A156649 -id:A156649 - cp's OEIS frontend

A077443 Numbers k such that (k^2 - 7)/2 is a square.

3, 5, 13, 27, 75, 157, 437, 915, 2547, 5333, 14845, 31083, 86523, 181165, 504293, 1055907, 2939235, 6154277, 17131117, 35869755, 99847467, 209064253, 581953685, 1218515763, 3391874643, 7102030325, 19769294173, 41393666187, 115223890395, 241259966797, 671574048197

Offset: 1

Gregory V. Richardson, Nov 06 2002

Lim_{n -> inf} a(n)/a(n-2) = 3 + 2*sqrt(2) = R1*R2. Lim_{k -> inf} a(2*k-1)/a(2*k) = (9 + 4*sqrt(2))/7 = R1 = A156649 (ratio #1). Lim_{k -> inf} a(2*k)/a(2*k-1) = (11 + 6*sqrt(2))/7 = R2 (ratio #2).
Also gives solutions > 3 to the equation x^2-4 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004
From Paul Curtz, Dec 15 2012: (Start)
a(n-1) and A006452(n) are companions. Like A000129 and A001333.
Reduced mod 10 this is a sequence of period 12: 3, 5, 3, 7, 5, 7, 7, 5, 7, 3, 5, 3.
(End)
The Pisano periods (periods of the sequence reducing a(n) modulo m) for m>=1 are 1,  1,  8,  4, 12,  8,  6,  4, 24, 12, 24,  8, 28, ... R. J. Mathar, Dec 15 2012
Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 56 = 0. - Colin Barker, Feb 08 2014
From Wolfdieter Lang, Feb 05 2015: (Start)
a(n+1) gives for n >= 0 all positive x solutions of the (generalized) Pell equation x^2 - 2*y^2 = +7.
The corresponding y solutions are given in A077442(n), n >= 0. The, e.g., the Nagell reference for finding all solutions.
Because the primitive Pythagorean triangle (3,4,5) is the only one with the sum of legs equal to 7 all positive solutions (x(n),y(n)) = (a(n+1),A077442(n)) of the Pell equation x^2 - 2*y^2 = +7 satisfy x(n) - y(n) < y(n) if n >= 1; only the first solution (x(0),y(0)) = (3,2) satisfies 3-1 > 1. Proof: Primitive Pythagorean triangles are characterized by the positive integer pairs [u,v] with  u+v odd, gcd(u,v) = 1 and u > v. See the Niven et al. reference, Theorem 5.5, p. 232. The leg sum is L = (u+v)^2 - 2*v^2. With L = 7, x = u+v and y = v, every solution (x(n),y(n)) with x(n)-y(n) = u(n) > v(n) = y(n) will correspond to a primitive Pythagorean triangle. Note that because of gcd(x,y) = 1 also gcd(u,v) = 1. But there is only one such triangle with L=7, namely the one with  [u(0),v(0)] = [2,1]. All other solutions with n >= 1 must therefore satisfy x(n)-y(n) < y(n). (End)
For n > 0, a(n+1) is the n-th almost Lucas-cobalancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

			a(3)^2 - 2*A077442(2)^2 = 13^2 - 2*9^2 = +7. - _Wolfdieter Lang_, Feb 05 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.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.
Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
J. J. O'Connor and E. F. Robertson, History of Pell's Equation
J. P. Robertson, Solving the Generalized Pell Equation
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A000129, A001333, A006452, A038761, A038762, A077442, A101386, A124124, A156649, A176981, A216134, A253811.

LinearRecurrence[{0,6,0,-1},{3,5,13,27},50] (* Sture Sjöstedt, Oct 09 2012 *)

a(2n+1) = A038762(n). a(2n) = A101386(n-1).
The same recurrences hold for the odd and the even indices: a(n+2) = 6*a(n) - a(n-2), a(n+1) = 3*a(n) + 2*(2*a(n)^2-14)^0.5 - Richard Choulet, Oct 11 2007
O.g.f.: -x*(x-1)*(3*x^2+8*x+3) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Nov 23 2007
If n is even a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^(-n/2) + (1/2)*(3-sqrt(2))*(3-2*sqrt(2))^(-n/2); if n is odd a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^((n-1)/2) + (1/2)*(3-sqrt(2))*(3-2*sqrt(2))^((n-1)/2). - Antonio Alberto Olivares, Apr 20 2008
a(n) = A000129(n+1) + (-1)^n*A176981(n-1), n>1. - R. J. Mathar, Jul 03 2011
a(n) = A000129(n+1) -(-1)^n*A000129(n-2), rephrasing the formula above. - Paul Curtz, Dec 07 2012
a(n) = sqrt(8*A216134(n)^2 + 8*A216134(n) + 9) = 2*A124124(n) + 1. - Raphie Frank, May 24 2013
E.g.f.: cosh(sqrt(2)*x)*(3*cosh(x) - sinh(x)) + sqrt(2)*(2*cosh(x) - sinh(x))*sinh(sqrt(2)*x) - 3. - Stefano Spezia, Nov 25 2022

More terms from Richard Choulet, Oct 11 2007

Edited: replaced n by a(n) in the name. Moved Pell remarks to the comment section. Added cross references. - Wolfdieter Lang, Feb 05 2015

A076296 Consider all Pythagorean triples (X,X+7,Z); sequence gives X values.

Original entry on oeis.org

-3, 0, 5, 8, 21, 48, 65, 140, 297, 396, 833, 1748, 2325, 4872, 10205, 13568, 28413, 59496, 79097, 165620, 346785, 461028, 965321, 2021228, 2687085, 5626320, 11780597, 15661496, 32792613, 68662368, 91281905, 191129372, 400193625, 532029948, 1113983633

Offset: 0

Henry Bottomley, Oct 05 2002

First two terms included for consistency with A076293.
From Klaus Brockhaus, Feb 18 2009: (Start)
Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
Lim_{n -> infinity} a(n)/a(n-1) = (9+4*sqrt(2))/7 for n mod 3 = {1, 2}.
Lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2)) / ((9+4*sqrt(2))/7)^2 for n mod 3 = 0. (End)
For the generic case x^2 + (x+p)^2 = y^2 with p=2*m^2-1 a prime number in A066436, m >= 2, the x values are given by the sequence defined by: a(n) = 6*a(n-3) - a(n-6) + 2p with a(1)=0, a(2)=2m+1, a(3)=6m^2-10m+4, a(4)=3p, a(5)=6m^2+10m+4, a(6)=40m^2-58m+21. Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=2m^2+2m+1, b(3)=10m^2-14m+5, b(4)=5p, b(5)=10m^2+14m+5, b(6)=58m^2-82m+29. - Mohamed Bouhamida, Sep 09 2009
For the generic case x^2 + (x + p)^2 = y^2 with p = 2*m^2 - 1 a prime number, m>=2, the first three consecutive solutions are: (0;p), (2*m+1; 2*m^2+2*m+1), (6*m^2-10*m+4; 10*m^2-14*m+5) and the other solutions are defined by: (X(n); Y(n))= (3*X(n-3)+2*Y(n-3)+p; 4*X(n-3)+3*Y(n-3)+2*p). - Mohamed Bouhamida, Aug 20 2019

			8 is in the sequence as the shorter leg of the (8,15,17) triangle.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A001652, A076293, A076294, A076295.

Cf. A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7). - Klaus Brockhaus, Feb 18 2009

I:=[-3,0,5,8,21,48,65]; [n le 7 select I[n] else Self(n-1) +6*Self(n-3) -6*Self(n-4) -Self(n-6) +Self(n-7): n in [1..30]]; // G. C. Greubel, May 04 2018

CoefficientList[Series[(3-3x-5x^2-21x^3+5x^4+3x^5+4x^6)/(-1+x+6x^3-6x^4-x^6+x^7),{x,0,50}],x] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2012 *)
LinearRecurrence[{1,0,6,-6,0,-1,1}, {-3,0,5,8,21,48,65}, 50] (* T. D. Noe, Feb 07 2012 *)

x='x+O('x^30); Vec((-3+3*x+5*x^2+21*x^3-5*x^4-3*x^5-4*x^6)/((1-x)*(1-6*x^3 +x^6))) \\ G. C. Greubel, May 04 2018

a(n) = 6a(n-3) - a(n-6) + 14 = (A076293(n) - 7)/2.
a(n) = sqrt(A076294(n)^2 - A076295(n)^2) = A076295(n) - 7.
a(3*n+1) = 7*A001652(n).
From Mohamed Bouhamida, Jul 06 2007: (Start)
a(n) = 5*(a(n-3) + a(n-6)) - a(n-9) + 28.
a(n) = 7*(a(n-3) - a(n-6)) + a(n-9). (End)
G.f.: (-3 + 3*x + 5*x^2 + 21*x^3 - 5*x^4 - 3*x^5 - 4*x^6)/((1-x)*(1 - 6*x^3 + x^6)). - Klaus Brockhaus, Feb 18 2009

More terms from Klaus Brockhaus, Feb 18 2009

A077446 Numbers k such that 2*k^2 + 14 is a square.

Original entry on oeis.org

1, 5, 11, 31, 65, 181, 379, 1055, 2209, 6149, 12875, 35839, 75041, 208885, 437371, 1217471, 2549185, 7095941, 14857739, 41358175, 86597249, 241053109, 504725755, 1404960479, 2941757281, 8188709765, 17145817931, 47727298111

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "2*n^2 + 14 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = 14.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = (7+n^2)/2. - Ctibor O. Zizka, Nov 09 2009
From Wolfdieter Lang, Feb 26 2015: (Start)
This sequence gives all positive solutions x = a(n+1), n >= 0, of the Pell equation x^2 - 2*y^2 = -7. For the corresponding y-solutions see y(n) = 2*A006452(n+2) = A077447(n+1)/2. This implies that X^2 - 2*Y^2 = 14 has the general solutions (X(n),Y(n)) = (2*y(n), x(n)). See the first comment above.
For the positive first class solutions see (A054490(n), 2*A038723(n)) and for the second class solutions (A255236(n), 2*A038725(n+1)). (End)
For n > 0, a(n) is the n-th almost Lucas-balancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

			n = 3: (A077447(3))^2 - 2*a(3)^2 = 16^2 - 2*11^2  = 14;
a(3)^2 - 2*(2*A006452(3+1))^2 = 11^2 - 2*(2*4)^2 = -7. - _Wolfdieter Lang_, Feb 26 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.

Michael De Vlieger, Table of n, a(n) for n = 1..2612
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
J. J. O'Connor and E. F. Robertson, Pell's Equation
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Eric Weisstein's World of Mathematics, Pell Equation
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A001109, A006452, A038723, A038725, A049310, A054490, A077447, A155765, A156035, A156649, A255236.

LinearRecurrence[{0,6,0,-1},{1,5,11,31},50] (* Sture Sjöstedt, Oct 08 2012 *)

2*(a(n))^2 + 14 = (A077447(n))^2.
Lim. n-> Inf. a(n)/a(n-2) = 5.8284271247461... = 3 + 2*sqrt(2) = A156035 = RG (Great Ratio).
Lim. k-> Inf. a(2*k+1)/a(2*k) = 2.09383632135605... = (9 + 4*sqrt(2))/7 = A156649 = R1 (Ratio 1).
Lim. k -> Inf. a(2*k)/a(2*k-1) = 2.78361162489122432754 = (11 + 6*sqrt(2))/7 = R2 (Ratio 2); RG = R1*R2.
a(2*k-1) = [ 2*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] - [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] + [(3+2*Sqrt(2))^(n-2) - (3-2*Sqrt(2))^(n-2)] ] / (4*Sqrt(2)) a(2*k) = [ 5*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] + [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] ] / (4*Sqrt(2)).
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1+x)*(x^2+4*x+1) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Jul 03 2011
a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=5, a(3)=11, a(4)=31. - Sture Sjöstedt, Oct 08 2012
Bisection: a(2*k+1) = S(k, 6) + 5*S(k-1, 6), a(2*k) = 5*S(k-1, 6) + S(k-2, 6), with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1). See A054490 and A255236, and the given g.f.s. - Wolfdieter Lang, Feb 26 2015
E.g.f.: 1 - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x). - Stefano Spezia, Nov 26 2022
a(n) = a(n-1) + 2*A217975(n-1) + A123335(n-2) - A123335(n-3) for n > 1 and with A123335(-1) = 1. - Vladimir Pletser, Aug 30 2025

A118611 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+343)^2 = y^2.

Original entry on oeis.org

0, 77, 132, 245, 392, 585, 728, 1029, 1428, 1725, 2352, 3185, 4292, 5117, 6860, 9177, 10904, 14553, 19404, 25853, 30660, 40817, 54320, 64385, 85652, 113925, 151512, 179529, 238728, 317429, 376092, 500045, 664832, 883905, 1047200, 1392237, 1850940, 2192853

Offset: 1

Mohamed Bouhamida, May 08 2006

Also values x of Pythagorean triples (x, x+343, y); 343=7^3.
Corresponding values y of solutions (x, y) are in A157246.
Limit_{n -> oo} a(n)/a(n-7) = 3+2*sqrt(2).
Limit_{n -> oo} a(n)/a(n-1) = (3+2*sqrt(2)) / ((9+4*sqrt(2))/7)^2 for n mod 7 = {1, 2, 4, 5, 6}.
Limit_{n -> oo} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^5 / (3+2*sqrt(2))^2 for n mod 7 = {0, 3}.

			132^2+(132+343)^2 = 17424+225625 = 243049 = 493^2.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,6,-6,0,0,0,0,0,-1,1).

Cf. A157246, A001652, A118576, A118554, A118611, A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7).

LinearRecurrence[{1, 0, 0, 0, 0, 0, 6, -6, 0, 0, 0, 0, 0, -1, 1}, {0, 77, 132, 245, 392, 585, 728, 1029, 1428, 1725, 2352, 3185, 4292, 5117, 6860}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 13 2012 *)

{forstep(n=0, 1400000, [1, 3], if(issquare(n^2+(n+343)^2), print1(n, ",")))}

a(n) = 6*a(n-7)-a(n-14)+686 for n > 14; a(1)=0, a(2)=77, a(3)=132, a(4)=245, a(5)=392, a(6)=585, a(7)=728, a(8)=1029, a(9)=1428, a(10)=1725, a(11)=2352, a(12)=3185, a(13)=4292, a(14)=5117.
G.f.: x*(77+55*x+113*x^2+147*x^3+193*x^4+143*x^5+301*x^6-63*x^7 -33*x^8-51*x^9-49*x^10-51*x^11-33*x^12-63*x^13)/((1-x)*(1-6*x^7+x^14)).
a(7*k+1) = 343*A001652(k) for k >= 0.

Edited by Klaus Brockhaus, Feb 25 2009

A118576 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+16807)^2 = y^2.

Original entry on oeis.org

0, 2145, 3773, 6468, 8540, 12005, 19208, 24521, 28665, 35672, 41148, 50421, 61388, 69972, 84525, 95921, 115248, 156065, 186480, 210308, 250733, 282405, 336140, 399797, 449673, 534296, 600600, 713097, 950796, 1127973, 1266797, 1502340

Offset: 1

Mohamed Bouhamida, May 16 2006

Also values x of Pythagorean triples (x, x+16807, y); 16807 = 7^5.
Corresponding values y of solutions (x, y) are in A156713.
Limit_{n -> oo} a(n)/a(n-11) = 3+2*sqrt(2).
Limit_{n -> oo} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^5 / (3+2*sqrt(2))^2 for n mod 11 = {1, 2, 4, 6, 8, 10}.
Limit_{n -> oo} a(n)/a(n-1) = (3+2*sqrt(2))^3 / ((9+4*sqrt(2))/7)^7 for n mod 11 = {0, 3, 5, 9}.
Limit_{n -> oo} a(n)/a(n-1) = (3+2*sqrt(2)) / ((9+4*sqrt(2))/7)^2 for n mod 11 = 7.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,6,-6,0,0,0,0,0,0,0,0,0,-1,1).

Cf. A156713, A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7).

{forstep(n=0, 1600000, [1, 3], if(issquare(2*n^2 + 33614*n + 282475249), print1(n, ",")))}

a(n) = 6*a(n-11)-a(n-22)+33614 for n > 22; a(1) = 0, a(2) = 2145, a(3) = 3773, a(4) = 6468, a(5) = 8540, a(6) = 12005, a(7) = 19208, a(8) = 24521, a(9) = 28665, a(10) = 35672, a(11) = 41148, a(12) = 50421, a(13) = 61388, a(14) = 69972, a(15) = 84525, a(16) = 95921, a(17) = 115248, a(18) = 156065, a(19) = 186480, a(20) = 210308, a(21) = 250733, a(22) = 282405.

G.f.: x*(2145+1628*x+2695*x^2+2072*x^3+3465*x^4+7203*x^5+5313*x^6+4144*x^7+7007*x^8+5476*x^9+9273*x^10-1903*x^11-1184*x^12-1617*x^13-1036*x^14-1463*x^15-2401*x^16-1463*x^17 -1036*x^18-1617*x^19-1184*x^20-1903*x^21 )/((1-x)*(1-6*x^11+x^22)).

Edited by Klaus Brockhaus, Feb 14 2009

A118630 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+2401)^2 = y^2.

Original entry on oeis.org

0, 539, 924, 1220, 1715, 2744, 3503, 4095, 5096, 7203, 9996, 12075, 13703, 16464, 22295, 26640, 30044, 35819, 48020, 64239, 76328, 85800, 101871, 135828, 161139, 180971, 214620, 285719, 380240, 450695, 505899, 599564, 797475, 944996, 1060584

Offset: 1

Mohamed Bouhamida, May 09 2006

Also values x of Pythagorean triples (x, x+2401, y); 2401=7^4.
Corresponding values y of solutions (x, y) are in A157247.
Limit_{n -> oo} a(n)/a(n-9) = 3+2*sqrt(2).
Limit_{n -> oo} a(n)/a(n-1) = (3+2*sqrt(2)) / ((9+4*sqrt(2))/7)^2 for n mod 9 = {1, 2, 6}.
Limit_{n -> oo} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^5 / (3+2*sqrt(2))^2 for n mod 9 = {0, 3, 5, 7}.
Limit_{n -> oo} a(n)/a(n-1) = (3+2*sqrt(2))^3 / ((9+4*sqrt(2))/7)^7 for n mod 9 = {4, 8}.

			924^2+(924+2401)^2 = 853776+11055625 = 11909401 = 3451^2.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,6,-6,0,0,0,0,0,0,0,-1,1).

Cf. A157247, A001652, A118576, A118554, A118611, A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7).

{forstep(n=0, 1100000, [3 ,1], if(issquare(n^2+(n+2401)^2), print1(n, ",")))}

a(n) = 6*a(n-9)-a(n-18)+4802 for n > 18; a(1)=0, a(2)=539, a(3)=924, a(4)=1220, a(5)=1715, a(6)=2744, a(7)=3503, a(8)=4095, a(9)=5096, a(10)=7203, a(11)=9996, a(12)=12075, a(13)=13703, a(14)=16464,a (15)=22295, a(16)=26640, a(17)=30044, a(18)=35819.
G.f.: x*(539+385*x+296*x^2+495*x^3+1029*x^4+759*x^5+592*x^6 +1001*x^7+2107*x^8-441*x^9-231*x^10-148*x^11-209*x^12-343*x^13 -209*x^14-148*x^15-231*x^16-441*x^17) / ((1-x)*(1-6*x^9+x^18)).
a(9*k+1) = 2401*A001652(k) for k >= 0.

Edited by Klaus Brockhaus, Feb 25 2009

A129837 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+119)^2 = y^2.

Original entry on oeis.org

0, 24, 49, 57, 85, 136, 180, 196, 261, 357, 481, 616, 660, 816, 1105, 1357, 1449, 1824, 2380, 3100, 3885, 4141, 5049, 6732, 8200, 8736, 10921, 14161, 18357, 22932, 24424, 29716, 39525, 48081, 51205, 63940, 82824, 107280, 133945, 142641, 173485, 230656

Offset: 1

Mohamed Bouhamida, May 21 2007

nonn

Also values x of Pythagorean triples (x, x+119, y).
Corresponding values y of solutions (x, y) are in A156650.
lim_{n -> infinity} a(n)/a(n-9) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = ((9+4*sqrt(2))/7)/((19+6*sqrt(2))/17) for n mod 9 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = ((19+6*sqrt(2))/17)^2/((9+4*sqrt(2))/7) for n mod 9 = {0, 3}.
lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2))/(((9+4*sqrt(2))/7)*((19+6*sqrt(2))/17)^2) for n mod 9 = {4, 8}.
lim_{n -> infinity} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^2*((19+6*sqrt(2))/17)/(3+2*sqrt(2)) for n mod 9 = {5, 7}.
lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2))/((9+4*sqrt(2))/7)^2 for n mod 9 = 6.

Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 6, -6, 0, 0, 0, 0, 0, 0, 0, -1, 1).

Cf. A156650, A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7), A156163 (decimal expansion of (19+6*sqrt(2))/17), A118630.

LinearRecurrence[{1,0,0,0,0,0,0,0,6,-6,0,0,0,0,0,0,0,-1,1}, {0,24,49,57,85,136,180,196,261,357,481,616,660,816,1105,1357,1449,1824,2380}, 140] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2012 *)

{forstep(n=0, 240000, [1, 3], if(issquare(n^2+(n+119)^2), print1(n, ",")))}

a(n) = 6*a(n-9)-a(n-18)+238 for n > 18; a(1)=0, a(2)=24, a(3)=49, a(4)=57, a(5)=85, a(6)=136, a(7)=180, a(8)=196, a(9)=261, a(10)=357, a(11)=481, a(12)=616, a(13)=660, a(14)=816, a(15)=1105, a(16)=1357, a(17)=1449, a(18)=1824.

G.f.: x*(24+25*x+8*x^2+28*x^3+51*x^4+44*x^5+16*x^6+65*x^7+96*x^8-20*x^9-15*x^10-4*x^11-12*x^12-17*x^13-12*x^14-4*x^15-15*x^16-20*x^17 )/((1-x)*(1-6*x^9+x^18))

Edited and extended by Klaus Brockhaus, Feb 13 2009

A077447 Numbers k such that (k^2 - 14)/2 is a square.

Original entry on oeis.org

4, 8, 16, 44, 92, 256, 536, 1492, 3124, 8696, 18208, 50684, 106124, 295408, 618536, 1721764, 3605092, 10035176, 21012016, 58489292, 122467004, 340900576, 713790008, 1986914164, 4160273044, 11580584408, 24247848256, 67496592284

Offset: 1

Gregory V. Richardson, Nov 09 2002

nonn

The equation "(n^2 - 14)/2 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = 14.

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.

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 (0,6,0,-1).

Cf. A156649 (R1).

LinearRecurrence[{0,6,0,-1},{4,8,16,44},40] (* Harvey P. Dale, Jul 22 2013 *)

Limit_{k -> oo} a(2*k+1)/a(2*k) = 2.09383632135605431360 = (9 + 4*sqrt(2))/7 = R1.
Limit_{k -> oo} a(2*k)/a(2*k-1) = 2.78361162489122432754 = (11 + 6*sqrt(2))/7 = R2.
Limit_{k -> oo} a(n)/a(n-2) = 3 + 2*sqrt(2) = RG (Grand Ratio); RG = R1*R2.
For n = 2*k-1, a(n) = [ 2*[(3+2*sqrt(2))^n + (3-2*sqrt(2))^n] - [(3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1)] + [(3+2*sqrt(2))^(n-2) + (3-2*sqrt(2))^(n-2)] ] / 4.
For n = 2*k, a(n) = [ 5*[(3+2*sqrt(2))^n + (3-2*sqrt(2))^n] + [(3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1)] ] / 4.
a(n) = 6*a(n-2) - a(n-4) = 4*A006452(n).
G.f.: -4*x*(x-1)*(x^2+3*x+1) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Jul 03 2011

A129010 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+833)^2 = y^2.

Original entry on oeis.org

0, 124, 168, 187, 343, 399, 595, 624, 915, 952, 1260, 1372, 1768, 1827, 1975, 2499, 3135, 3367, 3468, 4312, 4620, 5712, 5875, 7524, 7735, 9499, 10143, 12427, 12768, 13624, 16660, 20352, 21700, 22287, 27195, 28987, 35343, 36292, 45895, 47124

Offset: 1

Mohamed Bouhamida, May 27 2007

nonn

Also values x of Pythagorean triples (x, x+833, y); 833=7^2*17.
Corresponding values y of solutions (x, y) are in A156835.
lim_{n -> infinity} a(n)/a(n-15) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^2*((19+6*sqrt(2))/17)/(3+2*sqrt(2)) for n mod 15 = {1, 2, 5, 7, 11, 13}.
lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2))/(((9+4*sqrt(2))/7)*((19+6*sqrt(2))/17)^2) for n mod 15 = {0, 3, 6, 12}.
lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2))*((19+6*sqrt(2))/17)/((9+4*sqrt(2))/7)^3 for n mod 15 = {4, 8, 10, 14}.
lim_{n -> infinity} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^4/((3+2*sqrt(2))*((19+6*sqrt(2))/17)^2) for n mod 15 = 9.

			124^2+(124+833)^2 = 15376+915849 = 931225 = 965^2.

Cf. A156835, A076296, A118120, A118554, A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7), A156163 (decimal expansion of (19+6*sqrt(2))/17).

{forstep(n=0, 50000, [3, 1], if(issquare(2*n^2+1666*n+693889), print1(n, ",")))}

a(n) = 6*a(n-15)-a(n-30)+1666 for n > 30; a(1) = 0, a(2) = 124, a(3) = 168, a(4) = 187, a(5) = 343, a(6) = 399, a(7) = 595, a(8) = 624, a(9) = 915, a(10) = 952, a(11) = 1260, a(12) = 1372, a(13) = 1768, a(14) = 1827, a(15) = 1975, a(16) = 2499, a(17) = 3135, a(18) = 3367, a(19) = 3468, a(20) = 4312, a(21) = 4620, a(22) = 5712, a(23) = 5875, a(24) = 7524, a(25) = 7735, a(26) = 9499, a(27) = 10143, a(28) = 12427, a(29) = 12768, a(30) = 13624.

G.f.: x*(124+44*x+19*x^2+156*x^3+56*x^4+196*x^5+29*x^6+291*x^7+37*x^8+308*x^9+112*x^10+396*x^11+59*x^12+148*x^13+524*x^14-108*x^15-32*x^16-13*x^17-92*x^18-28*x^19-84*x^20-11*x^21-97*x^22-11*x^23-84*x^24-28*x^25-92*x^26-13*x^27-32*x^28-108*x^29)/((1-x)*(1-6*x^15+x^30)).

Edited by Klaus Brockhaus, Feb 16 2009

A156650 Positive numbers y such that y^2 is of the form x^2+(x+119)^2 with integer x.

Original entry on oeis.org

85, 89, 91, 101, 119, 145, 175, 185, 221, 289, 349, 371, 461, 595, 769, 959, 1021, 1241, 1649, 2005, 2135, 2665, 3451, 4469, 5579, 5941, 7225, 9605, 11681, 12439, 15529, 20111, 26045, 32515, 34625, 42109, 55981, 68081, 72499, 90509, 117215, 151801

Offset: 1

Klaus Brockhaus, Feb 17 2009

nonn

(-51, a(1)), (-39, a(2)), (-35, a(3)), (-20, a(4)) and (A129837(n), a(n+4)) are solutions (x, y) to the Diophantine equation x^2+(x+119)^2 = y^2.
lim_{n -> infinity} a(n)/a(n-9) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2))/((9+4*sqrt(2))/7)^2 for n mod 9 = 1.
lim_{n -> infinity} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^2*((19+6*sqrt(2))/17)/(3+2*sqrt(2)) for n mod 9 = {0, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2))/(((9+4*sqrt(2))/7)*((19+6*sqrt(2))/17)^2) for n mod 9 = {3, 8}.
lim_{n -> infinity} a(n)/a(n-1) = ((19+6*sqrt(2))/17)^2/((9+4*sqrt(2))/7) for n mod 9 = {4, 7}.
lim_{n -> infinity} a(n)/a(n-1) = ((9+4*sqrt(2))/7)/((19+6*sqrt(2))/17) for n mod 9 = {5, 6}.

			(-51, a(1)) = (-51, 85) is a solution: (-51)^2+(-51+119)^2 = 2601+4624 = 7225 = 85^2.
(A129837(1), a(5)) = (0, 119) is a solution: 0^2+(0+119)^2 = 14161 = 119^2.
(A129837(3), a(7)) = (49, 175) is a solution: 49^2+(49+119)^2 = 2401+28224 = 30625 = 175^2.

Cf. A129837, A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7), A156163 (decimal expansion of (19+6*sqrt(2))/17).

upto=200000; With[{max=Ceiling[(Sqrt[2*upto^2]-119)/2]},Union[ Sqrt[#]&/@ Select[Table[x^2+(x+119)^2,{x,-250,max}],IntegerQ[Sqrt[#]]&]]](* Harvey P. Dale, Aug 11 2011 *)

{forstep(n=-52, 120000, [1, 3], if(issquare(n^2+(n+119)^2, &k), print1(k, ",")))}

a(n) = 6*a(n-9)-a(n-18) for n > 18; a(1)=85, a(2)=89, a(3)=91, a(4)=101, a(5)=119, a(6)=145, a(7)=175, a(8)=185, a(9)=221, a(10)=289, a(11)=349, a(12)=371, a(13)=461, a(14)=595, a(15)=769, a(16)=959, a(17)=1021, a(18)=1241.

G.f.: x * (1-x) * (85 +174*x +265*x^2 +366*x^3 +485*x^4 +630*x^5 +805*x^6 +990*x^7 +1211*x^8 +990*x^9 +805*x^10 +630*x^11 +485*x^12 +366*x^13 +265*x^14 +174*x^15 +85*x^16) / (1 -6*x^9 +x^18). [adapted to the offset by Bruno Berselli, Apr 01 2011]

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A077443 Numbers k such that (k^2 - 7)/2 is a square.

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A076296 Consider all Pythagorean triples (X,X+7,Z); sequence gives X values.

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A077446 Numbers k such that 2*k^2 + 14 is a square.

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A118611 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+343)^2 = y^2.

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A118576 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+16807)^2 = y^2.

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A118630 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+2401)^2 = y^2.

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A129837 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+119)^2 = y^2.

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