A075839 -id:A075839 - cp's OEIS frontend

A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 13, 1, 1, 5, 19, 41, 34, 1, 1, 6, 29, 91, 153, 89, 1, 1, 7, 41, 169, 436, 571, 233, 1, 1, 8, 55, 281, 985, 2089, 2131, 610, 1, 1, 9, 71, 433, 1926, 5741, 10009, 7953, 1597, 1, 1, 10, 89, 631, 3409, 13201, 33461, 47956, 29681

Offset: 1

Ralf Stephan, May 31 2004

Also, values of polynomials with coefficients in A098493 (see Fink et al.). See A098495 for negative k.

Number of dimer tilings of the graph S_{k-1} X P_{2n-2}.

			1,1,1,1,1,1,1,1,1,1,1,1,1,1, ...
1,2,5,13,34,89,233,610,1597, ...
1,3,11,41,153,571,2131,7953, ...
1,4,19,91,436,2089,10009,47956, ...
1,5,29,169,985,5741,33461,195025, ...
1,6,41,281,1926,13201,90481,620166, ...

Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114.
Elizabeth Wilmer, A note on Stephan's conjecture 87
Elizabeth Wilmer, A note on Stephan's conjecture 87 [cached copy]

Rows are first differences of rows in array A073134.
Rows 2-14 are A000012, A001519, A079935/A001835, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570. Other rows: A007805 (k=18), A075839 (k=20), A077420 (k=34), A078988 (k=66).
Columns include A028387. Diagonals include A094955, A094956. Antidiagonal sums are A094957.

max = 14; row[k_] := Rest[ CoefficientList[ Series[ x*(1-x)/(1-k*x+x^2), {x, 0, max}], x]]; t = Table[ row[k], {k, 2, max+1}]; Flatten[ Table[ t[[k-n+1, n]], {k, 1, max}, {n, 1, k}]] (* Jean-François Alcover, Dec 27 2011 *)

PARI
```
T(k,n)=polcoeff(x*(1-x)/(1-k*x+x*x),n)
```

Recurrence: T(k, 1) = 1, T(k, 2) = k-1, T(k, n) = kT(k, n-1) - T(k, n-2).
For n>3, T(k, n) = [k(k-2) + T(k, n-1)T(k, n-2)] / T(k, n-3).
T(k, n+1) = S(n, k) - S(n-1, k) = U(n, k/2) - U(n-1, k/2), with S, U = Chebyshev polynomials of second kind.
T(k+2, n+1) = Sum[i=0..n, k^(n-i) * C(2n-i, i)] (from comments by Benoit Cloitre).

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

Original entry on oeis.org

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449

Offset: 0

Bruno Berselli, Feb 25 2014

First bisection of A041611.

Bruno Berselli, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (36,-1).

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

[n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020

CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]

a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016

m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
print(f.coefficients())

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

A083043 Integers y such that 11x^2 - 9y^2 = 2 for some integer x.

Original entry on oeis.org

1, 21, 419, 8359, 166761, 3326861, 66370459, 1324082319, 26415275921, 526981436101, 10513213446099, 209737287485879, 4184232536271481, 83474913437943741, 1665314036222603339, 33222805811014123039

Offset: 1

Michael Somos, Apr 17 2003

G. C. Greubel, Table of n, a(n) for n = 1..750
Tanya Khovanova, Recursive Sequences
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (20,-1).

Cf. A075839, A075843.

a:=[1,21];; for n in [3..20] do a[n]:=20*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

I:=[1,21]; [n le 2 select I[n] else 20*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 06 2019

seq(coeff(series( x*(1+x)/(1-20*x+x^2), x, n+1), x, n), n = 1..20); # G. C. Greubel, Dec 06 2019

LinearRecurrence[{20,-1},{1,21},20] (* Harvey P. Dale, Jun 02 2014 *)

a(n)=subst(poltchebi(n+1)-poltchebi(n),x,10)/9

def A083043_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1+x)/(1-20*x+x^2) ).list()
a=A083043_list(20); a[1:] # G. C. Greubel, Dec 06 2019

G.f.: x*(1+x)/(1-20*x+x^2).
a(n) = 20*a(n-1) - a(n-2).
a(1-n) = -a(n).
11*A075839(n)^2 - 9*a(n)^2 = 2.
a(n+1) = 10*a(n) + sqrt(99*a(n)^2 + 22). - Richard Choulet, Sep 27 2007
a(n) = ((3 + sqrt(11))*(10 + 3*sqrt(11))^(n-1) + (3 - sqrt(11))*(10 - 3*sqrt(11))^(n-1))/6. - G. C. Greubel, Dec 06 2019
E.g.f.: 1 + (1/3)*exp(10*x)*(-3*cosh(3*sqrt(11)*x) + sqrt(11)*sinh(3*sqrt(11)*x)). - Stefano Spezia, Dec 06 2019 after G. C. Greubel

Corrected by T. D. Noe, Nov 07 2006

Offset changed to 1 by G. C. Greubel, Dec 06 2019

A221762 Numbers m such that 11*m^2 + 5 is a square.

Original entry on oeis.org

1, 2, 22, 41, 439, 818, 8758, 16319, 174721, 325562, 3485662, 6494921, 69538519, 129572858, 1387284718, 2584962239, 27676155841, 51569671922, 552135832102, 1028808476201, 11015040486199, 20524599852098, 219748673891878, 409463188565759

Offset: 1

Bruno Berselli, Jan 24 2013

Corresponding squares are: 16, 49, 5329, 18496, 2119936, 7360369, 843728209, 2929407376, ... (subsequence of A016778).
The Diophantine equation 11*x^2+k = y^2, for |k|<11, has integer solutions with the following k values:
k = -10, the nonnegative x values are in A198947;
k =  -8,            "                    2*A075839;
k =  -7,            "                    A221763;
k =  -2,            "                    A075839;
k =   1,            "                    A001084;
k =   4,            "                    A075844;
k =   5,            "                    this sequence;
k =   9,            "                    3*A001084.
Also, the Diophantine equation h*x^2+5 = y^2 has infinitely many integer solutions for h = 5, 11, 19, 20, 29, 31, 41, 44, 55, 59, ...
a(n+1)/a(n) tends alternately to (1+sqrt(11))^2/10 and (4+sqrt(11))^2/5.
a(n+2)/a(n) tends to A176395^2/2.

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

Cf. A001084, A075839, A075844, A198947, A198949, A221763.

Cf. A049629 (numbers m such that 20*m^2 + 5 is a square), A075796 (numbers m such that 5*m^2 + 5 is a square).

m:=24; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+2*x^2+x^3)/(1-20*x^2+x^4)));

I:=[1,2,22,41]; [n le 4 select I[n] else 20*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Aug 18 2013

A221762:=proc(q)
local n;
for n from 1 to q do if type(sqrt(11*n^2+5), integer) then print(n);
fi; od; end:
A221762(1000); # Paolo P. Lava, Feb 19 2013

LinearRecurrence[{0, 20, 0, -1}, {1, 2, 22, 41}, 24]
CoefficientList[Series[(1 + 2 x + 2 x^2 + x^3)/(1 - 20 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(expand(((-11*(-1)^n+4*sqrt(11))*(10+3*sqrt(11))^floor(n/2)-(11*(-1)^n+4*sqrt(11))*(10-3*sqrt(11))^floor(n/2))/22), n, 1, 24);

G.f.: x*(1+2*x+2*x^2+x^3)/(1-20*x^2+x^4).
a(n) = -a(1-n) = ((-11*(-1)^n+4*t)*(10+3*t)^floor(n/2)-(11*(-1)^n+4*t)*(10-3*t)^floor(n/2))/22, where t=sqrt(11).
a(n) = 20*a(n-2) - a(n-4) for n>4, a(1)=1, a(2)=2, a(3)=22, a(4)=41.
a(n)*a(n-3)-a(n-1)*a(n-2) = -(3/2)*(9-7*(-1)^n).
a(n+1) + a(n-1) =  A198949(n), with a(0)=-1.
2*a(n-1) - a(n) =  A001084(n/2-1) for even n.

A075844 Numbers k such that 11*k^2 + 4 is a square.

Original entry on oeis.org

0, 6, 120, 2394, 47760, 952806, 19008360, 379214394, 7565279520, 150926376006, 3010962240600, 60068318435994, 1198355406479280, 23907039811149606, 476942440816512840, 9514941776519107194, 189821893089565631040

Offset: 0

Gregory V. Richardson, Oct 14 2002

Lim. n-> Inf. a(n)/a(n-1) = 10 + 3*sqrt(11).

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.

Harvey P. Dale, Table of n, a(n) for n = 0..750
Tanya Khovanova, Recursive Sequences
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 (20,-1).

Cf. A221762.

a:=[0,6];; for n in [3..20] do a[n]:=20*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

R:=PowerSeriesRing(Integers(), 20); [0] cat Coefficients(R!( 6*x/(1 - 20*x + x^2) )); // G. C. Greubel, Dec 06 2019

seq(coeff(series(6*x/(1-20*x+x^2), x, n+1), x, n), n = 0..20); # G. C. Greubel, Dec 06 2019

LinearRecurrence[{20,-1},{0,6},20] (* Harvey P. Dale, May 28 2012 *)

my(x='x+O('x^20)); concat([0], Vec(6*x/(1-20*x+x^2))) \\ G. C. Greubel, Dec 06 2019

def A075844_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 6*x/(1-20*x+x^2) ).list()
A075844_list(20) # G. C. Greubel, Dec 06 2019

a(n) = ((10+3*sqrt(11))^n - (10-3*sqrt(11))^n) / sqrt(11).
a(n) = 20*a(n-1) - a(n-2).
G.f.: 6*x/(1 - 20*x + x^2).
a(n) = (1/3)*(A075839(n+1) - A075839(n)), n>=1. - N. J. A. Sloane, Sep 22 2004
a(n) = 6*A075843(n). - R. J. Mathar, Jul 03 2011

A238240 Positive integers n such that x^2 - 20xy + y^2 + n = 0 has integer solutions.

Original entry on oeis.org

18, 35, 50, 63, 72, 74, 83, 90, 95, 98, 99, 107, 140, 162, 171, 200, 215, 227, 252, 266, 275, 288, 296, 315, 332, 347, 359, 360, 362, 371, 380, 387, 392, 395, 396, 407, 428, 450, 491, 495, 530, 539, 560, 567, 602, 623, 626, 635, 648, 666, 684, 695, 711, 722, 743, 747, 755, 770, 791, 794, 800, 810

Offset: 1

Colin Barker, Feb 20 2014

nonn

Positive integers n such that x^2 - 99 y^2 + n = 0 has integer solutions. - Robert Israel, Oct 22 2024

			63 is in the sequence because x^2 - 20xy + y^2 + 63 = 0 has integer solutions, for example (x, y) = (1, 16).

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A075839 (n = 18), A221763 (n = 63), A198947 (n = 90), A001085 (n = 99).

Cf. A031363, A084917, A237351, A237599, A237606, A237609, A237610, A236330, A236331.

filter:= t -> [isolve(99*y^2 - z^2 = t)] <> []:
select(filter, [$1..1000]); # Robert Israel, Oct 22 2024

Corrected by Robert Israel, Oct 22 2024

A131216 Numbers X such that 99*X^2 - 2178 is a square.

Original entry on oeis.org

11, 209, 4169, 83171, 1659251, 33101849, 660377729, 13174452731, 262828676891, 5243399085089, 104605153024889, 2086859661412691, 41632588075228931, 830564901843165929, 16569665448788089649, 330562744073918627051

Offset: 1

Richard Choulet, Sep 27 2007

G. C. Greubel, Table of n, a(n) for n = 1..750
Index entries for linear recurrences with constant coefficients, signature (20,-1).

Cf. A083043.

a:=[11,209];; for n in [3..20] do a[n]:=20*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( 11*x*(1-x)/(1-20*x+x^2) )); // G. C. Greubel, Dec 06 2019

seq(coeff(series(11*x*(1-x)/(1-20*x+x^2), x, n+1), x, n), n = 0..20); # G. C. Greubel, Dec 06 2019

LinearRecurrence[{20, -1}, {11, 209}, 20] (* G. C. Greubel, Dec 06 2019 *)

my(x='x+O('x^20)); Vec(11*x*(1-x)/(1-20*x+x^2)) \\ G. C. Greubel, Dec 06 2019

def A131216_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 11*x*(1-x)/(1-20*x+x^2) ).list()
A131216_list(20) # G. C. Greubel, Dec 06 2019

a(n+2) = 20*a(n+1) - a(n).
a(n+1) = 10*a(n+1)+ sqrt(99*a(n)^2 -2178).
G.f.: 11*z*(1-z)/(1-20*z+z^2) - Richard Choulet, Oct 09 2007
a(n) = 11*A075839(n). - R. J. Mathar, Aug 22 2012

A159680 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 9n(j) + 1 = a(j)a(j) and 11n(j) + 1 = b(j)b(j) with positive integer numbers.

Original entry on oeis.org

0, 40, 15960, 6352080, 2528111920, 1006182192120, 400457984351880, 159381271589856160, 63433345634778399840, 25246312181370213280200, 10047968814839710107119800, 3999066341994023252420400240, 1591618356144806414753212175760, 633460106679290959048526025552280

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..350
Index entries for linear recurrences with constant coefficients, signature (399,-399,1).

Cf. A075839, A083043, A157456.

R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients(R!(40*x^2/((1-x)*(1-398*x+x^2)))); // G. C. Greubel, Jun 03 2018

for a from 1 by 2 to 100000 do b:=sqrt((9*a*a-2)/7): if (trunc(b)=b) then
n:=(a*a-1)/7: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:

LinearRecurrence[{399,-399,1}, {0,40,15960}, 50] (* G. C. Greubel, Jun 03 2018 *)

a(n) = round((-20+(10+3*sqrt(11))*(199+60*sqrt(11))^(-n)+(10-3*sqrt(11))*(199+60*sqrt(11))^n)/198) \\ Colin Barker, Jul 26 2016

concat(0, Vec(-40*x^2/((x-1)*(x^2-398*x+1)) + O(x^20))) \\ Colin Barker, Jul 26 2016

[(10/99)*(chebyshev_U(n, 199) -397*chebyshev_U(n-1, 199) -1) for n in (1..30)] # G. C. Greubel, Jun 26 2022

The a(j) recurrence is a(1)=1; a(2)=19; a(t+2) = 20*a(t+1) - a(t) resulting in terms 1, 19, 379, 7561, ... (A075839).
The b(j) recurrence is b(1)=1; b(2)=21; b(t+2) = 20*b(t+1) - b(t) resulting in terms 1, 21, 419, 8359, ... (A083043).
The n(j) recurrence is n(0)=n(1)=0; n(2)=40; n(t+3) = 399*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 40, 15960, 6352080 as listed above
G.f.: 40*x^2/((1-x)*(1-398*x+x^2)). - R. J. Mathar, Apr 20 2009
a(n) = (-20 + (10 + 3*sqrt(11))*(199 + 60*sqrt(11))^(-n) + (10 - 3*sqrt(11))*(199 + 60*sqrt(11))^n)/198. - Colin Barker, Jul 26 2016
From G. C. Greubel, Jun 26 2022: (Start)
a(n) = (10/99)*( ChebyshevU(n, 199) - 397*ChebyshevU(n-1, 199) - 1 ).
E.g.f.: (10/99)*(exp(199*x)*( (3*sqrt(11)/10)*sinh(60*sqrt(11)*x) + cosh(60*sqrt(11)*x) ) - exp(x)). (End)

More terms from R. J. Mathar, Apr 20 2009

A269028 a(n) = 40*a(n - 1) - a(n - 2) for n>1, a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 39, 1559, 62321, 2491281, 99588919, 3981065479, 159143030241, 6361740144161, 254310462736199, 10166056769303799, 406387960309415761, 16245352355607326641, 649407706263983649879, 25960062898203738668519, 1037753108221885563090881

Offset: 0

Ilya Gutkovskiy, Feb 18 2016

In general, the ordinary generating function for the recurrence relation b(n) = k*b(n - 1) - b(n - 2) with n>1 and b(0)=1, b(1)=1, is (1 - (k - 1)*x)/(1 - k*x +x^2). This recurrence gives the closed form b(n) = (2^( -n - 1)*((k - 2)*(k - sqrt(k^2 - 4))^n + sqrt(k^2 - 4)*(k - sqrt(k^2 - 4))^n - (k - 2)*(sqrt(k^2 - 4) + k)^n + sqrt(k^2 - 4)*(sqrt(k^2 - 4) + k)^n))/sqrt(k^2 - 4).

Index entries for linear recurrences with constant coefficients, signature (40,-1).

Cf. A001519, A001835, A001653, A049685, A070997, A070998, A072256, A078922, A160682, A007805, A075839, A157014, A159664, A159668, A157877, A238379, A097315.

[n le 2 select 1 else 40*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 19 2016

Table[Cosh[n Log[20 + Sqrt[399]]] - Sqrt[19/21] Sinh[n Log[20 + Sqrt[399]]], {n, 0, 17}]
Table[(2^(-n - 2) (38 (40 - 2 Sqrt[399])^n + 2 Sqrt[399] (40 - 2 Sqrt[399])^n - 38 (40 + 2 Sqrt[399])^n + 2 Sqrt[399] (40 + 2 Sqrt[399])^n))/Sqrt[399], {n, 0, 17}]
LinearRecurrence[{40, -1}, {1, 1}, 17]

G.f.: (1 - 39*x)/(1 - 40*x + x^2).
a(n) = cosh(n*log(20 + sqrt(399))) - sqrt(19/21)*sinh(n*log(20 + sqrt(399))).
a(n) = (2^(-n - 2)*(38*(40 - 2*sqrt(399))^n + 2*sqrt(399)*(40 - 2*sqrt(399))^n - 38*(40 + 2*sqrt(399))^n + 2*sqrt(399)*(40 + 2*sqrt(399))^n))/sqrt(399).
Sum_{n>=0} 1/a(n) = 2.0262989201139499769986...

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A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

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A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

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A083043 Integers y such that 11*x^2 - 9*y^2 = 2 for some integer x.

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A221762 Numbers m such that 11*m^2 + 5 is a square.

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A075844 Numbers k such that 11*k^2 + 4 is a square.

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GAP

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A238240 Positive integers n such that x^2 - 20xy + y^2 + n = 0 has integer solutions.

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A083043 Integers y such that 11x^2 - 9y^2 = 2 for some integer x.

A159680 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 9n(j) + 1 = a(j)a(j) and 11n(j) + 1 = b(j)b(j) with positive integer numbers.