A031396 -id:A031396 - cp's OEIS frontend

A031397 Nonsquarefree n such that Pell equation x^2 - n y^2 = -1 is soluble.

50, 125, 250, 325, 338, 425, 845, 925, 1025, 1250, 1325, 1445, 1450, 1525, 1625, 1682, 1825, 1850, 2050, 2125, 2197, 2425, 2725, 2738, 2825, 2873, 2890, 3050, 3125, 3250, 3425, 3625, 3725, 3925, 4250, 4325, 4394, 4625, 4825, 4901, 4913

Offset: 1

David W. Wilson

nonn

Harvey Cohn, Advanced Number Theory, Dover Publications, New York, N.Y. (1980).
S Vidhyalakshmi, V Krithika, K Agalya, On The Negative Pell Equation, International Journal of Emerging Technologies in Engineering Research (IJETER), Volume 4, Issue 2, February (2016) www.ijeter.everscience.org,

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 500 terms from Robert Israel)

Equals {A003814} minus {A003654}, cf. A031396.

filter:= t -> not numtheory:-issqrfree(t) and [isolve(x^2 - t*y^2 = -1)]<>[]:
select(filter, [$1..10000]); # Robert Israel, Jul 10 2018

r[n_] := Reduce[x>0 && y>0 && x^2 - n y^2 == -1, {x, y}, Integers];
Reap[For[n = 1, n <= 5000, n++, If[!SquareFreeQ[n], If[r[n] =!= False, Print[n]; Sow[n]]]]][[2, 1]] (* Jean-François Alcover, Mar 05 2019 *)

Offset changed by Robert Israel, Jul 10 2018

A157757 a(n) = 2809n^2 - 4618n + 1898.

Original entry on oeis.org

89, 3898, 13325, 28370, 49033, 75314, 107213, 144730, 187865, 236618, 290989, 350978, 416585, 487810, 564653, 647114, 735193, 828890, 928205, 1033138, 1143689, 1259858, 1381645, 1509050, 1642073, 1780714, 1924973, 2074850

Offset: 1

Vincenzo Librandi, Mar 06 2009

The identity (15780962*n^2-25943924*n+10662963)^2-(2809*n^2-4618*n+1898)*(297754*n-244754)^2=1 can be written as A157759(n)^2-a(n)*A157758(n)^2=1.
From Klaus Purath, Mar 31 2025: (Start)
Numbers k such that k*53^2-1 is a square, and k is the sum of two squares (see FORMULA).
All a(n) = D satisfy the Pell equation (k*x)^2 - D*(53*y)^2 = -1 for any integer n where a(1-n) = A157760(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*53^2 - 1), x(0) = 1, x(1) = 4*D*53^2 - 1, y(0) = 1, y(1) = 4*D*53^2 - 3. The two recurrences are of the form (4*D*53^2 - 2, -1).
It follows from the above that this sequence and A157760 are subsequences of A031396. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Subsequence of A031396.

Cf. A157758, A157759.

I:=[89, 3898, 13325]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];

LinearRecurrence[{3,-3,1},{89,3898,13325},40]
Table[2809n^2-4618n+1898,{n,40}] (* Harvey P. Dale, Aug 02 2024 *)

PARI
```
a(n) = 2809*n^2 - 4618*n + 1898;
```

a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
G.f.: x*(-89-3631*x-1898*x^2)/(x-1)^3.
a(n) = (28*n - 23)^2 + (45*n - 37)^2. - Klaus Purath, Mar 31 2025
53^2*a(n) - 1 = (2809*n-2309)^2. - Klaus Purath, Mar 31 2025

A157760 a(n) = 2809n^2 - 1000n + 89.

Original entry on oeis.org

1898, 9325, 22370, 41033, 65314, 95213, 130730, 171865, 218618, 270989, 328978, 392585, 461810, 536653, 617114, 703193, 794890, 892205, 995138, 1103689, 1217858, 1337645, 1463050, 1594073, 1730714, 1872973, 2020850, 2174345

Offset: 1

Vincenzo Librandi, Mar 06 2009

The identity (15780962*n^2-5618000*n+500001)^2-(2809*n^2-1000*n+89)*( 297754*n-53000)^2=1 can be written as A157762(n)^2-a(n)*A157761(n)^2=1.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Subsequence of A031396, see A157757.

Cf. A157761, A157762.

I:=[1898, 9325, 22370]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];

LinearRecurrence[{3,-3,1},{1898,9325,22370},50]

PARI
```
a(n) = 2809*n^2 - 1000*n + 89;
```

G.f.: x*(1898 + 3631*x + 89*x^2)/(1 - x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

A322781 Numbers of the form p*q where p, q are distinct primes congruent to 1 mod 4 such that Legendre(p/q) = -1.

Original entry on oeis.org

65, 85, 185, 265, 365, 481, 485, 493, 533, 565, 629, 685, 697, 785, 865, 949, 965, 985, 1037, 1073, 1157, 1165, 1189, 1241, 1261, 1285, 1385, 1417, 1465, 1565, 1585, 1649, 1685, 1765, 1769, 1781, 1853, 1865, 1921, 1937, 1985, 2117, 2165, 2173, 2257, 2285, 2509, 2561, 2581, 2785, 2813, 2885, 2929, 2941

Offset: 1

N. J. A. Sloane, Jan 11 2019

nonn

If k is a term, the Pell equation x^2 - k*y^2 = -1 has a solution [Dirichlet, Newman (1977)]. This is only a sufficient condition, there are many other solutions, see A031396.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Morris Newman, A note on an equation related to the Pell equation, The American Mathematical Monthly 84.5 (1977): 365-366.

Cf. A002144, A031396, A323271, A323272.

isok(n) = my (f=factor(n)); omega(f)==2 && bigomega(f)==2 && f[1,1]%4==1 && f[2,1]%4==1 && kronecker(f[1,1], f[2,1])==-1 \\ Rémy Sigrist, Jan 11 2019

from sympy.ntheory import legendre_symbol, factorint
A322781_list, k = [], 1
while len(A322781_list) < 10000:
    fk, fv = zip(*list(factorint(4*k+1).items()))
    if sum(fv) == len(fk) == 2 and fk[0] % 4 == fk[1] % 4 == 1 and legendre_symbol(fk[0],fk[1]) == -1:
            A322781_list.append(4*k+1)
    k += 1 # Chai Wah Wu, Jan 11 2019

A323271 Numbers of the form pqr where p, q, r are distinct primes congruent to 1 mod 4 such that Legendre(p/q) = Legendre(p/r) = Legendre(q/r) = -1.

Original entry on oeis.org

2405, 3145, 4745, 6205, 6305, 8245, 8905, 9605, 12545, 12805, 14705, 16405, 16745, 17945, 18241, 19045, 19345, 19805, 20213, 20605, 20905, 22945, 23545, 25805, 26605, 26945, 28645, 29705, 30073, 33745, 35705, 35989, 36205, 36305, 37505, 38369, 38545

Offset: 1

N. J. A. Sloane, Jan 11 2019

nonn

If k is a term, the Pell equation x^2 - k*y^2 = -1 has a solution [Dirichlet, Newman (1977)]. This is only a sufficient condition, there are many other solutions, see A031396.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Morris Newman, A note on an equation related to the Pell equation, The American Mathematical Monthly 84.5 (1977): 365-366.

Cf. A002144, A031396, A322781, A323272.

from sympy.ntheory import legendre_symbol, factorint
A323271_list, k = [], 1
while len(A323271_list) < 10000:
    fk, fv = zip(*list(factorint(4*k+1).items()))
    if sum(fv) == len(fk) == 3 and fk[0] % 4 == fk[1] % 4 == fk[2] % 4 == 1 and legendre_symbol(fk[0],fk[1]) == legendre_symbol(fk[0],fk[2]) == legendre_symbol(fk[1],fk[2]) == -1:
            A323271_list.append(4*k+1)
    k += 1 # Chai Wah Wu, Jan 11 2019

A323272 Numbers of the form p_1p_2p_3...p_r where r is 2 or an odd number > 2, and the p_i are distinct primes congruent to 1 mod 4 such that Legendre(p_i/p_j) = -1 for all i != j.

Original entry on oeis.org

65, 85, 185, 265, 365, 481, 485, 493, 533, 565, 629, 685, 697, 785, 865, 949, 965, 985, 1037, 1073, 1157, 1165, 1189, 1241, 1261, 1285, 1385, 1417, 1465, 1565, 1585, 1649, 1685, 1765, 1769, 1781, 1853, 1865, 1921, 1937, 1985, 2117, 2165, 2173

Offset: 1

N. J. A. Sloane, Jan 11 2019

nonn

If k is a term, the Pell equation x^2 - k*y^2 = -1 has a solution [Dirichlet, Newman (1977)]. This is only a sufficient condition, there are many other solutions, see A031396.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Morris Newman, A note on an equation related to the Pell equation, The American Mathematical Monthly 84.5 (1977): 365-366.

Cf. A002144, A031396. Includes the union of A322781 and A323271.

A062769 Smallest number m such that the continued fraction expansion of sqrt(m) has period 2n + 1.

Original entry on oeis.org

2, 41, 13, 58, 106, 61, 193, 109, 157, 337, 181, 586, 457, 949, 821, 601, 613, 1061, 421, 541, 1117, 1153, 1249, 1069, 1021, 1201, 1669, 2381, 1453, 2137, 2053, 1801, 2293, 1381, 1549, 3733, 3541, 3217, 5857, 1621, 3169, 4657, 2689, 3049, 2389, 4057, 4549

Offset: 0

Lekraj Beedassy, Jul 17 2001

nonn

If the continued fraction for sqrt(N) has period (2k + 1) and k-th convergent P(k)/Q(k) [taking P(-1)=1; Q(-1)=0 where necessary], then the i-th positive solution V(i) = [x(i),y(i)] to the Pell equation x^2 - N*y^2 = 1 satisfies the recurrence V(i+2) = 2*A*V(i+1) - V(i) starting with V(0)=(1,0); V(1) = (A,B) where A = 2*S^2 + 1; B = 2*S*T and S = P(k)*Q(k) + P(k-1)*Q(k-1); T = Q(k)^2 + Q(k-1)^2.

			For n = 2, 2n+1 = 5. a(2) = 13 and we indeed have sqrt(13) = [3; 1, 1, 1, 1, 6] with period 5, the first one in the sequence sqrt(29) = [5; 2, 1, 1, 2, 10], sqrt(53) = [7; 3, 1, 1, 3, 14], sqrt(74) = [8; 1, 1, 1, 1, 16], sqrt(85) = [9; 4, 1, 1, 4, 18], sqrt(89) = [9; 2, 3, 3, 2, 18], ...

Michael De Vlieger, Table of n, a(n) for n = 0..1000 (Terms 0..999 from T. D. Noe)
Dario Alpern, Continued Fraction calculator.
Keith Matthews, Calculating the simple continued fraction of a quadratic irrational.
Ulrich Sondermann, Continued Fractions.
Gang Xiao, Contfrac,continued fraction expansion server with k-th convergent calculator.

Cf. A013646, A003814, A031396, A003654.

nn = 50; t = Table[0, {nn}]; n = 1; found = 0; While[found < nn, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[OddQ[len] && (len + 1)/2 <= nn && t[[(len + 1)/2]] == 0, t[[(len + 1)/2]] = n; found++]]]; t (* T. D. Noe, Apr 04 2014 *)

More terms from Naohiro Nomoto, Jan 01 2002

A157110 a(n) = 1681n^2 - 2606n + 1010.

Original entry on oeis.org

85, 2522, 8321, 17482, 30005, 45890, 65137, 87746, 113717, 143050, 175745, 211802, 251221, 294002, 340145, 389650, 442517, 498746, 558337, 621290, 687605, 757282, 830321, 906722, 986485, 1069610, 1156097, 1245946, 1339157, 1435730, 1535665

Offset: 1

Vincenzo Librandi, Feb 23 2009

The identity (5651522*n^2 - 8761372*n + 3395619)^2 - (1681*n^2 - 2606*n + 1010)*(137842*n - 106846)^2 = 1 can be written as A157112(n)^2 - a(n)*A157111(n)^2 = 1. - Vincenzo Librandi, Jan 25 2012
The continued fraction expansion of sqrt(a(n)) is [41n-32; {4, 1, 1, 4, 82n-64}]. - Magus K. Chu, Oct 03 2022
From Klaus Purath, Apr 18 2025: (Start)
a(n)*41^2-1 is a square, and a(n) is the sum of two squares (see FORMULA). There are no squares in this sequence. The odd prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*(41*y)^2 = -1 for any integer n where a(1-n) = A157010(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*41^2 - 1), x(0) = 1, x(1) = 4*D*41^2 - 1, y(0) = 1, y(1) = 4*D*41^2 - 3. The two recurrences are of the form (4*D*41^2 - 2, -1).
It follows from the above that the terms of this sequence and of A157010 belong to A031396. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1 [Broken link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A157111, A157112.

I:=[85, 2522, 8321]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jan 25 2012

LinearRecurrence[{3,-3,1},{85,2522,8321},40] (* Vincenzo Librandi, Jan 25 2012 *)
Table[1681*n^2-2606*n+1010,{n,40}] (* Harvey P. Dale, Nov 24 2024 *)

for(n=1, 22, print1(1681*n^2 - 2606*n + 1010", ")); \\ Vincenzo Librandi, Jan 25 2012

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jan 25 2012
G.f.: x*(-85 - 2267*x - 1010*x^2)/(x-1)^3. - Vincenzo Librandi, Jan 25 2012
From Klaus Purath, Apr 18 2025: (Start)
a(n) = (9*n - 7)^2 + (40*n - 31)^2 for any integer n.
1681*a(n) - 1 = (1681*n - 1303)^2 for any integer n. (End)

A202155 x-values in the solution to x^2 - 13*y^2 = -1.

Original entry on oeis.org

18, 23382, 30349818, 39394040382, 51133434066018, 66371158023650982, 86149711981264908618, 111822259780523827735182, 145145207045407947135357618, 188398366922679734857866452982, 244540935120431250437563520613018, 317413945387952840388222591889244382

Offset: 1

Bruno Berselli, Dec 15 2011

The corresponding values of y of this Pell equation are in A202156.

A. H. Beiler, Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, Dover (New York), 1966, p. 264.

Bruno Berselli, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences.
A. M. S. Ramasamy, Polynomial solutions for the Pell's equation, Indian Journal of Pure and Applied Mathematics 25 (1994), p. 579 (Theorem 4, case t=1).
J. P. Robertson, Solving the generalized Pell equation x^2-D*y^2=N, pp. 9, 24.
Index entries for linear recurrences with constant coefficients, signature (1298,-1).

Cf. A002313, A003654, A031396, A114047, A202156.

m:=13; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(18*x*(1+x)/(1-1298*x+x^2)));

LinearRecurrence[{1298, -1}, {18, 23382}, 12]

makelist(expand(((18+5*sqrt(13))^(2*n-1)+(18-5*sqrt(13))^(2*n-1))/2), n, 1, 12);

G.f.: 18*x*(1+x)/(1-1298*x+x^2).

a(n) = -a(-n+1) = (r^(2n-1)-1/r^(2n-1))/2, where r=18+5*sqrt(13).

A202156 y-values in the solution to x^2 - 13*y^2 = -1.

Original entry on oeis.org

5, 6485, 8417525, 10925940965, 14181862955045, 18408047189707445, 23893631070377308565, 31013914721302556809925, 40256037414619648361974085, 52252305550261582271285552405, 67823452348202119168480285047605, 88034788895660800419105138706238885

Offset: 1

Bruno Berselli, Dec 15 2011

The corresponding values of x of this Pell equation are in A202155.

A. H. Beiler, Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, Dover Publications (New York), 1966, p. 264.

Bruno Berselli, Table of n, a(n) for n = 1..200
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences.
A. M. S. Ramasamy, Polynomial solutions for the Pell's equation, Indian Journal of Pure and Applied Mathematics 25 (1994), p. 579 (Theorem 4, case t=1).
J. P. Robertson, Solving the generalized Pell equation x^2-D*y^2=N, pp. 9, 24.
Index entries for linear recurrences with constant coefficients, signature (1298,-1).

Cf. A002313, A003654, A006191, A031396, A075871, A202155.

m:=13; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(5*x*(1-x)/(1-1298*x+x^2)));

LinearRecurrence[{1298, -1}, {5, 6485}, 12]

makelist(expand(((18+5*sqrt(13))^(2*n-1)-(18-5*sqrt(13))^(2*n-1))/(2*sqrt(13))), n, 1, 12);

G.f.: 5*x*(1-x)/(1-1298*x+x^2).
a(n) = a(-n+1) = 5*(r^(2n-1)+1/r^(2n-1))/(r+1/r), where r=18+5*sqrt(13).
a(n) = A006191(6*n - 3). - Michael Somos, Feb 24 2023

cp's OEIS Frontend

A031397 Nonsquarefree n such that Pell equation x^2 - n y^2 = -1 is soluble.

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Maple

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A157757 a(n) = 2809*n^2 - 4618*n + 1898.

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Magma

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A157760 a(n) = 2809*n^2 - 1000*n + 89.

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Magma

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A322781 Numbers of the form p*q where p, q are distinct primes congruent to 1 mod 4 such that Legendre(p/q) = -1.

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A323271 Numbers of the form p*q*r where p, q, r are distinct primes congruent to 1 mod 4 such that Legendre(p/q) = Legendre(p/r) = Legendre(q/r) = -1.

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A323272 Numbers of the form p_1*p_2*p_3*...*p_r where r is 2 or an odd number > 2, and the p_i are distinct primes congruent to 1 mod 4 such that Legendre(p_i/p_j) = -1 for all i != j.

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A062769 Smallest number m such that the continued fraction expansion of sqrt(m) has period 2n + 1.

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Mathematica

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A157110 a(n) = 1681*n^2 - 2606*n + 1010.

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A157757 a(n) = 2809n^2 - 4618n + 1898.

A157760 a(n) = 2809n^2 - 1000n + 89.

A323271 Numbers of the form pqr where p, q, r are distinct primes congruent to 1 mod 4 such that Legendre(p/q) = Legendre(p/r) = Legendre(q/r) = -1.

A323272 Numbers of the form p_1p_2p_3...p_r where r is 2 or an odd number > 2, and the p_i are distinct primes congruent to 1 mod 4 such that Legendre(p_i/p_j) = -1 for all i != j.

A157110 a(n) = 1681n^2 - 2606n + 1010.