Gregory V. Richardson - cp's OEIS frontend

A235230 Integer radii of circles tiled by square tiles such that the ratio of uncut tiles to cut tiles is an integer and four square tiles meet at the center of the circle.

1, 6, 15, 364, 585, 5052, 9573, 191714, 13682428

Offset: 1

Gregory V. Richardson, Jan 05 2014

It is my conjecture that there are an infinite number of solutions and that they occur by chance, accounting for the widening gaps between valid answers as the number of digits for the sums of tiles increases.
The inspiration for this problem came from Enigma #1686 of the New Scientist Magazine.
The values involved are the following {a(n), #uncut, cut, ratio} : {1, 0, 4, 0}, {6, 88, 44, 2}, {15, 648, 108, 6}, {364, 414700, 2900, 143}, {585, 1072764, 4644, 231}, {5052, 80161536, 40404, 1984}, {9573, 287864220, 76580, 3759},{191714, 115466138200, 1533700, 75286}, {13682428, 588133849050724, 109459412, 5373077}. No further terms up to 15*10^6. - Giovanni Resta, Jan 06 2014

			See picture in Links.

New Scientist Magazine, Enigma #1686, 22 February 2012.
Giovanni Resta, Illustration for a(2) and a(3)
Gregory V. Richardson, QuickBasic 64 program

QuickBASIC
```
' See Links.
```

A076218 Numbers n such that 2n^2 - 3n + 1 is a square.

Original entry on oeis.org

0, 1, 5, 145, 4901, 166465, 5654885, 192099601, 6525731525, 221682772225, 7530688524101, 255821727047185, 8690408031080165, 295218051329678401, 10028723337177985445, 340681375412721826705, 11573138040695364122501, 393146012008229658338305

Offset: 1

Gregory V. Richardson, Nov 03 2002

Limit_{n -> infinity} a(n)/a(n-1) = 33.970562748477140585620264690516... = 17 + 12*sqrt(2).
Conjecture: a nonzero number occurs twice in A055524 if and only if it's in this sequence. - J. Lowell, Jul 23 2016
Equivalently, n=0 or both n-1 and 2*n-1 are perfect squares. - Sture Sjöstedt, Feb 22 2017

			5 is in the sequence since 2*5^2 - 3*5 + 1 = 50 - 15 + 1 = 36 is a square. - _Michael B. Porter_, Jul 24 2016

Colin Barker, Table of n, a(n) for n = 1..650
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: A084703 (k=-1), this sequence (k=3), A278310 (k=-5).

Join[{0},LinearRecurrence[{35,-35,1},{1,5,145},20]] (* Harvey P. Dale, Nov 27 2012 *)

a(n)=if(n>1,([0,1,0;0,0,1;1,-35,35]^n*[145;5;1])[1,1],0) \\ Charles R Greathouse IV, Jul 24 2016

concat(0, Vec(x^2*(1-30*x+5*x^2) / ((1-x)*(1-34*x+x^2)) + O(x^30))) \\ Colin Barker, Nov 21 2016

From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 04 2002: (Start)
a(n) = ( (3+(17+12*sqrt(2))^(n-1)) + (3+(17-12*sqrt(2))^(n-1)) )/8 for n>=1.
a(n) = 35 * a(n-1) - 35 * a(n-2) + a(n-3).
G.f.: (x-30*x^2+5*x^3)/(1-35*x+35*x^2-x^3). (End)
Product of adjacent odd-indexed Pell numbers (A000129). - Gary W. Adamson, Jun 07 2003
sqrt(2) - 1 = 0.414213562... = 2/5 + 2/145 + 2/4901 + 2/166465 + ... = Sum_{n>=2} 2/a(n). - Gary W. Adamson, Jun 07 2003
For n > 0, one more than square of adjacent even-indexed Pell numbers (A000129). - Charlie Marion, Mar 09 2005
a(n) = A001652(n-1) + 2*A001652(n-1)*A001652(n-2) + A001652(n-2) + 2. - Charlie Marion, Nov 24 2018

A075871 Numbers k such that 13*k^2 + 1 is a square.

Original entry on oeis.org

0, 180, 233640, 303264540, 393637139280, 510940703520900, 663200639532988920, 860833919173116097260, 1117361763886065161254560, 1450334708690193406192321620, 1882533334518107155172472208200, 2443526817869794397220462733921980

Offset: 1

Gregory V. Richardson, Oct 16 2002

Limit_{n->infinity} a(n)/a(n-1) = 649 + 180*sqrt(13).

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 13*y^2 = 1. The corresponding x values are in A114047. - Vincenzo Librandi, Aug 08 2010, edited by Jon E. Schoenfield, May 04 2014

Colin Barker, Table of n, a(n) for n = 1..322
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1298,-1).

Cf. A114047.

Cf. A202156.

I:=[0,180]; [n le 2 select I[n] else 1298*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Jun 14 2015

LinearRecurrence[{1298, -1}, {0, 180}, 50] (* Vincenzo Librandi, Jun 14 2015 *)

concat(0, Vec(180*x^2/(1-1298*x+x^2) + O(x^20))) \\ Colin Barker, Jun 13 2015

a(n) = 1298*a(n-1) - a(n-2), n>1. - Michael Somos, Oct 30 2002
a(n) = ((649 + 180*sqrt(13))^n - (649 - 180*sqrt(13))^n) / (2*sqrt(13)).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 1297*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 1299*(a(n-1) - a(n-2)) + a(n-3). (End)
G.f.: 180*x^2/(1-1298*x+x^2). - Philippe Deléham, Nov 18 2008

A075886 Number of cubes at generation n when building fractal cube with edge ratio of 1/2.

Original entry on oeis.org

1, 6, 30, 150, 726, 3510, 16854, 80886, 387606, 1857078, 8894550, 42598518, 204000918, 976929462, 4678286550, 22403128566, 107282481942, 513746046774, 2460185076054, 11781130000758, 56416485185430, 270162504104886

Offset: 1

Gregory V. Richardson, Oct 17 2002

This sequence gives the number of cubes in each generation "#n" of the fractal cube with edge ratio of 1/2. To create the fractal cube, start with a cube of edge length 1 (the mother cube). Now attach 6 smaller cubes of edge length 1/2 to each free face of the mother cube. Keep edges parallel and place the smaller cubes centrally on each face of the mother cube. These 6 cubes are generation #2.
Next place 30 cubes of edge length 1/4 onto each free face of the 6 generation #2 cubes. Continue to place smaller cubes on each remaining free face. However, the generation #4 cubes touch the mother cube face-to-face such that some of the 5th generation cubes lie totally submerged within the mother cube. In later generations other cubes are submerged within cubes from their previous generations, too. Submerged cubes are not permitted, so these are not included in the totals.
The final displaced volume of the fractal cube (with R = 1/2) as n -> infinity = 2 22/23. If the intersections ARE allowed, the final total volume of all cubes is 3. In addition to the mother cube, there are 5 other types of cubes based on how many larger and smaller cubes are adjacent. The mother cube gives birth to 6 "A" cubes. An "A" cube begets another "A" and 4 "B" cubes. A "B" cube begets an "A", 3 "B" cubes and a "C". A "C" begets 3 "D" cubes and an "E". A "D" begets an "A", 2 "B" cubes and 2 "C" cubes. An "E" cube begets 2 "D" cubes and 2 "E" cubes.
Summing these up, the new number of "A" cubes = A + B + D. New "B" cubes = 4*A + 3*B + 2*D. New "C" cubes = B + 2*D. New "D" cubes = 3*C + 2*E. New "E" cubes = C + 2*E. In the final fractal cube, an "A" cube abuts one larger cube and its children abut no larger cubes. A "B" cube abuts one larger cube and its children abut one larger cube. A "C" cube abuts two larger cubes and its children abut one larger cube. A "D" cube abuts one larger cube and its children abut two larger cubes.
An "E" cube abuts two larger cubes and its children abut two larger cubes. The final fractal cube appears similar to a regular octahedron but apparently exhibits small indentations that differentiate it therefrom.

Index entries for linear recurrences with constant coefficients, signature (5,4,-24).

Rest[CoefficientList[Series[x (-4x^2+x+1)/(24x^3-4x^2-5x+1),{x,0,30}],x]] (* or *) LinearRecurrence[{5,4,-24},{1,6,30},30] (* Harvey P. Dale, Mar 17 2012 *)

50 word *: point 80
100 cls: print: print "Fractal Cube Ratio = 1/2": print
110 locate 3: print "1";: locate 9: print "1"
120 locate 3: print "2";: locate 9: print "6"
130 A=6: B=0: C=0: D=0: E=0: G=2: Cc=2: V=1.75: Dv=1/8
140 G=G+1: Cc=Cc+1: Dv=Dv/8
150 Na=A+B+D: Nb=4*A+3*B+2*D
160 Nc=B+2*D: Nd=3*C+2*E: Ne=C+2*E
170 Ncubes = Na+Nb+Nc+Nd+Ne
180 A=Na: B=Nb: C=Nc: D=Nd: E=Ne: V=V+Ncubes*Dv
190 Locate 2: print G;: locate 8: print Ncubes
200 if Cc<40 then 140
210 print: locate 4: print "V = ";: print using(2,60),V: print
220 print: print " Push 'S' to stop or space bar to continue": print
230 A$=inkey: if A$="" then 230
240 if A$="s" or A$="S" then print: end
250 Cc=0: goto 140

G.f.: x(-4*x^2 + x + 1)/(24*x^3 - 4*x^2 - 5*x + 1). - Ralf Stephan, May 09 2004

a(n) = 5*a(n-1) + 4*a(n-2) - 24*a(n-3). - Harvey P. Dale, Mar 17 2012

A077444 Numbers k such that (k^2 + 4)/2 is a square.

Original entry on oeis.org

2, 14, 82, 478, 2786, 16238, 94642, 551614, 3215042, 18738638, 109216786, 636562078, 3710155682, 21624372014, 126036076402, 734592086398, 4281516441986, 24954506565518, 145445522951122, 847718631141214, 4940866263896162, 28797478952235758, 167844007449518386

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "(k^2 + 4)/2 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = -4.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(2)). - Thomas Baruchel, Sep 15 2003
Equivalently, 2*n^2 + 8 is a square.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = 2 + n^2/2. - Ctibor O. Zizka, Nov 09 2009
The continued fraction [a(n);a(n),a(n),...] = (1 + sqrt(2))^(2*n-1). - Thomas Ordowski, Jun 07 2013
a((p+1)/2) == 2 (mod p) where p is an odd prime. - Altug Alkan, Mar 17 2016

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.

Amiram Eldar, Table of n, a(n) for n = 1..1306
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Tanya Khovanova, Recursive Sequences
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
J. J. O'Connor and E. F. Robertson, Pell's Equation. [Broken link]
Eric Weisstein's World of Mathematics, Pell Equation.
Eric Weisstein's World of Mathematics, NSW Number.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

(A077445(n))^2 - 2*a(n) = 8.
First differences of A001541.
Pairwise sums of A001542.
Bisection of A002203 and A080039.
Cf. A001653.

[n: n in [0..10^8] | IsSquare((n^2 + 4) div 2)]; // Vincenzo Librandi, Jun 20 2015

LinearRecurrence[{6,-1},{2,14},30] (* Harvey P. Dale, Jul 25 2018 *)

for(n=1,20,q=(1+sqrt(2))^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec(2*x*(1+x)/(1-6*x+x^2) + O(x^100)) \\ Altug Alkan, Mar 17 2016

a(n) = (((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) + ((3 + 2*sqrt(2))^(n-1) - (3 - 2*sqrt(2))^(n-1))) / (2*sqrt(2)).
a(n) = 2*A002315(n-1).
Recurrence: a(n) = 6*a(n-1) - a(n-2), starting 2, 14.
Offset 0, with a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = a^((2n+1)/2) - b^((2n+1)/2). a(n) = 2*(A001109(n+1) + A001109(n)) = (A003499(n+1) - A003499(n))/2 = 2*sqrt(A001108(2n+1)) = sqrt(A003499(2n+1)-2). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
Limit_{n->oo} a(n)/a(n-1) = 5.82842712474619009760... = 3 + 2*sqrt(2). See A156035.
From R. J. Mathar, Nov 16 2007: (Start)
G.f.: 2*x*(1+x)/(1-6*x+x^2).
a(n) = 2*(7*A001109(n) - A001109(n+1)). (End)
a(n) = (1+sqrt(2))^(2*n-1) - (1+sqrt(2))^(1-2*n). - Gerson Washiski Barbosa, Sep 19 2010
a(n) = floor((1 + sqrt(2))^(2*n-1)). - Thomas Ordowski, Jun 07 2013
a(n) = sqrt(2*A075870(n)^2-4). - Derek Orr, Jun 18 2015
a(n) = 2*sqrt((2*A001653(n)^2)-1). - César Aguilera, Jul 13 2023
E.g.f.: 2*(1 + exp(3*x)*(sqrt(2)*sinh(2*sqrt(2)*x) - cosh(2*sqrt(2)*x))). - Stefano Spezia, Aug 27 2024

A057174 a(n+3)=a(n)+a(n+1)-a(n+2), starting with 1,2,3.

Original entry on oeis.org

1, 2, 3, 0, 5, -2, 7, -4, 9, -6, 11, -8, 13, -10, 15, -12, 17, -14, 19, -16, 21, -18, 23, -20, 25, -22, 27, -24, 29, -26, 31, -28, 33, -30, 35, -32, 37, -34, 39, -36, 41, -38, 43, -40, 45, -42, 47, -44, 49, -46, 51, -48, 53, -50, 55, -52, 57, -54, 59, -56, 61, -58, 63, -60, 65, -62, 67, -64, 69, -66, 71, -68, 73, -70, 75, -72

Offset: 0

Gregory V. Richardson, Oct 17 2000

			a(4)=0 because 1+2-3=0

J. Brillhart et al., Factorizations of b^n +- 1, Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 3rd edition, 2002.
Index entries for linear recurrences with constant coefficients, signature (-1,1,1).

LinearRecurrence[{-1,1,1},{1,2,3},80] (* Harvey P. Dale, Jul 15 2023 *)

O.g.f.: (1+3x+4x^2)/((1-x)(1+x)^2). a(n)=2+(n-1)*(-1)^n. [From R. J. Mathar, Aug 22 2008]

More terms from James Sellers, Oct 18 2000

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

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

Original entry on oeis.org

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

A075841 Numbers k such that 2*k^2 - 9 is a square.

Original entry on oeis.org

3, 15, 87, 507, 2955, 17223, 100383, 585075, 3410067, 19875327, 115841895, 675176043, 3935214363, 22936110135, 133681446447, 779152568547, 4541233964835, 26468251220463, 154268273357943, 899141388927195

Offset: 1

Gregory V. Richardson, Oct 14 2002

Lim. n-> Inf. a(n)/a(n-1) = 3 + 2*sqrt(2).
Positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 36 = 0. - Colin Barker, Feb 08 2014
For each member t of the sequence there exists a nonnegative r such that t^2 = r^2 + (r+3)^2. The r values are in A241976. Example: 87^2 = 60^2 + 63^2. - Bruno Berselli, Jul 10 2017

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..200
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 (6,-1).

CoefficientList[Series[3 (1 - x)/(1 - 6 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 11 2014 *)
LinearRecurrence[{6,-1},{3,15},20] (* Harvey P. Dale, Jun 05 2023 *)

isok(n) = issquare(2*n^2-9); \\ Michel Marcus, Jul 10 2017

a(n) = 3*sqrt(2)/4*((1+sqrt(2))^(2*n-1)-(1-sqrt(2))^(2*n-1)) = 6*a(n-1) - a(n-2).
G.f.: 3*x*(1-x)/(1-6*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = 3*A001653(n). - R. J. Mathar, Sep 27 2014

A075843 Numbers k such that 99*k^2 + 1 is a square.

Original entry on oeis.org

0, 1, 20, 399, 7960, 158801, 3168060, 63202399, 1260879920, 25154396001, 501827040100, 10011386405999, 199725901079880, 3984506635191601, 79490406802752140, 1585823629419851199, 31636982181594271840

Offset: 0

Gregory V. Richardson, Oct 14 2002

From Wolfdieter Lang, Nov 08 2002: (Start)
Chebyshev's polynomials U(n,x) evaluated at x=10.
The a(n) give all (unsigned, integer) solutions of Pell equation b(n)^2 - 99*a(n)^2 = +1 with b(n)= A001085(n). (End)
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 20's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imagianry unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,19}. - Milan Janjic, Jan 25 2015

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 = 0..700
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 sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (20,-1).

Cf. A001084.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), this sequence (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

m:=10;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 22 2019

I:=[0,1]; [n le 2 select I[n] else 20*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012

seq( simplify(ChebyshevU(n-1, 10)), n=0..20); # G. C. Greubel, Dec 22 2019

Table[GegenbauerC[n-1, 1, 10], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
CoefficientList[Series[x/(1-20x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Dec 24 2012 *)
ChebyshevU[Range[22] -2, 10] (* G. C. Greubel, Dec 22 2019 *)
LinearRecurrence[{20,-1},{0,1},20] (* Harvey P. Dale, Dec 03 2023 *)

vector( 22, n, polchebyshev(n-2, 2, 10) ) \\ G. C. Greubel, Dec 22 2019

[lucas_number1(n,20,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,10) for n in (0..20)] # G. C. Greubel, Dec 22 2019

a(n) = ((10+3*sqrt(11))^n - (10-3*sqrt(11))^n) / (6*sqrt(11)).
a(n) = 20*a(n-1) - a(n-2), n>=1, a(0)=0, a(1)=1.
a(n) = S(n-1, 20), with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
G.f.: x/(1 - 20*x + x^2).
a(n) = sqrt((A001085(n)^2 - 1)/99).
Lim_{n->inf.} a(n)/a(n-1) = 10 + 3*sqrt(11).
a(n+1) = Sum_{k=0..n} A101950(n,k)*19^k. - Philippe Deléham, Feb 10 2012
Product_{n>=1} (1 + 1/a(n)) = 1/3*(3 + sqrt(11)). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = 3/20*(3 + sqrt(11)). - Peter Bala, Dec 23 2012

cp's OEIS Frontend

User: Gregory V. Richardson

A235230 Integer radii of circles tiled by square tiles such that the ratio of uncut tiles to cut tiles is an integer and four square tiles meet at the center of the circle.

Views

Author

Keywords

Comments

Examples

Links

Programs

QuickBASIC

A076218 Numbers n such that 2*n^2 - 3*n + 1 is a square.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A075871 Numbers k such that 13*k^2 + 1 is a square.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A075886 Number of cubes at generation n when building fractal cube with edge ratio of 1/2.

Views

Author

Keywords

Comments

Links

Programs

Mathematica

UBASIC

Formula

A077444 Numbers k such that (k^2 + 4)/2 is a square.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

A057174 a(n+3)=a(n)+a(n+1)-a(n+2), starting with 1,2,3.

Views

Author

Keywords

Examples

Links

Programs

Mathematica

Formula

Extensions

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

A076218 Numbers n such that 2n^2 - 3n + 1 is a square.