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If the continued fraction of sqrt(A002145(n)) is [a_0; {a_1, a_2, ..., a_(k-1), a_k, a_(k-1), ..., a_1, 2*a_0}], where {} is the periodic part, let x/y = [a_0; a_1, a_2, ..., a_(k-1)], gcd(x, y) = 1, then a(n) = x and A306566(n) = y.

The smallest solution to x^2 - p*y^2 = 2*(-1)^((p+1)/4) for the first primes congruent to 3 modulo 4:

Cf. A002145, A306566 (y-values).

1, 3, 3, 13, 5, 39, 59, 7, 23, 221, 59, 9, 9, 477, 31, 2175, 103, 8807, 41571, 8005, 13, 2047, 2999, 127539, 527593, 15, 15, 2489, 1917, 373, 340551, 11759, 9409, 4109, 52778687, 801, 19, 137913, 113759383, 137, 16437, 12311, 21, 21, 15732537, 1275, 1729, 7204587, 305987, 67

a(n) exists for all n.

X = a(n)^2 - (-1)^((p+1)/4), Y = a(n)*A306566(n) gives the smallest solution to x^2 - p*y^2 = 1, p = A002145(n). As a result, all the positive solutions to x^2 - p*y^2 = 2*(-1)^((p+1)/4) are given by (x_n, y_n) where x_n + (y_n)*sqrt(p) = (a(n) + A306566(n)*sqrt(p))*(X + Y*sqrt(p))^n.

n | Equation | x_min | y_min

1 | x^2 - 3*y^2 = -2 | 1 | 1

2 | x^2 - 7*y^2 = +2 | 3 | 1

3 | x^2 - 11*y^2 = -2 | 3 | 1

4 | x^2 - 19*y^2 = -2 | 13 | 3

5 | x^2 - 23*y^2 = +2 | 5 | 1

6 | x^2 - 31*y^2 = +2 | 39 | 7

7 | x^2 - 43*y^2 = -2 | 59 | 9

8 | x^2 - 47*y^2 = +2 | 7 | 1

9 | x^2 - 59*y^2 = -2 | 23 | 3

(PARI) b(p) = if(isprime(p)&&p%4==3, x=1; while(!issquare((x^2 - 2*(-1)^((p+1)/4))/p), x++); x)

forprime(p=3, 500, if(p%4==3, print1(b(p), ", ")))

Similar sequences: A094048, A094049 (x^2 - A002144(n)*y^2 = -1); A306618, A306619 (2*x^2 - A002145(n)*y^2 = (-1)^((p+1)/4))).

allocated for Jianing Songx-value of the smallest solution to x^2 - p*y^2 = 2*(-1)^((p+1)/4), p = A002145(n).

1, 3, 3, 13, 5, 39, 59, 7, 23

1,2

Cf. A306566 (y-values).

allocated

nonn

Jianing Song, Mar 25 2019

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Square array, read by antidiagonals, of integers N such that one or both of 6N + 1 and 6N + 5 is not prime.

4, 5, 5, 9, 8, 9, 10, 12, 12, 10, 14, 15, 20, 15, 14, 15, 19, 23, 23, 19, 15, 19, 22, 31, 28, 31, 22, 19, 20, 26, 34, 36, 36, 34, 26, 20, 24, 29, 42, 41, 48, 41, 42, 29, 24, 25, 33, 45, 49, 53, 53, 49, 45, 33, 25, 29, 36, 53, 54, 65, 60, 65, 54, 53, 36, 29, 30, 40, 56, 62, 70, 72, 72, 70, 62, 56, 40, 30

1,1

This sequence without duplicates is the complement of A056956, the positive integers n for which 6n + 1 and 6n + 5 are "cousin" primes, primes with gap 4.

This square array, which sieves for prime pairs with gap 4, is calculated without explicit reference to primes.

The array consists of 2 X 2 blocks, for c >= 1, d >= 1, which are

6cd - c - d 6cd + c - d - 1

6cd - c + d - 1 6cd + c + d

or

(6c - 1)d - c (6c - 1)d + c - 1

(6c + 1)d - c - 1 (6c + 1)d + c

That is, rows of the array are, alternately, positive multiples of 6c - 1 with -c or c - 1 added, or positive multiples of 6c + 1 with -c - 1 or c added.

The array arises because factors of 6N + 1 or 6N + 5 have the form (6c +- 1) * (6d +- 1).

The worksheet linked below marks multiples of 6c - 1 or 6c + 1 with a dot and the numbers eliminated by the sieve by X. It finds which of the numbers N = 1 to 68 give rise to cousin primes.

More generally, primes with gaps 6k - 2 of the form 6N + 1 and 6N + (6k - 1) can be found by using a sieve array with rows that are positive multiples of 6c - 1 with -c or c - k added, or positive multiples of 6c + 1 with -c - k or c added.

S. Lampret, <a href="http://nntdm.net/papers/nntdm-20/NNTDM-20-3-54-60.pdf">Sieving 2m-Prime Pairs</a>, Notes on Number Theory and Discrete Mathematics 20, no. 3 (2014), 54-60.

Sally Myers Moite, <a href="/A306529/a306529.jpg">Worksheet, N = 1 to 68</a>

a(m,n) = 6*floor((m+1)/2)*floor((n+1)/2) + ((-1)^n)*floor((m+1)/2) + ((-1)^m)*floor((n+1)/2) - (m+n) mod 2, m,n >= 1.

Square array begins:

4, 5, 9, 10, 14, 15, 19, 20, 24, 25, ...

5, 8, 12, 15, 19, 22, 26, 29, 33, 36, ...

9, 12, 20, 23, 31, 34, 42, 45, 53, 56, ...

10, 15, 23, 28, 36, 41, 49, 54, 62, 67, ...

14, 19, 31, 36, 48, 53, 65, 70, 82, 87, ...

15, 22, 34, 41, 53, 60, 72, 79, 91, 98, ...

19, 26, 42, 49, 65, 72, 88, 95, 111, 118, ...

20, 29, 45, 54, 70, 79, 95, 104, 120, 129, ...

24, 33, 53, 62, 82, 91, 111, 120, 140, 149, ...

25, 36, 56, 67, 87, 98, 118, 129, 149, 160, ...

...

Note that, for example, the third row (or column) contains numbers 2 less than and 1 more than multiples of 11 = 6*2 - 1, and the eighth row contains numbers 5 less than and 4 more than multiples of 25 = 6*4 + 1.

(PARI) a(m, n) = 6*floor((m+1)/2)*floor((n+1)/2) + ((-1)^n)*floor((m+1)/2) + ((-1)^m)*floor((n+1)/2) - (m+n)%2;

matrix(7, 7, n, k, a(n, k))

The first row is A047208.

The diagonal is A062717, the numbers m for which 6*m + 1 is a perfect square.

Cf. A007310, A087679, A157834, A047847, A215918.

Similar sequences for twin primes are A323674, A067611, A002822.

nonn,tabl,easy,changed

recycled

Sally Myers Moite, Feb 21 2019

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