A075844 -id:A075844 - cp's OEIS frontend

A075839 Numbers k such that 11*k^2 - 2 is a square.

1, 19, 379, 7561, 150841, 3009259, 60034339, 1197677521, 23893516081, 476672644099, 9509559365899, 189714514673881, 3784780734111721, 75505900167560539, 1506333222617099059, 30051158552174420641, 599516837820871313761, 11960285597865251854579

Offset: 1

Gregory V. Richardson, Oct 14 2002

Lim_{n -> infinity} a(n)/a(n-1) = 10 + 3*sqrt(11).

Positive values of x (or y) satisfying x^2 - 20xy + y^2 + 18 = 0. - Colin Barker, Feb 18 2014

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.

G. C. Greubel, Table of n, a(n) for n = 1..750 (terms 1..200 from Vincenzo Librandi)
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 two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (20,-1).

Row 20 of array A094954.
Cf. A075844, A221762, A041015.
Cf. similar sequences listed in A238379.

a:=[1,19];; for n in [3..20] do a[n]:=20*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

I:=[1,19]; [n le 2 select I[n] else 20*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 20 2014

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},{1,19},20] (* Harvey P. Dale, Apr 13 2012 *)
Rest@CoefficientList[Series[x*(1-x)/(1-20x+x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Feb 20 2014 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041015 *)
a[11, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)+poltchebi(n),x,10)/11

def A075839_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1-x)/(1-20*x+x^2) ).list()
a=A075839_list(20); a[1:] # G. C. Greubel, Dec 06 2019

11*a(n)^2 - 9*A083043(n)^2 = 2.
a(n) = ((3+sqrt(11))*(10+3*sqrt(11))^(n-1) - (3-sqrt(11))*(10-3*sqrt(11))^(n-1) )/(2*sqrt(11)). - Dean Hickerson, Dec 09 2002
From Michael Somos, Oct 29 2002: (Start)
G.f.: x*(1-x)/(1-20*x+x^2).
a(n) = 20*a(n-1) - a(n-2), n>1. (End)
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 18). - Benoit Cloitre, Dec 06 2002
a(-n+1) = a(n). - Michael Somos, Apr 18 2003
E.g.f.: (1/11)*exp(10*x)*(11*cosh(3*sqrt(11)*x) - 3*sqrt(11)*sinh(3*sqrt(11)*x)) - 1. - Stefano Spezia, Dec 06 2019

More terms from Colin Barker, Feb 18 2014

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.

A378908 Square array, read by descending antidiagonals, where each row n comprises the integers w >= 1 such that A000037(n)*w^2+4 is a square.

Original entry on oeis.org

4, 24, 2, 140, 8, 1, 816, 30, 3, 4, 4756, 112, 8, 40, 6, 27720, 418, 21, 396, 96, 2, 161564, 1560, 55, 3920, 1530, 12, 12, 941664, 5822, 144, 38804, 24384, 70, 456, 6, 5488420, 21728, 377, 384120, 388614, 408, 17316, 120, 1, 31988856, 81090, 987, 3802396

Offset: 1

Charles L. Hohn, Dec 10 2024

Also, integers w >= 1 for each row n >= 1 such that z+(1/z) is an integer, where x = A000037(n), y = w*sqrt(x), and z = (y+ceiling(y))/2.
All terms of row n are positive integer multiples of T(n, 1).
Limit_{k->oo} T(n, k+1)/T(n, k) = (sqrt(b^2-4)+b)/2 where b=T(n, 2)/T(n, 1).

			n=row index; x=nonsquare integer of index n (A000037(n)):
 n  x    T(n, k)
------+---------------------------------------------------------------------
 1  2 |  4,   24,   140,     816,      4756,       27720,        161564, ...
 2  3 |  2,    8,    30,     112,       418,        1560,          5822, ...
 3  5 |  1,    3,     8,      21,        55,         144,           377, ...
 4  6 |  4,   40,   396,    3920,     38804,      384120,       3802396, ...
 5  7 |  6,   96,  1530,   24384,    388614,     6193440,      98706426, ...
 6  8 |  2,   12,    70,     408,      2378,       13860,         80782, ...
 7 10 | 12,  456, 17316,  657552,  24969660,   948189528,   36006232404, ...
 8 11 |  6,  120,  2394,   47760,    952806,    19008360,     379214394, ...
 9 12 |  1,    4,    15,      56,       209,         780,          2911, ...
10 13 |  3,   33,   360,    3927,     42837,      467280,       5097243, ...
11 14 |  8,  240,  7192,  215520,   6458408,   193536720,    5799643192, ...
12 15 |  2,   16,   126,     992,      7810,       61488,        484094, ...
13 17 | 16, 1056, 69680, 4597824, 303386704, 20018924640, 1320945639536, ...
14 18 |  8,  272,  9240,  313888,  10662952,   362226480,   12305037368, ...
...

Rows from n = 1 (nonzero terms): A005319, A052530, A001906, A122652, A001080*2, A001542, A239365, A075844, A001353, A075835, A068204*2, A001090*2, A121740*2, A202299*2.

Cf. A000037, A002420, A033999, A048942, A180495, A298675.

row(n)={my(v=List()); for(t=3, oo, if((t^2-4)%x>0 || !issquare((t^2-4)/x), next); listput(v, sqrtint((t^2-4)/x)); break); listput(v, v[1]*sqrtint(v[1]^2*x+4)); while(#v<10, listput(v, v[#v]*(v[2]/v[1])-v[#v-1])); Vec(v)}
for(n=1, 20, x=n+floor(1/2+sqrt(n)); print (n, " ", x, " ", row(n)))

For x = A000037(n) (nonsquare integer of index n):
If x is not the sum of 2 squares, then T(n, 1) = A048942(n); otherwise, T(n, 1) is a positive integer multiple of A048942(n).
For j in {-2, 1, 2, 4}, if x-j is a square (except 2-2=0^2 or 5-1=2^2), then T(n, 1) = (4/abs(j))*sqrt(x-j) and T(n, 2) = T(n, 1)^3/(4/abs(j)) + sign(j)*2*T(n, 1).
For j in {1, 4}, if x+j is a square, then T(n, 1) = 2/sqrt(4/j) and T(n, 2) = (4/j)*sqrt(x+j).
For k >= 2, T(n, k) = T(n, k-1)*sqrt(T(n, 1)^2*x+4) - [k>=3]*T(n, k-2).
T(n, 2) = Sum_{i=0..oo}(T(n, 1)^(2-2*i) * x^((1-2*i)/2) * A002420(i) * A033999(i)).
If T(n, 1) is even, then T(n, 2) = T(n, 1)*A180495(n); if T(n, 1) is odd and x is even, then T(n, 2) = T(n, 1)*sqrt(A180495(n)+2); if T(n, 1) and x are both odd, then T(n, 2) is a factor of T(n, 1)*A180495(n).
For k >= 3, T(n, k) = T(n, k-1)*(T(n, 2)/T(n, 1)) - T(n, k-2) = T(n, 1)*A298675(T(n, 2)/T(n, 1), k-1) + T(n, k-2) = sqrt((A298675(T(n, 2)/T(n, 1), k)^2-4)/x).

cp's OEIS Frontend

A075839 Numbers k such that 11*k^2 - 2 is a square.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Magma

Maple

Mathematica

Maxima

Formula

A378908 Square array, read by descending antidiagonals, where each row n comprises the integers w >= 1 such that A000037(n)*w^2+4 is a square.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula