cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 100 results. Next

A254760 Fundamental positive solution x = x1(n) of the first class of the Pell equation x^2 - 2*y^2 = A007519(n), n>=1 (primes congruent to 1 mod 8).

Original entry on oeis.org

5, 7, 9, 11, 13, 11, 13, 15, 19, 21, 17, 17, 21, 25, 19, 23, 21, 21, 29, 23, 23, 31, 33, 25, 27, 25, 29, 31, 31, 29, 29, 37, 41, 31, 35, 31, 37, 39, 41, 43, 35, 39, 35, 35, 43, 35, 49, 41, 37
Offset: 1

Views

Author

Wolfdieter Lang, Feb 10 2015

Keywords

Comments

For the corresponding term y1(n) see 2*A254761(n).
For the positive fundamental proper (sometimes called primitive) solutions x2(n) and y2(n) of the second class of this (generalized) Pell equation see A254762(n) and 2*A254763(n).
The present solutions of this first class are the smallest positive ones.
See the Nagell reference Theorem 111, p. 210, for the proof of the existence of solutions (the discriminant of this binary quadratic form is +8 hence it is an indefinite form with an infinitude of solutions if there exists at least one).
See the Nagell reference Theorem 110, p. 208, for the proof that there are only two classes of solutions for this Pell equation, because the equation is solvable and the primes from A007519 do not divide 4.
The present fundamental solutions are found according to the Nagell reference Theorem 108, p. 205, adapted to the case at hand, by scanning the following two inequalities for solutions x1(n) = 2*X1(n) + 1 and y1(n) = 2*Y1(n). The intervals to be scanned are ceiling((sqrt(8 + p(n))-1)/2) <= X1(n) <= floor((sqrt(2*p(n))-1)/2), with p(n) = A007519(n), and
1 <= Y1(n) <= floor(sqrt(A005123(n))).
The general positive proper solutions are for both classes obtained by applying positive powers of the matrix M = [[3,4],[2,3]] on the positive fundamental column vectors (x(n),y(n))^T. The n-th power M^n = S(n-1, 6)*M - S(n-2, 6) 1_2 , where 1_2 is the 2 X 2 identity matrix and S(n, 6), with S(-2, 6) = -1 and S(-1, 6) = 0 is the Chebyshev S-polynomial evaluated at x = 6, given in A001109(n).
The least positive x solutions (that is the ones of the first class) for the primes +1 and -1 (mod 8) together (including in the first class also the prime 2) are given in A002334. - Wolfdieter Lang, Feb 12 2015

Examples

			The first pairs [x1(n), y1(n)] of the fundamental positive solutions of this first class are (we list the prime A007519(n) as first entry):
[17, [5, 2]], [41, [7, 2]], [73, [9, 2]], [89, [11, 4]], [97, [13, 6]], [113, [11, 2]], [137, [13, 4]], [193, [15, 4]], [233, [19, 8]], [241, [21, 10]], [257, [17, 4]], [281, [17, 2]], [313, [21, 8]], [337, [25, 12]], [353, [19, 2]], [401, [23, 8]], [409, [21, 4]], ...
n=1: 5^2 - 2*2^2 = 25 - 8 = 17, ...
		

References

  • T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

Crossrefs

Formula

a(n)^2 - 2*(2*A254760(n))^2 = A007519(n) gives the smallest positive (proper) solution of this (generalized) Pell equation.

A141174 Duplicate of A007519.

Original entry on oeis.org

17, 41, 73, 89, 97, 113, 137, 193, 233, 241, 257, 281, 313, 337, 353, 401, 409, 433, 449, 457, 521, 569, 577, 593, 601, 617, 641, 673, 761, 769, 809, 857, 881, 929, 937, 953, 977, 1009, 1033, 1049, 1097, 1129, 1153, 1193, 1201, 1217, 1249, 1289, 1297, 1321
Offset: 1

Views

Author

Laura Caballero Fernandez, Lourdes Calvo Moguer, Maria Josefa Cano Marquez, Oscar Jesus Falcon Ganfornina and Sergio Garrido Morales (oscfalgan(AT)yahoo.es), Jun 12 2008

Keywords

Comments

Originally "Primes of the form x^2 + 4xy - 4y^2 (as well as of the form x^2 + 6xy + y^2)."
R. J. Mathar was the first to wonder whether these are also primes of the form 8k + 1. I did the easy part, proving that all primes of the form x^2 + 4xy - 4y^2 are congruent to 1 mod 8. Since x^2 + 4xy - 4y^2 = 2 or -2 is impossible, x must be odd. And since x is odd, x^2 = 1 mod 8.
If y is even, then both 4xy and 4y^2 are multiples of 8. If y is odd, then 4xy = 4 mod 8, but so is 4y^2, cancelling out the effect and leaving x^2 = 1 mod 8.
It remains to prove that every prime of the form 8k + 1 has a representation as x^2 + 4xy - 4y^2. - Alonso del Arte, Jan 28 2017
A necessary and sufficient condition of representation of p = 8n + 1 in your quadratic form is {8y^2 + 8n + 1 is perfect square}, since only in this case solving square equation for x, we have x = -2y + sqrt(8y^2 + 8n + 1) is [an] integer. For this a sufficient condition is { n has a form n = k^2 - k + i(4k + i - 1)/2, i >= 0, k >= 1}. In this case x = 2i + 2k - 1. y = k." - Vladimir Shevelev, Jan 26 2017

Extensions

More terms from Michel Marcus, Feb 01 2014

A254762 Fundamental positive solution x = x2(n) of the second class of the Pell equation x^2 - 2*y^2 = A007519(n), n >= 1 (primes congruent to 1 mod 8).

Original entry on oeis.org

7, 13, 19, 17, 15, 25, 23, 29, 25, 23, 35, 43, 31, 27, 49, 37, 47, 55, 31, 45, 61, 37, 35, 59, 49, 67, 47, 45, 53, 63, 71, 47, 43, 77, 57, 85, 55, 53, 51, 49, 73, 61, 81, 89, 57, 97, 51, 67, 87
Offset: 1

Views

Author

Wolfdieter Lang, Feb 10 2015

Keywords

Comments

The corresponding term y = y2(n) of this fundamental solution of the second class of the (generalized) Pell equation x^2 - 2*y^2 = A007519(n) = 1 + 8*A005123(n) is given in 2*A254763(n).
For comments and the Nagell reference see A254760.

Examples

			The first pairs [x2(n), y2(n)] of the fundamental positive solutions of the second class are (we list the prime A007519(n) as first entry):
  [17, [7, 4]], [41, [13, 8]], [73, [19, 12]], [89, [17, 10]], [97, [15, 8]], [113, [25, 16]], [137, [23, 14]], [193, [29, 18]], [233, [25, 14]], [241, [23, 12]], [257, [35, 22]], [281, [43, 28]], [313, [31, 18]], [337, [27, 14]], [353, [49, 32]], [401, [37, 22]], [409, [47, 30]], ...
a(4) = 3*11 - 8*2 = 17.
		

Crossrefs

Formula

a(n)^2 - 2*(2*A254763(n))^2 = A007519(n) gives the second smallest positive (proper) solution of this (generalized) Pell equation.
a(n) = 3*A254760(n) - 8*A254761(n), n >= 1.

A254763 One half of the fundamental positive solution y = y2(n) of the second class of the Pell equation x^2 - 2*y^2 = A007519(n), n>=1 (primes congruent to 1 mod 8).

Original entry on oeis.org

2, 4, 6, 5, 4, 8, 7, 9, 7, 6, 11, 14, 9, 7, 16, 11, 15, 18, 8, 14, 20, 10, 9, 19, 15, 22, 14, 13, 16, 20, 23, 13, 11, 25, 17, 28, 16, 15, 14, 13, 23, 18, 26, 29, 16, 32, 13, 20, 28, 24
Offset: 1

Views

Author

Wolfdieter Lang, Feb 10 2015

Keywords

Comments

The corresponding fundamental solution x2(n) of this second class of positive solutions is given in A254762(n).
See the comments and the Nagell reference in A254760.

Examples

			n = 2: 13^2 - 2*(2*4)^2 = 169 - 128 = 41.
The smallest positive solution is (x1(2), y1(2)) = (7, 2) from (A254760(2), 2*A254761(2)).
See also A254762.
a(4) = 11 - 3*2 = 5.
		

Crossrefs

Formula

A254762(n)^2 - 2*(2*a(n))^2 = A007519(n) gives the second smallest positive (proper) solution of this (generalized) Pell equation.
a(n) = A254760(n) - 3*A254761(n), n >= 1.

A254934 Fundamental positive solution x = x1(n) of the first class of the Pell equation x^2 - 2*y^2 = -A007519(n), n>=1 (primes congruent to 1 mod 8).

Original entry on oeis.org

1, 3, 5, 3, 1, 7, 5, 7, 3, 1, 9, 13, 5, 1, 15, 7, 13, 17, 1, 11, 19, 3, 1, 17, 11, 21, 9, 7, 11, 17, 21, 5, 1, 23, 11, 27, 9, 7, 5, 3, 19, 11, 23, 27, 7, 31, 1, 13, 25, 19, 33, 7, 5, 21, 31, 25, 9, 29, 5, 15, 27, 13, 31, 11, 7, 17, 3, 37, 41, 31, 19, 25, 7, 35, 5, 17, 33, 13, 21, 19, 45, 25, 3
Offset: 1

Views

Author

Wolfdieter Lang, Feb 18 2015

Keywords

Comments

For the corresponding term y1(n) see A254935(n).
For the positive fundamental proper (sometimes called primitive) solutions x2(n) and y2(n) of the second class of this (generalized) Pell equation see A254936(n) and A254937(n).
The present solutions of this first class are the smallest positive ones.
See the Nagell reference Theorem 111, p. 210, for the proof of the existence of solutions (the discriminant of this binary quadratic form is +8 hence it is an indefinite form with an infinitude of solutions if there exists at least one).
See the Nagell reference Theorem 110, p. 208, for the proof that there are only two classes of solutions for this Pell equation, because the equation is solvable and each prime from A007519 does not divide 4.
The present fundamental solutions are found according to the Nagell reference Theorem 108a, p. 206-207, adapted to the case at hand, by scanning the following two inequalities for solutions x1(n) = 2*X1(n) + 1 and y1(n) = 2*Y1(n) + 1 (because even y is out in this Pell equation). The intervals to be scanned are identical for X1(n) and Y1(n), namely [0, floor((sqrt(p(n) - 1)/2)] with p(n) = A007519(n).
The general positive proper solutions are for both classes obtained by applying positive powers of the matrix M = [[3,4],[2,3]] on the fundamental positive column vectors (x(n),y(n))^T. The n-th power M^n = S(n-1, 6)*M - S(n-2, 6) 1_2 , where 1_2 is the 2 X 2 identity matrix and S(n, 6), with S(-2, 6) = -1 and S(-1, 6) = 0 is the Chebyshev S-polynomial evaluated at x = 6, given in A001109(n).
The least positive x solutions (that is the ones of the first class) for the primes +1 and -1 (mod 8) together (including prime 2) are given in A255235.

Examples

			The first pairs [x1(n), y1(n)] of the fundamental positive solutions of this first class are (the prime A007519(n) is listed as first entry):
[17, [1, 3]], [41, [3, 5]], [73, [5, 7]],
[89, [3, 7]], [97, [1, 7]], [113, [7, 9]],
[137, [5, 9]], [193, [7, 11]], [233, [3, 11]],
[241, [1, 11]], [257, [9, 13]], [281, [13, 15]],
[313, [5, 13]], [337, [1, 13]], [353, [15, 17]],
[401, [7, 15]], [409, [13, 17]], [433, [17, 19]],
[449, [1, 15]], [457, [11, 17]], [521, [19, 21]],
[569, [3, 17]], [577, [1, 17]], [593, [17, 21]],
[601, [11, 19]], [617, [21, 23]], [641, [9, 19]],
[673, [7, 19]], [761, [11, 21]], [769, [17, 23]],
...
n=1: 1^2 - 2*3^2 = 1 - 18 = -17, ...
		

References

  • T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

Crossrefs

Cf. A007519 (primes == 1 mod 8), A005123 (8k+1 is prime).
Cf. A254935 (corresponding y1 values), A254936 (x2 values), A254937 (y2 values), A254938, A255232, A255235, A254760.

Programs

  • PARI
    apply( {A254934(n, p=A007519(n))=Set(abs(qfbsolve(Qfb(-1,0,2), p,1)))[1][1]}, [1..77]) \\ The 2nd optional arg allows to directly specify the prime. - M. F. Hasler, May 22 2025

Formula

a(n)^2 - 2*A254935(n)^2 = -A007519(n), n >=1, gives the smallest positive (proper) solution of this (generalized) Pell equation.

Extensions

More terms from M. F. Hasler, May 22 2025

A254761 One half of the fundamental positive solution y = y1(n) of the first class of the Pell equation x^2 - 2*y^2 = A007519(n), n >= 1 (primes congruent to 1 mod 8).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 2, 2, 4, 5, 2, 1, 4, 6, 1, 4, 2, 1, 7, 3, 1, 7, 8, 2, 4, 1, 5, 6, 5, 3, 2, 8, 10, 2, 6, 1, 7, 8, 9, 10, 4, 7, 3, 2, 9, 1, 12, 7, 3, 5
Offset: 1

Views

Author

Wolfdieter Lang, Feb 12 2015

Keywords

Comments

For the corresponding term x1(n) see A254760(n).
See A254760 also for the Nagell reference.
The least positive y solutions (that is the ones of the first class) for the primes +1 and -1 (mod 8) together (including in the first class also prime 2) are given in A002335.

Examples

			See A254760.
n = 3: 9^2 - 2*(2*1)^2 = 81 - 8 = 73.
		

Crossrefs

Formula

A254760(n)^2 - 2*(2*a(n))^2 = A007519(n) gives the smallest positive (proper) solution of this (generalized) Pell equation.

A254935 Fundamental positive solution y = y1(n) of the first class of the Pell equation x^2 - 2*y^2 = -A007519(n), n>=1 (primes congruent to 1 mod 8).

Original entry on oeis.org

3, 5, 7, 7, 7, 9, 9, 11, 11, 11, 13, 15, 13, 13, 17, 15, 17, 19, 15, 17, 21, 17, 17, 21, 19, 23, 19, 19, 21, 23, 25, 21, 21, 27, 23, 29, 23, 23, 23, 23, 27, 25, 29, 31, 25, 33, 25, 27, 31, 29, 35, 27, 27, 31, 35, 33, 29, 35, 29, 31, 35, 31, 37, 31, 31, 33, 31, 41, 43, 39, 35, 37, 33, 41, 33, 35, 41
Offset: 1

Views

Author

Wolfdieter Lang, Feb 18 2015

Keywords

Comments

For the corresponding term x1(n) see A254934(n).
See A254934 also for the Nagell reference.
The least positive y solutions (that is the ones of the first class) for the primes +1 and -1 (mod 8) together (including prime 2) are given in A255246.

Examples

			See A254934.
n = 3: 5^2 - 2*7^2 = 25 - 98 = -73.
		

Crossrefs

Cf. A007519 (primes == 1 mod 8), A005123 (8k+1 is prime).
Cf. A254934 (corresponding x1 values), A254936 (x2 values), A254937 (y2 values), A254938 (same for primes == 7 mod 8), A255232 (y2 values, halved).

Programs

Formula

A254934(n)^2 - 2*a(n)^2 = -A007519(n) gives the smallest positive (proper) solution of this (generalized) Pell equation.

Extensions

More terms from M. F. Hasler, May 22 2025

A254937 Fundamental positive solution y = y2(n) of the second class of the Pell equation x^2 - 2*y^2 = -A007519(n), n>=1 (primes congruent to 1 mod 8).

Original entry on oeis.org

7, 9, 11, 15, 19, 13, 17, 19, 27, 31, 21, 19, 29, 37, 21, 31, 25, 23, 43, 29, 25, 45, 49, 29, 35, 27, 39, 43, 41, 35, 33, 53, 61, 35, 47, 33, 51, 55, 59, 63, 43, 53, 41, 39, 61, 37, 73, 55, 43, 49, 39, 67, 71, 51, 43, 49, 69, 47, 77, 63, 51, 67, 49, 71, 79, 65, 87, 49, 47, 55, 67, 61, 85, 53
Offset: 1

Views

Author

Wolfdieter Lang, Feb 18 2015

Keywords

Comments

The corresponding positive fundamental solution x2(n) of this second class solutions is given in A254936(n).
See the comments and the Nagell reference in A254934.

Examples

			n = 2: 11^2 - 2*9^2 = 121 - 162 = -41.
a(2) = -(2*3 - 3*5) = 9.
See also A254936.
		

Crossrefs

Cf. A007519 (primes == 1 mod 8), A254936 (x2-values), A254934 (first (class) solution x1), A254935 (y1), A255234 (y2/2 for primes == 7 mod 8), A255248, A254760.

Programs

  • PARI
    apply( {A254937(n, p=A007519(n))=n=Set(abs(qfbsolve(Qfb(-1, 0, 2), p, 1)))[1]*[-2, 3]~}, [1..77]) \\ The 2nd optional arg allows to directly specify the prime. - M. F. Hasler, May 22 2025

Formula

A254936(n)^2 - 2*a(n)^2 = -A007519(n) gives the second smallest positive (proper) solution of this (generalized) Pell equation.
a(n) = -(2*A254934(n) - 3*A254935(n)), n >= 1.

Extensions

More terms from M. F. Hasler, May 22 2025

A254936 Fundamental positive solution x = x2(n) of the second class of the Pell equation x^2 - 2*y^2 = -A007519(n), n >= 1 (primes congruent to 1 mod 8).

Original entry on oeis.org

9, 11, 13, 19, 25, 15, 21, 23, 35, 41, 25, 21, 37, 49, 23, 39, 29, 25, 57, 35, 27, 59, 65, 33, 43, 29, 49, 55, 51, 41, 37, 69, 81, 39, 59, 35, 65, 71, 77, 83, 51, 67, 47, 43, 79, 39, 97, 69, 49, 59, 41, 87, 93, 61, 47, 57, 89, 53, 101, 79, 59, 85, 55, 91, 103, 81, 115, 53, 49, 63, 83, 73, 111, 59
Offset: 1

Views

Author

Wolfdieter Lang, Feb 18 2015

Keywords

Comments

The corresponding term y = y2(n) of this fundamental solution of the second class of the (generalized) Pell equation x^2 - 2*y^2 = -A007519(n) = -(1 + 8*A005123(n)) is given in A254937(n).
For comments and the Nagell reference see A254934.

Examples

			The first pairs [x2(n), y2(n)] of the fundamental positive solutions of this second class are (the prime A007519(n) appears as first entry):
  [17, [9, 7]], [41, [11, 9]], [73, [13, 11]],
  [89, [19, 15]], [97, [25, 19]], [113, [15, 13]],
  [137, [21, 17]], [193, [23, 19]], [233, [35, 27]],
  [241, [41, 31]], [257, [25, 21]], [281, [21, 19]],
  [313, [37, 29]], [337, [49, 37]], [353, [23, 21]],
  [401, [39, 31]], [409, [29, 25]], [433, [25, 23]],
  [449, [57, 43]], [457, [35, 29]], [521, [27, 25]],
  [569, [59, 45]], [577, [65, 49]], [593, [33, 29]],
  [601, [43, 35]], [617, [29, 27]], [641, [49, 39]], ...
a(4) = -(3*3 - 4*7) = 28 - 9 = 19.
		

Crossrefs

Cf. A007519 (primes == 1 mod 8), A005123 (8k+1 is prime).
Cf. A254937 (corresponding y2-values), A254934 (x1 values), A254935 (y1 values), A255233 (same for primes == 7 mod 8), A255247.

Programs

  • PARI
    apply( {A254936(n, p=A007519(n))=n=Set(abs(qfbsolve(Qfb(-1, 0, 2), p, 1)))[1]*[-3,4]~}, [1..77]) \\ The 2nd optional arg allows to directly specify the prime. - M. F. Hasler, May 22 2025

Formula

a(n)^2 - 2*A254937(n)^2 = -A007519(n) gives the second smallest positive (proper) solution of this (generalized) Pell equation.
a(n) = -(3*A254934(n) - 4*A254935(n)), n >= 1.

Extensions

More terms from M. F. Hasler, May 22 2025

A094928 Let p = n-th prime == 1 mod 8 (A007519); a(n) = smallest prime q such that p is not a square mod q.

Original entry on oeis.org

3, 3, 5, 3, 5, 3, 3, 5, 3, 7, 3, 3, 5, 5, 3, 3, 7, 5, 3, 5, 3, 3, 5, 3, 7, 3, 3, 5, 3, 7, 3, 3, 3, 3, 5, 3, 3, 11, 5, 3, 3, 11, 5, 3, 11, 3, 7, 3, 5, 7, 3, 3, 3, 3, 7, 3, 3, 7, 5, 3, 3, 5, 5, 11, 5, 3, 3, 5, 5, 3, 7, 5, 3, 5, 3, 7, 3, 7, 3, 5, 3, 3, 3, 5, 11, 5, 3, 5, 3, 3, 13, 5, 3, 3, 3, 3, 5, 5, 3, 5, 3, 7
Offset: 1

Views

Author

N. J. A. Sloane, Jun 19 2004

Keywords

Examples

			n=3, p = 73, a(3) = q = 5: Legendre(73,5) = -1.
		

References

  • M. Kneser, Quadratische Formen, Springer, 2002; see Hilfssatz 18.3.

Crossrefs

Subsequence of A094929.

Programs

  • Maple
    f:= proc(p) local q;
         q:= 3:
         do
          if numtheory:-quadres(p,q) = -1 then return q fi;
          q:= nextprime(q);
         od;
    end proc:
    map(f, select(isprime, [seq(p,p=1..10000,8)])); # Robert Israel, May 06 2019
  • Mathematica
    f[n_] := Prime[ Position[ JacobiSymbol[n, Select[Range[3, n - 1], PrimeQ[ # ] &]], -1][[1, 1]] + 1]; f /@ Select[ Prime[ Range[435]], Mod[ #, 8] == 1 &] (* Robert G. Wilson v, Jun 23 2004 *)

Formula

a(n) = A094929(A269704(n)). - Robert Israel, May 06 2019

Extensions

More terms from Robert G. Wilson v, Jun 23 2004
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