A221874 -id:A221874 - cp's OEIS frontend

A097315 Pell equation solutions (3b(n))^2 - 10a(n)^2 = -1 with b(n) = A097314(n), n >= 0.

1, 37, 1405, 53353, 2026009, 76934989, 2921503573, 110940200785, 4212806126257, 159975692596981, 6074863512559021, 230684837784645817, 8759948972303982025, 332647376109766671133, 12631840343198829521029, 479677285665445755127969, 18215105014943739865341793, 691694313282196669127860165

Offset: 0

Wolfdieter Lang, Aug 31 2004

Hypotenuses of primitive Pythagorean triples in A195616 and A195617. - Clark Kimberling, Sep 22 2011

			(x,y) = (3,1), (117,37), (4443,1405), ... give the positive integer solutions to x^2 - 10*y^2 = -1.
G.f. = 1 + 37*x + 1405*x^2 + 53353*x^3 + ... - _Michael Somos_, Feb 24 2023

Indranil Ghosh, Table of n, a(n) for n = 0..631
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A078987 (partial sums), A097314, A195616, A195617, A049310, A053120, A005668.
Row 3 of array A188647.
Cf. A221874.
Cf. similar sequences listed in A238379.

a:=[1,37];; for n in [3..20] do a[n]:=38*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

I:=[1, 37]; [n le 2 select I[n] else 38*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Aug 01 2019

CoefficientList[Series[(1-x)/(1-38x+x^2), {x,0,20}], x] (* Michael De Vlieger, Feb 04 2017 *)
LinearRecurrence[{38,-1}, {1,37}, 21] (* G. C. Greubel, Aug 01 2019 *)

Vec((1-x)/(1-38*x+x^2) + O(x^20)) \\ Michel Marcus, Jun 04 2015

from itertools import islice
def A097315_gen(): # generator of terms
    x, y = 30, 10
    while True:
        yield y//10
        x, y = x*19+y*60, x*6+y*19
A097315_list = list(islice(A097315_gen(),20)) # Chai Wah Wu, Apr 24 2025

((1-x)/(1-38*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

a(n) = S(n, 38) - S(n-1, 38) = T(2*n+1, sqrt(10))/sqrt(10), with Chebyshev polynomials of the second and first kind. See A049310 for the triangle of S(n, x)= U(n, x/2) coefficients. S(-1, x) := 0 =: U(-1, x); and A053120 for the T-triangle.
a(n) = ((-1)^n)*S(2*n, 6*i) with the imaginary unit i and Chebyshev polynomials S(n, x) with coefficients shown in A049310.
G.f.: (1-x)/(1-38*x+x^2).
a(n) = 38*a(n-1) - a(n-2) for n > 1. - Philippe Deléham, Nov 18 2008
a(n) = sqrt(2+(19-6*sqrt(10))^(1+2*n) + (19+6*sqrt(10))^(1+2*n))/(2*sqrt(10)). - Gerry Martens, Jun 04 2015
a(n) = A078987(n) - A078987(n-1). - R. J. Mathar, Dec 05 2015
a(n) = A005668(2*n+1). - Michael Somos, Feb 24 2023
E.g.f.: exp(19*x)*(10*cosh(6*sqrt(10)*x) + 3*sqrt(10)*sinh(6*sqrt(10)*x))/10. - Stefano Spezia, Apr 24 2025

Typo in recurrence formula corrected by Laurent Bonaventure (bonave(AT)free.fr), Oct 03 2010

More terms added by Indranil Ghosh, Feb 04 2017

A129556 Numbers k such that the k-th centered pentagonal number A005891(k) = (5k^2 + 5k + 2)/2 is a square.

Original entry on oeis.org

0, 2, 21, 95, 816, 3626, 31005, 137711, 1177392, 5229410, 44709909, 198579887, 1697799168, 7540806314, 64471658493, 286352060063, 2448225223584, 10873837476098, 92968086837717, 412919472031679, 3530339074609680, 15680066099727722, 134059916748330141

Offset: 1

Alexander Adamchuk, Apr 20 2007

Corresponding numbers m > 0 such that m^2 is a centered pentagonal number are listed in A129557 = {1, 4, 34, 151, 1291, 5734, 49024, ...}.
From Andrea Pinos, Nov 02 2022: (Start)
By definition: 5*T(a(n)) = A129557(n)^2 - 1 where triangular number T(j) = j*(j+1)/2. This implies:
Every odd prime factor of a(n) and d(n)=a(n)+1 is present in b(n)=A129557(n)+1 or in c(n)=A129557(n)-1. (End)
From the law of cosines the non-Pythagorean triple {a(n), a(n)+1=A254332(n), A129557(n+1)} forms a near-isosceles triangle whose angle between the consecutive integer sides is equal to the central angle of the regular pentachoron polytope (4-simplex) (see A140244 and A140245). This implies that the terms {a(n)} are also those numbers k such that 1 + 5*A000217(k) is a square. - Federico Provvedi, Apr 04 2023

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Centered Pentagonal Number
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A005891 (centered pentagonal numbers), A129557 (numbers k>0 such that k^2 is a centered pentagonal number), A221874.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; this sequence for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013

A005891 := proc(n) (5*n^2+5*n+2)/2 ; end: n := 0 : while true do if issqr(A005891(n)) then print(n) ; fi ; n := n+1 ; od : # R. J. Mathar, Jun 06 2007

Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[n] ], {n,1,40000} ]
LinearRecurrence[{1,38,-38,-1,1},{0,2,21,95,816},30] (* Harvey P. Dale, Nov 09 2017 *)
Table[(((x^(n+2))+(((-1)^n*(x^(2*n+1)+1)-x)/(x^n)))/(x^2+1)-1)/2/.x->3+Sqrt[10],{n,0,50}]//Round (* Federico Provvedi, Apr 04 2023 *)

a(n)=([0,1,0,0,0; 0,0,1,0,0; 0,0,0,1,0; 0,0,0,0,1; 1,-1,-38,38,1]^(n-1)*[0;2;21;95;816])[1,1] \\ Charles R Greathouse IV, Feb 11 2019

For n >= 5, a(n) = 38*a(n-2) - a(n-4) + 18. - Max Alekseyev, May 08 2009
G.f.: x^2*(x^3+2*x^2-19*x-2) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)). - Colin Barker, Feb 21 2013
a(n) = (A221874(n) - 1) / 2. - Bruno Berselli, Feb 21 2013
From Andrea Pinos, Oct 24 2022: (Start)
The ratios of successive terms converge to two different limits:
lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7+2*sqrt(10))/3;
upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13+4*sqrt(10))/3.
So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).  (End)
a(n) = (x^(2*(n+1)) + (-1)^n*(x^(2*n+1)+1) - x) / (2*x^n*(x^2 + 1)) - (1/2), with x=3+sqrt(10). - Federico Provvedi, Apr 04 2023

More terms from R. J. Mathar, Jun 06 2007
Further terms from Max Alekseyev, May 08 2009
a(22)-a(23) from Colin Barker, Feb 21 2013

A084070 a(n) = 38*a(n-1) - a(n-2), with a(0)=0, a(1)=6.

Original entry on oeis.org

0, 6, 228, 8658, 328776, 12484830, 474094764, 18003116202, 683644320912, 25960481078454, 985814636660340, 37434995712014466, 1421544022419889368, 53981237856243781518, 2049865494514843808316, 77840907553707820934490, 2955904621546382351702304

Offset: 0

Benoit Cloitre, May 10 2003

nonn

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 10*y^2 = 1. The corresponding x values are in A078986. - Vincenzo Librandi, Aug 08 2010 [edited by Jon E. Schoenfield, May 04 2014]

			G.f. = 6*x + 228*x^2 + 8658*x^3 + 328776*x^4 + ... - _Michael Somos_, Feb 24 2023

Indranil Ghosh, Table of n, a(n) for n = 0..632
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A001078, A001109, A001353, A001653, A005668, A060645, A084068, A084069, A221874.

Cf. A078986. - Vincenzo Librandi, Apr 14 2010

a:=[0,6];; for n in [3..20] do a[n]:=38*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2020

I:=[0,6]; [n le 2 select I[n] else 38*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Jan 12 2020

seq( simplify(6*ChebyshevU(n-1, 19)), n=0..20); # G. C. Greubel, Jan 12 2020

LinearRecurrence[{38,-1},{0,6},30] (* Harvey P. Dale, Nov 01 2011 *)
6*ChebyshevU[Range[20]-2, 19] (* G. C. Greubel, Jan 12 2020 *)

u=0; v=6; for(n=2,20, w=38*v-u; u=v; v=w; print1(w,","))

vector(21, n, 6*polchebyshev(n-2, 2, 19) ) \\ G. C. Greubel, Jan 12 2020

[6*chebyshev_U(n-1, 19) for n in (0..20)] # G. C. Greubel, Jan 12 2020

Numbers k such that 10*k^2 = floor(k*sqrt(10)*ceiling(k*sqrt(10))).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 37*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 39*(a(n-1) - a(n-2)) + a(n-3). (End)
From R. J. Mathar, Feb 19 2008: (Start)
O.g.f.: 6*x/(1 - 38*x + x^2).
a(n) = 6*A078987(n-1). (End)
a(n) = 6*ChebyshevU(n-1, 19). - G. C. Greubel, Jan 12 2020
a(n) = A005668(2*n). - Michael Somos, Feb 24 2023

A075836 Numbers k such that 10*k^2 + 9 is a square.

Original entry on oeis.org

0, 2, 4, 18, 80, 154, 684, 3038, 5848, 25974, 115364, 222070, 986328, 4380794, 8432812, 37454490, 166354808, 320224786, 1422284292, 6317101910, 12160109056, 54009348606, 239883517772, 461763919342, 2050932962736

Offset: 1

Gregory V. Richardson, Oct 14 2002

(5/4)*a(n)^2 +1 is a triangular number. - Bruno Berselli, Aug 17 2013

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (0,0,38,0,0,-1).

Cf. A221874.

I:=[0,2,4,18,80,154]; [n le 6 select I[n] else 38*Self(n-3)-Self(n-6): n in [1..30]]; // Vincenzo Librandi, Aug 17 2013

CoefficientList[Series[2 x (1 + 2 x + 9 x^2 + 2 x^3 + x^4) / (1 - 38 x^3 + x^6), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 17 2013 *)

From Gregory V. Richardson, Oct 16 2002: (Start)
Limit_{n->oo} a(n)/a(n-3) = 19 + 6*sqrt(10).
Limit_{n->oo} a(3*n)/a(3*n-1) = (11 + 2*sqrt(10))/9.
Limit_{n->oo} a(3*n+1)/a(3*n) = (7 + 2*sqrt(10))/3.
Limit_{n->oo} a(3*n+2)/a(3*n+1) = (7 + 2*sqrt(10))/3. (End)
G.f.: 2*x^2*(1+2*x+9*x^2+2*x^3+x^4) / ( 1-38*x^3+x^6 ). - R. J. Mathar, Jul 03 2011
a(n) = 2*A075873(n). - R. J. Mathar, Jul 03 2011

A052454 Positive integer values of k such that 10*k^2 - 9 is a square.

Original entry on oeis.org

1, 3, 13, 25, 111, 493, 949, 4215, 18721, 36037, 160059, 710905, 1368457, 6078027, 26995669, 51965329, 230804967, 1025124517, 1973314045, 8764510719, 38927735977, 74933968381, 332820602355, 1478228842609, 2845517484433, 12638418378771

Offset: 1

John W. Layman, May 20 2003

			25 is a term of the sequence since 10*25^2-9 = 6241 = 79^2.

Paolo Xausa, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,38,0,0,-1).

Cf. A077442, A082651, A221874.

LinearRecurrence[{0, 0, 38, 0, 0, -1}, {1, 3, 13, 25, 111, 493}, 50] (* Paolo Xausa, Mar 18 2024 *)

a(n) = 38*a(n-3)-a(n-6).

G.f.: x*(1-x)*(1+4*x+17*x^2+4*x^3+x^4)/(1-38*x^3+x^6). [Colin Barker, Jun 13 2012]

More terms from Bruno Berselli, Jan 29 2013

A133283 Numbers k such that 30*k^2 + 6 is a square.

Original entry on oeis.org

1, 23, 505, 11087, 243409, 5343911, 117322633, 2575754015, 56549265697, 1241508091319, 27256628743321, 598404324261743, 13137638505015025, 288429642786068807, 6332314502788498729, 139022489418560903231, 3052162452705551372353, 67008551470103569288535

Offset: 1

Richard Choulet, Oct 16 2007

From Klaus Purath, Apr 19 2025: (Start)
Nonnegative solutions to the Diophantine equation 5*a(n)^2 - 6*b(n)^2 = -1. The corresponding b(n) are A157014(n). Note that (a(n+1)^2 - a(n)*a(n+2))/4 = 6 and (b(n)*b(n+2) - b(n+1)^2)/4 = 5.
(a(n) + b(n))/2 = (a(n+1) - b(n+1))/2 = A077421(n-1) = Lucas U(22,1). Also a(n)*b(n+1) - a(n+1)*b(n) = -2.
a(n+1) = (t(i+2*n+1) - t(i))/(t(i+n+1) - t(i+n)) as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0 where (t) is a sequence satisfying t(i+3) = 23*t(i+2) - 23*t(i+1) + t(i) or t(i+2) = 22*t(i+1) - t(i), regardless of the initial values and including this sequence itself. (End)

Colin Barker, Table of n, a(n) for n = 1..700
Index entries for linear recurrences with constant coefficients, signature (22,-1).

Cf. A221874.

a:=[1,23];; for n in [3..20] do a[n]:=22*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2020

I:=[1,23]; [n le 2 select I[n] else 22*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Jan 13 2020

a[1]:=1: a[2]:=23: for n to 14 do a[n+2]:=22*a[n+1]-a[n] end do: seq(a[n],n= 1..16); # Emeric Deutsch, Oct 24 2007

Table[n /. {ToRules[Reduce[n > 0 && k >= 0 && 30*n^2+6 == k^2, n, Integers] /. C[1] -> c]} // Simplify, {c, 1, 20}] // Flatten // Union (* Jean-François Alcover, Dec 19 2013 *)
Rest@ CoefficientList[Series[x(1+x)/(1-22x+x^2), {x,0,20}], x] (* Michael De Vlieger, Jul 14 2016 *)
LinearRecurrence[{22,-1},{1,23},20] (* Harvey P. Dale, Sep 22 2017 *)
Table[ChebyshevU[n-1, 11] + ChebyshevU[n-2, 11], {n, 20}] (* G. C. Greubel, Jan 13 2020 *)

Vec(x*(1+x)/(1-22*x+x^2) + O(x^20)) \\ Colin Barker, Jul 14 2016

vector(20, n, polchebyshev(n-1,2,11) + polchebyshev(n-2,2,11) ) \\ G. C. Greubel, Jan 13 2020

[chebyshev_U(n-1,11) + chebyshev_U(n-2,11) for n in (1..20)] # G. C. Greubel, Jan 13 2020

a(n+2) = 22*a(n+1) - a(n); a(n+1) = 11*a(n) + 2*sqrt(30*a(n)^2 + 6).
a(n) = (sqrt(30)/10 - 1/2)*(11 + 2*sqrt(30))^n - (sqrt(30)/10 + 1/2) * (11 - 2*sqrt(30))^n. - Emeric Deutsch, Oct 24 2007
G.f.: x*(1+x)/(1-22*x+x^2). - R. J. Mathar, Nov 14 2007
a(n) = A077421(n) + A077421(n-1). - R. J. Mathar, Feb 19 2016
a(n) = Chebyshev(n-1, 11) + Chebyshev(n-2, 11). - G. C. Greubel, Jan 13 2020

More terms from Emeric Deutsch, Oct 24 2007

A221875 Numbers m such that 10*m^2 - 6 is a square.

Original entry on oeis.org

1, 7, 31, 265, 1177, 10063, 44695, 382129, 1697233, 14510839, 64450159, 551029753, 2447408809, 20924619775, 92937084583, 794584521697, 3529161805345, 30173287204711, 134015211518527, 1145790329257321, 5089048875898681, 43509859224573487

Offset: 1

Bruno Berselli, Jan 28 2013

See the first two comments on A221874.

For the corresponding numbers whose square is 10*m^2 - 6, see A281647. - Jon E. Schoenfield, Aug 05 2023

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,38,0,-1).

Cf. A221874, A281647.

m:=22; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-x)*(1+8*x+x^2)/((1-6*x-x^2)*(1+6*x-x^2))));

I:=[1, 7, 31, 265]; [n le 4 select I[n] else 38*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Mar 25 2013

LinearRecurrence[{0, 38, 0, -1}, {1, 7, 31, 265}, 22]

makelist(expand(((5-sqrt(10)*(-1)^n)*(3+sqrt(10))^(2*floor(n/2))+(5+sqrt(10)*(-1)^n)*(3-sqrt(10))^(2*floor(n/2)))/10), n, 1, 22);

G.f.: x*(1-x)*(1+8*x+x^2)/((1-6*x-x^2)*(1+6*x-x^2)).
a(n) = ((5-t*(-1)^n)*(3+t)^(2*floor(n/2)) + (5+t*(-1)^n)*(3-t)^(2*floor(n/2)))/10, where t=sqrt(10).
a(n)*a(n-3) - a(n-1)*a(n-2) = 36 + 12(-1)^n.

A359800 a(n) is the least m such that the concatenation of n^2 and m is a square.

Original entry on oeis.org

6, 9, 61, 9, 6, 1, 284, 516, 225, 489, 104, 4, 744, 249, 625, 3201, 444, 9, 201, 689, 4201, 416, 984, 4801, 681, 5201, 316, 996, 5801, 601, 6201, 144, 936, 6801, 449, 7201, 7401, 804, 7801, 225, 8201, 8401, 6, 8801, 9001, 9201, 9401, 324, 9801, 19344, 769, 38025

Offset: 1

Mohammed Yaseen, Jan 13 2023

The only one-digit terms are 1, 4, 6 and 9. Proof: Squares mod 10 are 0, 1, 4, 5, 6 and 9. Concatenation of a square and 0 is 10 times that square, which is not a square. So 0 is ruled out. Squares with last digit 5 have second last digit 2. Since no square ends in 2, 5 is also ruled out.
From Thomas Scheuerle, Jan 14 2023: (Start)
The only term with two digits is a(3) = 61.
Some terms with an odd number of digits appear infinitely often, for example, 516 for n = 8, 1352, 632958674, ... .
If a term has an even number of digits and is of the form 1+2*k*10^(d+1) with 10^d <= 2*k < 10^(d+1), then it appears only once at k = n in this sequence. Are terms with an even number of digits possible which are not of this form? (End)

			For n=3, 61 is the least number m such that the concatenation of 3^2 and m is a square: 961 = 31^2. So a(3) = 61.
For n=7, 284 is the least number m such that the concatenation of 7^2 and m is a square: 49284 = 222^2. So a(7) = 284.

Cf. A000290, A071176, A075836, A084070, A221874, A246560.

a(n)={my(m=n^2, b=1); while(1, m*=10; my(r=(sqrtint(m+b-1)+1)^2-m); b*=10; if(rAndrew Howroyd, Jan 13 2023

from math import isqrt
def a(n):
    t, k = str(n*n), isqrt(10*n**2)
    while not (s:=str(k*k)).startswith(t) or s[len(t)]=="0": k += 1
    return int(s[len(t):])
print([a(n) for n in range(1, 53)]) # Michael S. Branicky, Jan 15 2023

from math import isqrt
from sympy.ntheory.primetest import is_square
def A359800(n):
    m = 10*n*n
    if is_square(m): return 0
    a = 1
    while (k:=(isqrt(a*(m+1)-1)+1)**2-m*a)>=10*a:
        a *= 10
    return k # Chai Wah Wu, Feb 15 2023

a(n) = A071176(n^2) = A071176(A000290(n)).
From Thomas Scheuerle, Jan 13 2023: (Start)
a(A084070(n)) = 1.
a(2*A084070(n)) = 4.
a(A221874(n)) = 6.
a(A075836(n)) = 9. (End)

cp's OEIS Frontend

A097315 Pell equation solutions (3*b(n))^2 - 10*a(n)^2 = -1 with b(n) = A097314(n), n >= 0.

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A129556 Numbers k such that the k-th centered pentagonal number A005891(k) = (5k^2 + 5k + 2)/2 is a square.

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A084070 a(n) = 38*a(n-1) - a(n-2), with a(0)=0, a(1)=6.

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A075836 Numbers k such that 10*k^2 + 9 is a square.

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A052454 Positive integer values of k such that 10*k^2 - 9 is a square.

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A133283 Numbers k such that 30*k^2 + 6 is a square.

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Programs

GAP

A097315 Pell equation solutions (3b(n))^2 - 10a(n)^2 = -1 with b(n) = A097314(n), n >= 0.