A156035 -id:A156035 - cp's OEIS frontend

A129993 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x+199)^2 = y^2.

0, 21, 504, 597, 704, 3441, 3980, 4601, 20540, 23681, 27300, 120197, 138504, 159597, 701040, 807741, 930680, 4086441, 4708340, 5424881, 23818004, 27442697, 31619004, 138821981, 159948240, 184289541, 809114280, 932247141, 1074118640

Offset: 1

Mohamed Bouhamida, Jun 14 2007

Also values x of Pythagorean triples (x, x+199, y).
Corresponding values y of solutions (x, y) are in A159548.
For the generic case x^2+(x+p)^2 = y^2 with p = 2*m^2-1 a (prime) number in A066436 see A118673 or A129836.
Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
Lim_{n -> infinity} a(n)/a(n-1) = (201+20*sqrt(2))/199 for n mod 3 = {1, 2}.
Lim_{n -> infinity} a(n)/a(n-1) = (91443+58282*sqrt(2))/199^2 for n mod 3 = 0.

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

Cf. A159548, A066436, A118673, A118674, A129836, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A159549 (decimal expansion of (201+20*sqrt(2))/199), A159550 (decimal expansion of (91443+58282*sqrt(2))/199^2).

I:=[0,21,504,597,704,3441,3980]; [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..50]]; // G. C. Greubel, Mar 31 2018

LinearRecurrence[{1,0,6,-6,0,-1,1},{0,21,504,597,704,3441,3980},30] (* Harvey P. Dale, Jun 03 2012 *)

{forstep(n=0, 500000000, [1, 3], if(issquare(2*n^2+398*n+39601), print1(n, ",")))};

a(n) = 6*a(n-3) - a(n-6) + 398 for n > 6; a(1)=0, a(2)=21, a(3)=504, a(4)=597, a(5)=704, a(6)=3441.
G.f.: x*(21+483*x+93*x^2-19*x^3-161*x^4-19*x^5) / ((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 199*A001652(k) for k >= 0.
a(1)=0, a(2)=21, a(3)=504, a(4)=597, a(5)=704, a(6)=3441, a(7)=3980, a(n)=a(n-1)+6*a(n-3)-6*a(n-4)-a(n-6)+a(n-7). - Harvey P. Dale, Jun 03 2012

Edited and two terms added by Klaus Brockhaus, Apr 14 2009

A157258 Decimal expansion of 7 + 2*sqrt(2).

Original entry on oeis.org

9, 8, 2, 8, 4, 2, 7, 1, 2, 4, 7, 4, 6, 1, 9, 0, 0, 9, 7, 6, 0, 3, 3, 7, 7, 4, 4, 8, 4, 1, 9, 3, 9, 6, 1, 5, 7, 1, 3, 9, 3, 4, 3, 7, 5, 0, 7, 5, 3, 8, 9, 6, 1, 4, 6, 3, 5, 3, 3, 5, 9, 4, 7, 5, 9, 8, 1, 4, 6, 4, 9, 5, 6, 9, 2, 4, 2, 1, 4, 0, 7, 7, 7, 0, 0, 7, 7, 5, 0, 6, 8, 6, 5, 5, 2, 8, 3, 1, 4, 5, 4, 7, 0, 0, 2

Offset: 1

Klaus Brockhaus, Feb 26 2009

lim_{n -> infinity} b(n)/b(n-1) = (7+2*sqrt(2))/(7-2*sqrt(2)) for n mod 3 = {1, 2}, b = A129288.

lim_{n -> infinity} b(n)/b(n-1) = (7+2*sqrt(2))/(7-2*sqrt(2)) for n mod 3 = {0, 2}, b = A157257.

			7 + 2*sqrt(2) = 9.82842712474619009760...

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for algebraic numbers, degree 2.

Cf. A129288, A157257, A157259 (decimal expansion of 7-2*sqrt(2)), A157260 (decimal expansion of (7+2*sqrt(2))/(7-2*sqrt(2))).

[7+2*Sqrt(2)]; // G. C. Greubel, Nov 28 2017

RealDigits[7+2Sqrt[2],10,120][[1]]  (* Harvey P. Dale, Mar 20 2011 *)

7+2*sqrt(2) \\ G. C. Greubel, Nov 28 2017

Equals 4 + A156035. - R. J. Mathar, Feb 27 2009

A163960 Decimal expansion of 2*(sqrt(2) - 1).

Original entry on oeis.org

8, 2, 8, 4, 2, 7, 1, 2, 4, 7, 4, 6, 1, 9, 0, 0, 9, 7, 6, 0, 3, 3, 7, 7, 4, 4, 8, 4, 1, 9, 3, 9, 6, 1, 5, 7, 1, 3, 9, 3, 4, 3, 7, 5, 0, 7, 5, 3, 8, 9, 6, 1, 4, 6, 3, 5, 3, 3, 5, 9, 4, 7, 5, 9, 8, 1, 4, 6, 4, 9, 5, 6, 9, 2, 4, 2, 1, 4, 0, 7, 7, 7, 0, 0, 7, 7, 5, 0, 6, 8, 6, 5, 5, 2, 8, 3, 1, 4, 5

Offset: 0

N. J. A. Sloane, Oct 02 2010

Decimal expansion of shortest length, (B), of segment from side BC through incenter to side BA in right triangle ABC with sidelengths (a,b,c)=(1,1,sqrt(2)). (See A195284.) - Clark Kimberling, Sep 14 2011

			0.82842712474619009760337744841939615713934375075389614635335...

J. M. Steele, Probability Theory and Combinatorial Optimization, SIAM, 1997, p. 3.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for algebraic numbers, degree 2.

Cf. A084158, A156035, A195284.

Essentially the same digit sequence as A010466, A086178, A090488 and A157258.

RealDigits[2(Sqrt[2]-1),10,120][[1]] (* Harvey P. Dale, May 27 2016 *)

2*(sqrt(2)-1) \\ G. C. Greubel, Aug 13 2017

Equals Sum_{k>=0} (-1)^k * binomial(2*k,k)/((k+1) * 4^k). - Amiram Eldar, May 06 2022

Equals Sum_{k>=1} (-1)^(k+1)/A084158(k). - Amiram Eldar, Dec 02 2024

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

A115135 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+617)^2 = y^2.

Original entry on oeis.org

0, 108, 1407, 1851, 2407, 9768, 12340, 15568, 58435, 73423, 92235, 342076, 429432, 539076, 1995255, 2504403, 3143455, 11630688, 14598220, 18322888, 67790107, 85086151, 106795107, 395111188, 495919920, 622448988, 2302878255, 2890434603

Offset: 1

Mohamed Bouhamida, Jun 03 2007

Also values x of Pythagorean triples (x, x+617, y).
Corresponding values y of solutions (x, y) are in A160176.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (633+100*sqrt(2))/617 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (755667+461578*sqrt(2))/617^2 for n mod 3 = 0.

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

Cf. A160176, A001652, A111258, A156035 (decimal expansion of 3+2*sqrt(2)), A160177 (decimal expansion of (633+100*sqrt(2))/617), A160178 (decimal expansion of (755667+461578*sqrt(2))/617^2).

I:=[0,108,1407,1851,2407,9768,12340]; [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

LinearRecurrence[{1,0,6,-6,0,-1,1}, {0,108,1407,1851,2407,9768,12340}, 50] (* G. C. Greubel, May 04 2018 *)

{forstep(n=0, 10000000, [3, 1], if(issquare(2*n^2+1234*n+380689), print1(n, ",")))}

x='x+O('x^30); Vec(x*(108 +1299*x +444*x^2 -92*x^3 -433*x^4 -92*x^5)/((1-x)*(1 -6*x^3 +x^6))) \\ G. C. Greubel, May 04 2018

a(n) = 6*a(n-3) -a(n-6) +1234 for n > 6; a(1)=0, a(2)=108, a(3)=1407, a(4)=1851, a(5)=2407, a(6)=9768.
G.f.: x*(108 +1299*x +444*x^2 -92*x^3 -433*x^4 -92*x^5)/((1-x)*(1 -6*x^3 +x^6)).
a(3*k+1) = 617*A001652(k) for k >= 0.

Edited and two terms added by Klaus Brockhaus, May 18 2009

A129992 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x+127)^2 = y^2.

Original entry on oeis.org

0, 17, 308, 381, 468, 2117, 2540, 3045, 12648, 15113, 18056, 74025, 88392, 105545, 431756, 515493, 615468, 2516765, 3004820, 3587517, 14669088, 17513681, 20909888, 85498017, 102077520, 121872065, 498319268, 594951693, 710322756, 2904417845, 3467632892

Offset: 1

Mohamed Bouhamida, Jun 14 2007

nonn

Also values x of Pythagorean triples (x, x+127, y).
Corresponding values y of solutions (x, y) are in A159466.
For the generic case x^2+(x+p)^2 = y^2 with p = 2*m^2-1 a (prime) number in A066436 see A118673 or A129836.
Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
Lim_{n -> infinity} a(n)/a(n-1) = (129+16*sqrt(2))/127 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (34947+21922*sqrt(2))/127^2 for n mod 3 = 0.

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

Cf. A159466, A066436, A118673, A118674, A129836, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A159467 (decimal expansion of (129+16*sqrt(2))/127), A159468 (decimal expansion of (34947+21922*sqrt(2))/127^2).

I:=[0,17,308,381,468,2117,2540]; [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..50]]; // G. C. Greubel, Mar 31 2018

LinearRecurrence[{1,0,6,-6,0,-1,1},{0,17,308,381,468,2117,2540},80] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2012 *)

{forstep(n=0, 500000000, [1, 3], if(issquare(2*n^2+254*n+16129), print1(n, ",")))};

a(n) = 6*a(n-3) - a(n-6) + 254 for n > 6; a(1)=0, a(2)=17, a(3)=308, a(4)=381, a(5)=468, a(6)=2117.
G.f.: x*(17+291*x+73*x^2-15*x^3-97*x^4-15*x^5) / ((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 127*A001652(k) for k >= 0.

Edited and two more terms added by Klaus Brockhaus, Apr 13 2009

A155923 Positive numbers y such that y^2 is of the form x^2+(x+17)^2 with integer x.

Original entry on oeis.org

13, 17, 25, 53, 85, 137, 305, 493, 797, 1777, 2873, 4645, 10357, 16745, 27073, 60365, 97597, 157793, 351833, 568837, 919685, 2050633, 3315425, 5360317, 11951965, 19323713, 31242217, 69661157, 112626853, 182092985, 406014977, 656437405

Offset: 1

Klaus Brockhaus, Feb 09 2009

(-5,a(1)) and (A118120(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+17)^2 = y^2. (Offset 1 is assumed for A118120.)
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (19+6*sqrt(2))/17 for n mod 3 = {0, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (387+182*sqrt(2))/17^2 for n mod 3 = 1.
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)=2*m^2+2m+1, b(3)=10m^2-14m+5, b(4)=5p, b(5)=10m^2+14m+5, b(6)=58m^2-82m+29. [From Mohamed Bouhamida, Sep 09 2009]

			(-5,a(1)) = (-5,13) is a solution: (-5)^2+(-5+17)^2 = 25+144 = 169 = 13^2;
(A118120(1), a(2)) = (0, 17) is a solution: 0^2+(0+17)^2 = 289 = 17^2;
(A118120(2), a(3)) = (7, 25) is a solution: 7^2+(7+17)^2 = 49+576 = 625 = 25^2.

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

Cf. A118120, A156035 (decimal expansion of 3+2*sqrt(2)), A156163 (decimal expansion of (19+6*sqrt(2))/17), A157649 (decimal expansion of (387+182*sqrt(2))/17^2).

Cf. A156156 (first trisection), A156157 (second trisection), A156158 (third trisection).

LinearRecurrence[{0,0,6,0,0,-1},{13,17,25,53,85,137},50] (* Harvey P. Dale, Feb 11 2015 *)

{forstep(n=-5, 660000000, [1,3], if(issquare(2*n*(n+17)+289, &k), print1(k, ",")))}

a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1) = 13, a(2) = 17, a(3) = 25, a(4) = 53, a(5) = 85, a(6) = 137.

G.f.: x*(1-x)*(13+30*x+55*x^2+30*x^3+13*x^4)/(1-6*x^3+x^6).

G.f. corrected, first and fourth comment and examples edited, cross-reference added by Klaus Brockhaus, Sep 22 2009

A156567 Positive numbers y such that y^2 is of the form x^2+(x+23)^2 with integer x.

Original entry on oeis.org

17, 23, 37, 65, 115, 205, 373, 667, 1193, 2173, 3887, 6953, 12665, 22655, 40525, 73817, 132043, 236197, 430237, 769603, 1376657, 2507605, 4485575, 8023745, 14615393, 26143847, 46765813, 85184753, 152377507, 272571133, 496493125

Offset: 1

Klaus Brockhaus, Feb 11 2009 , Feb 16 2009

nonn

(-8, a(1)) and(A118337(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+23)^2 = y^2.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (27+10*sqrt(2))/23 for n mod 3 = {0, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (627+238*sqrt(2))/23^2 for n mod 3 = 1.
For the generic case x^2+(x+p)^2=y^2 with p=m^2-2 a prime number in A028871, 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+2, a(3)=3m^2-10m+8, a(4)=3p, a(5)=3m^2+10m+8, a(6)=20m^2-58m+42.Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=m^2+2m+2, b(3)=5m^2-14m+10, b(4)=5p, b(5)=5m^2+14m+10, b(6)=29m^2-82m+58. [From Mohamed Bouhamida, Sep 09 2009]

			(-8, a(1)) = (-8, 17) is a solution: (-8)^2+(-8+23)^2 = 64+225 = 289 = 17^2.
(A118337(1), a(2)) = (0, 23) is a solution: 0^2+(0+23)^2 = 529 = 23^2.
(A118337(3), a(4)) = (33, 65) is a solution: 33^2+(33+23)^2 = 1089+3136 = 4225 = 65^2.

Cf. A118337, A156035 (decimal expansion of 3+2*sqrt(2)), A156571 (decimal expansion of (27+10*sqrt(2))/23), A157472 (decimal expansion of (627+238*sqrt(2))/23^2).

A156570 (first trisection), A156568 (second trisection), A156569 (third trisection).

{forstep(n=-8, 360000000, [1,3], if(issquare(2*n*(n+23)+529, &k), print1(k, ",")))}

a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1)=17, a(2)=23, a(3)=37, a(4)=65, a(5)=115, a(6)=205.

G.f.: x*(1-x)*(17+40*x+77*x^2+40*x^3+17*x^4)/(1-6*x^3+x^6).

G.f. corrected, third and fourth comment edited, cross-reference added by Klaus Brockhaus, Sep 18 2009

A101152 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+569)^2 = y^2.

Original entry on oeis.org

0, 111, 1260, 1707, 2280, 8791, 11380, 14707, 52624, 67711, 87100, 308091, 396024, 509031, 1797060, 2309571, 2968224, 10475407, 13462540, 17301451, 61056520, 78466807, 100841620, 355864851, 457339440, 587749407, 2074133724, 2665570971

Offset: 1

Mohamed Bouhamida, Jun 03 2007

Also values x of Pythagorean triples (x, x+569, y).
Corresponding values y of solutions (x, y) are in A160090.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (587+102*sqrt(2))/569 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (617139+371510*sqrt(2))/569^2 for n mod 3 = 0.

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

Cf. A160090, A129298, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A160091 (decimal expansion of (587+102*sqrt(2))/569), A160092 (decimal expansion of (617139+371510*sqrt(2))/569^2).

I:=[0,111,1260,1707,2280,8791,11380]; [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, Apr 21 2018

LinearRecurrence[{1,0,6,-6,0,-1,1}, {0,111,1260,1707,2280,8791,11380}, 50] (* G. C. Greubel, Apr 21 2018 *)

{forstep(n=0, 10000000, [3, 1], if(issquare(2*n^2+1138*n+323761), print1(n, ",")))}

x='x+O('x^30); concat([0], Vec(x*(111 +1149*x +447*x^2 -93*x^3 -383*x^4 -93*x^5)/((1-x)*(1-6*x^3 +x^6)))) \\ G. C. Greubel, Apr 21 2018

a(n) = 6*a(n-3) - a(n-6) + 1138 for n > 6; a(1)=0, a(2)=111, a(3)=1260, a(4)=1707, a(5)=2280, a(6)=8791.
G.f.: x*(111 +1149*x +447*x^2 -93*x^3 -383*x^4 -93*x^5)/((1-x)*(1-6*x^3 +x^6)).
a(3*k+1) = 569*A001652(k) for k >= 0.

Edited and two terms added by Klaus Brockhaus, May 04 2009

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