0]^(p-kronecker(k^2-4,p)))[1,2])%(p^2)
forprime(p=2, 50, for(k=0, p^2-1, if(!B(k, p), print1(k, ", ")));print)

cp's OEIS Frontend

A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3(ny)^2 = 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Python

Formula

A306825 Primitive part of A001353(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A320161 Irregular triangle read by rows: row n lists 0 <= k < p^2 such that p^2 divides A316269(k, p-Kronecker(k^2-4, p)), p = prime(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

cp's OEIS Frontend

A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Python

Formula

A306825 Primitive part of A001353(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A320161 Irregular triangle read by rows: row n lists 0 <= k < p^2 such that p^2 divides A316269(k, p-Kronecker(k^2-4, p)), p = prime(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3(ny)^2 = 1.