A221762 -id:A221762 - 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

A198947 x values in the solution to 11*x^2 - 10 = y^2.

Original entry on oeis.org

1, 7, 13, 139, 259, 2773, 5167, 55321, 103081, 1103647, 2056453, 22017619, 41025979, 439248733, 818463127, 8762957041, 16328236561, 174819892087, 325746268093, 3487634884699, 6498597125299, 69577877801893, 129646196237887, 1388069921153161, 2586425327632441

Offset: 1

Sture Sjöstedt, Oct 31 2011

When are n and 11*n+1 perfect squares? This problem gives rise to the Diophantine equation 11*x^2 - 10 = y^2.

Positive values of x (or y) satisfying x^2 - 20xy + y^2 + 90 = 0. - Colin Barker, Feb 18 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0, 20, 0, -1).

Cf. A198949, A221762.

m:=26; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1-x)*(1+8*x+x^2)/(1-20*x^2+x^4))); // Bruno Berselli, Nov 07 2011

LinearRecurrence[{0,20,0,-1},{1,7,13,139},30] (* Vincenzo Librandi, Feb 06 2012 *)

makelist(expand(((11+(-1)^n*sqrt(11))*(10-3*sqrt(11))^floor(n/2)+(11-(-1)^n*sqrt(11))*(10+3*sqrt(11))^floor(n/2))/22), n, 1, 25); /* Bruno Berselli, Nov 07 2011 */

v=vector(25); v[1]=1; v[2]=7; v[3]=13; v[4]=139; for(i=5, #v, v[i]=20*v[i-2]-v[i-4]); v \\ Bruno Berselli, Nov 07 2011

a(n+4) = 20*a(n+2) - a(n) with a(1)=1, a(2)=7, a(3)=13, a(4)=139.
From Bruno Berselli, Nov 06 2011:  (Start)
G.f.: x*(1-x)*(1+8*x+x^2)/(1-20*x^2+x^4).
a(n) = ((11+(-1)^n*t)*(10-3*t)^floor(n/2)+(11-(-1)^n*t)*(10+3*t)^floor(n/2))/22 with t=sqrt(11).  (End).

Terms a(1)-a(7) confirmed, a(8)-a(15) added by John W. Layman, Nov 04 2011

a(16)-a(25) from Bruno Berselli, Nov 06 2011

A001084 a(n) = 20*a(n-1) - a(n-2) with a(0) = 0, a(1) = 3.

Original entry on oeis.org

0, 3, 60, 1197, 23880, 476403, 9504180, 189607197, 3782639760, 75463188003, 1505481120300, 30034159217997, 599177703239640, 11953519905574803, 238471220408256420, 4757470888259553597, 94910946544782815520, 1893461460007396756803, 37774318253603152320540

Offset: 0

N. J. A. Sloane

Also 11*x^2+1 is a square. n=11 in PARI script below. - Cino Hilliard, Mar 08 2003

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 11*y^2 = 1; the corresponding x values are in A001085. - Vincenzo Librandi, Nov 12 2010 [edited by Jon E. Schoenfield, May 04 2014]

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
"Questions D'Arithmetique", Question 3686, Solution by H.L. Mennessier, Mathesis, 65(4, Supplement) 1956, pp. 1-12.

T. D. Noe, Table of n, a(n) for n = 0..200
H. Brocard, Notes élémentaires sur le problème de Peel, Nouvelle Correspondance Mathématique, 4 (1878), 161-169.
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (20,-1).

Equals 3 * A075843.

Cf. A001085, A221762.

I:=[0,3]; [n le 2 select I[n] else 20*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001084:=3*z/(1-20*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

LinearRecurrence[{20, -1}, {0, 3}, 20] (* T. D. Noe, Dec 19 2011 *)
CoefficientList[Series[3*x/(1 - 20*x + x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
Table[3 ChebyshevU[-1 + n, 10], {n, 0, 18}] (* Herbert Kociemba, Jun 05 2022 *)

nxsqp1(m,n) = { for(x=1,m, y = n*x*x+1; if(issquare(y),print1(x" ")) ) }

x='x+O('x^30); concat([0], Vec(3*x/(1 - 20*x + x^2))) \\ G. C. Greubel, Dec 20 2017

Limit_{n->oo} a(n)/a(n-1) = 10 + 3*sqrt(11); for all n in the sequence, 11*n^2 + 1 is a perfect square. - Gregory V. Richardson, Oct 06 2002
a(n) = ((10 + 3*sqrt(11))^n - (10 - 3*sqrt(11))^n) / (2*sqrt(11)). - Gregory V. Richardson, Oct 06 2002
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 19*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 21*(a(n-1) - a(n-2)) + a(n-3). (End)
G.f.: 3*x/(1 - 20*x + x^2). - G. C. Greubel, Dec 20 2017
E.g.f.: exp(10*x)*sinh(3*sqrt(11)*x)/sqrt(11). - Stefano Spezia, Aug 16 2024

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

A221763 Numbers m such that 11*m^2 - 7 is a square.

Original entry on oeis.org

1, 4, 16, 79, 319, 1576, 6364, 31441, 126961, 627244, 2532856, 12513439, 50530159, 249641536, 1008070324, 4980317281, 20110876321, 99356704084, 401209456096, 1982153764399, 8004078245599, 39543718583896, 159680355455884, 788892217913521

Offset: 1

Bruno Berselli, Jan 24 2013

See the first comment of A221762.
a(n) == 1 (mod 3).
a(n+1)/a(n) tends alternately to (2+sqrt(11))^2/7 and (5+sqrt(11))^2/14; a(n+2)/a(n) tends to A176395^2/2.
Positive values of x (or y) satisfying x^2 - 20xy + y^2 + 63 = 0. - Colin Barker, Feb 18 2014

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. A221762.

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

I:=[1,4,16,79]; [n le 4 select I[n] else 20*Self(n-2)-Self(n-4): n in [1..25]]; // Vincenzo Librandi, Aug 18 2013

A221763:=proc(q)
local n;
for n from 1 to q do if type(sqrt(11*n^2-7), integer) then print(n);
fi; od; end:
A221763(1000); # Paolo P. Lava, Feb 19 2013

LinearRecurrence[{0, 20, 0, -1}, {1, 4, 16, 79}, 24]
CoefficientList[Series[(1 + 4 x - 4 x^2 - x^3)/(1 - 20 x^2 + x^4), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 18 2013 *)

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

G.f.: x*(1+4*x-4*x^2-x^3)/(1-20*x^2+x^4).
a(n) = ((11+2*t*(-1)^n)*(10-3*t)^floor(n/2)+(11-2*t*(-1)^n)*(10+3*t)^floor(n/2))/22, where t=sqrt(11).
a(n)*a(n-3)-a(n-1)*a(n-2) = (3/2)*(9+(-1)^n).

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

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A198947 x values in the solution to 11*x^2 - 10 = y^2.

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A001084 a(n) = 20*a(n-1) - a(n-2) with a(0) = 0, a(1) = 3.

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

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GAP

Magma

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A221763 Numbers m such that 11*m^2 - 7 is a square.

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Comments

Links

Crossrefs

Programs

Magma

Magma

Maple

Mathematica

Maxima

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