Sture Sjöstedt - cp's OEIS frontend

A293846 Numbers such that k is the altitude of a Heronian triangle with sides m-13, m, m+13.

9, 24, 39, 60, 105, 156, 231, 396, 585, 864, 1479, 2184, 3225, 5520, 8151, 12036, 20601, 30420, 44919, 76884, 113529, 167640, 286935, 423696, 625641, 1070856, 1581255, 2334924, 3996489, 5901324, 8714055, 14915100, 22024041, 32521296, 55663911, 82194840

Offset: 0

Sture Sjöstedt, Dec 27 2017

a(n) gives the values of y satifacting 3*x^2 - y^2 = 507; corresponding x values are given by A293817.

a(n)/3 is the radius of the inscribed circle.

			If the sides are 15, 28, 41 the triangle has the altitude 9 and is a part of the Pythagorean triangle with the sides 9, 40, 41, so 9 is a term.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-1).

Cf. A003500, A005320, A296795, A296796.

CoefficientList[ Series[ 3(3x^4 +8x^3 +13x^2 +8x +3)/(x^6 -4x^3 +1), {x, 0, 35}], x] (* or *)
LinearRecurrence[{0, 0, 4, 0, 0, -1}, 3 {3, 8, 13, 20, 35, 52}, 36] (* Robert G. Wilson v, Dec 27 2017 *)

Vec(3*(3 + 8*x + 13*x^2 + 8*x^3 + 3*x^4) / (1 - 4*x^3 + x^6) + O(x^40)) \\ Colin Barker, Dec 27 2017

a(n) = 4*a(n-3) - a(n-6), a(1)=9, a(2)=24, a(3)=39, a(4)=60, a(5)=105, a(6)=156.

G.f.: 3*(3 + 8*x + 13*x^2 + 8*x^3 + 3*x^4) / (1 - 4*x^3 + x^6). - Colin Barker, Dec 27 2017

A293817 Numbers k such that m=2*k is the middle side in a Heronian triangle with sides m-13, m , m+13.

Original entry on oeis.org

13, 14, 19, 26, 37, 62, 91, 134, 229, 338, 499, 854, 1261, 1862, 3187, 4706, 6949, 11894, 17563, 25934, 44389, 65546, 96787, 165662, 244621, 361214, 618259, 912938, 1348069, 2307374, 3407131, 5031062, 8611237, 12715586, 18776179, 32137574, 47455213, 70073654

Offset: 0

Sture Sjöstedt, Dec 27 2017

a(n) gives values of x satisfying 3*x^2 - y^2 = 507; corresponding y values are given by A293846.

			The smallest triangle of this type with 3 acute angles has the sides: 61, 74, 87.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-1).

Cf. A003500, A005320, A296795, A296796.

LinearRecurrence[{0,0,4,0,0,-1},{13,14,19,26,37,62},40] (* Harvey P. Dale, Oct 10 2023 *)

Vec((13 + 14*x + 19*x^2 - 26*x^3 - 19*x^4 - 14*x^5) / (1 - 4*x^3 + x^6) + O(x^40)) \\ Colin Barker, Dec 27 2017

a(n) = 4*a(n-3)-a(n-6), a(1)= 13, a(2)= 14, a(3)= 19, a(4)= 26, a(5)= 37, a(6)= 62.

G.f.: (13 + 14*x + 19*x^2 - 26*x^3 - 19*x^4 - 14*x^5) / (1 - 4*x^3 + x^6). - Colin Barker, Dec 27 2017

A296796 Numbers k such that k is the altitude of a Heronian triangle with sides m - 11, m, m + 11.

Original entry on oeis.org

12, 15, 33, 63, 72, 132, 240, 273, 495, 897, 1020, 1848, 3348, 3807, 6897, 12495, 14208, 25740, 46632, 53025, 96063, 174033, 197892, 358512, 649500, 738543, 1337985, 2423967, 2756280, 4993428, 9046368, 10286577, 18635727, 33761505, 38390028, 69549480

Offset: 0

Sture Sjöstedt, Dec 20 2017

a(n) gives the values of y satisfying 3*x^2 - y^2 = 363; corresponding x values are given by A296795.

a(n)/3 is the radius of the inscribed circle.

			If the sides are 17, 28, 39 the triangle has the altitude 15 against 28 and is a part of the Pythagorean triangle with the sides 15, 36, 39, so 15 is a term.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-1).

CoefficientList[Series[3 (4 + 5 x + 11 x^2 + 5 x^3 + 4 x^4)/(1 - 4 x^3 + x^6), {x, 0, 35}], x] (* Michael De Vlieger, Dec 22 2017 *)

Vec(3*(4 + 5*x + 11*x^2 + 5*x^3 + 4*x^4) / (1 - 4*x^3 + x^6) + O(x^40)) \\ Colin Barker, Dec 22 2017

From Colin Barker, Dec 22 2017: (Start)
G.f.: 3*(4 + 5*x + 11*x^2 + 5*x^3 + 4*x^4) / (1 - 4*x^3 + x^6).
a(n) = 4*a(n-3) - a(n-6) for n>5.
(End)

More terms from Colin Barker, Dec 22 2017

A296795 Numbers k such that m = 2*k is the middle side in a Heronian triangle with sides m - 11, m, m + 11.

Original entry on oeis.org

13, 14, 22, 38, 43, 77, 139, 158, 286, 518, 589, 1067, 1933, 2198, 3982, 7214, 8203, 14861, 26923, 30614, 55462, 100478, 114253, 206987, 374989, 426398, 772486, 1399478, 1591339, 2882957, 5222923, 5938958, 10759342, 19492214, 22164493, 40154411, 72745933

Offset: 0

Sture Sjöstedt, Dec 20 2017

a(n) gives values of x satisfying 3*x^2 - y^2 = 363; the corresponding y values are given by A296796.

			The smallest triangle of this type has following sides: 15, 26, 37 has the altitude 12 and is a part of the Pythagorean triangle with sides : 12, 35, 37.

Colin Barker, Table of n, a(n) for n = 0..1000
Wikipedia, Heronian triangle
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-1).

Cf. A003500, A005320, A296796.

CoefficientList[Series[(13 + 14 x + 22 x^2 - 14 x^3 - 13 x^4 - 11 x^5)/(1 - 4 x^3 + x^6), {x, 0, 36}], x] (* Michael De Vlieger, Dec 22 2017 *)

Vec((13 + 14*x + 22*x^2 - 14*x^3 - 13*x^4 - 11*x^5) / (1 - 4*x^3 + x^6) + O(x^40)) \\ Colin Barker, Dec 22 2017

From Colin Barker, Dec 22 2017: (Start)
G.f.: (13 + 14*x + 22*x^2 - 14*x^3 - 13*x^4 - 11*x^5) / (1 - 4*x^3 + x^6).
a(n) = 4*a(n-3) - a(n-6) for n>5.
(End)

More terms from Colin Barker, Dec 22 2017

A291591 Numbers k such that there exist exactly five distinct Pythagorean triangles, at least one of them primitive, with area k.

Original entry on oeis.org

71831760, 73513440, 1675212000, 6913932480, 4323749790360, 2678930100000, 175434192299520, 503151375767040

Offset: 1

Sture Sjöstedt, Aug 27 2017

I solve x^2 + 3*y^2 = (2*r)^2 over the positive integers. q, r, q-p and p are the y-coordinates in the first quadrant. Area = q*r*(q-p)*p. There are three Pythagorean triangles with this area. j, x, y with x > y and Area = j^2*x*y*(x-y)*(x+y) gives the area of an Pythagorean triangle.
Example: r = 169 in x^2 + 3*y^2 = (2*169)^2 gives q = 176, r = 169, q-p = 161 and p = 15;
k = q*r*(q-p)*p = 176*169*161*15 = 71831760.
j = 26, x = 23, y = 12 and j = 26, x = 28, y = 5 gives two Pythagorean triangles with k = 71831760;
k = 676*23*12*11*35 = 71831760 and k = 676*28*5*23*33 = 71831760.

			p^2 - p*q + q^2 = r^2;
p = 115, q = 448, q-p = 333, r = 403;
k = p*q*(q-p)*r = 115*448*333*403 = 6913932480.
x = 414, y = 104 and x = 558, y = 40 gives the same area.
k = x*y*(x-y)*(x+y) = 414*104*310*518 = 6913932480.
k = x*y*(x-y)*(x+y) = 558*40*518*598 = 6913932480.

Cf. A055193.

a(2), a(4)-a(7) from Giovanni Resta, Aug 28 2017

Missing term 73513440 inserted by Miguel-Ángel Pérez García-Ortega, Jul 19 2021

A291420 Numbers n such that there exist exactly four distinct Pythagorean triangles, at least one of them primitive, with area n.

Original entry on oeis.org

341880, 8168160, 14636160, 17957940, 52492440, 116396280, 1071572040, 1187525640, 1728483120, 5988702720, 6609482880, 22539095040, 29239970760, 136496680320, 258670630680, 398648544840, 494892478080, 592003418160, 1329673884000, 1343798407560, 2190884461920

Offset: 1

Sture Sjöstedt, Aug 23 2017

nonn

Numbers n such that there exist positive integers x, y with x > y and n = x*y*(x-y)*(x+y).

Many of them consist of a Pythagorean triangle plus a triple which is a solution to Carroll's problem: Find three Pythagorean triangles with the same area.

			p^2 - p*q + q^2 = r^2;
p = 208, q = 418, r = 362, q - p = 210;
n = p*r*q*(q-p) = 208*418*362*210 = 6609482880.
x = 640, y = 627 gives the same area:
n = x*y*(x-y)*(x+y) = 640*627*13*1267 = 6609482880.

Cf. A009127, A024407, A055193, A088513, A088977, A089025, A177021, A291591.

a(12)-a(21) from Giovanni Resta, Aug 28 2017

A266698 x-values of solutions to the Diophantine equation x^2 - 7*y^2 = 2.

Original entry on oeis.org

3, 45, 717, 11427, 182115, 2902413, 46256493, 737201475, 11748967107, 187246272237, 2984191388685, 47559815946723, 757972863758883, 12080006004195405, 192522123203367597, 3068273965249686147, 48899861320791610755, 779329507167416085933, 12420372253357865764173, 197946626546558436140835

Offset: 1

Sture Sjöstedt, Jan 03 2016

A159678 gives the y-values of solutions to the Diophantine equation x^2 - 7*y^2 = 2.

G. C. Greubel, Table of n, a(n) for n = 1..750
Index entries for linear recurrences with constant coefficients, signature (16,-1).

Cf. A041008, A077412, A157456, A159678.

[n: n in [1..2*10^7] | IsSquare((n^2-2)/7)]; // Vincenzo Librandi, Jan 06 2016

Mathematica
```
LinearRecurrence[{16,-1}, {3, 45}, 20 ]
```

lista(nn) = {print1(x = 3, ", "); print1(y = 45, ", "); for (n=2, nn, z = 16*y - x; print1(z, ", "); x = y; y = z;);} \\ Michel Marcus, Jan 05 2016

[3*(chebyshev_U(n-1, 8) - chebyshev_U(n-2, 8)) for n in (1..30)] # G. C. Greubel, Jun 25 2022

a(1)=3, a(2)=45, a(n) = 16*a(n-1) - a(n-2).
a(n) = A041008(4n-2). - Robert Israel, Jan 05 2016
From R. J. Mathar, Jan 12 2016: (Start)
G.f.: 3*x*(1-x) / ( 1-16*x+x^2 ).
a(n) = 3*A157456(n). (End)
From G. C. Greubel, Jun 25 2022: (Start)
a(n) = 3*(ChebyshevU(n-1, 8) - ChebyshevU(n-2, 8)).
E.g.f.: exp(8*x)*(3*cosh(3*sqrt(7)*x) - sqrt(7)*sinh(3*sqrt(7)*x)) - 3. (End)

A217975 Integers k such that 2*k^2 - 7 is a square.

Original entry on oeis.org

2, 4, 8, 22, 46, 128, 268, 746, 1562, 4348, 9104, 25342, 53062, 147704, 309268, 860882, 1802546, 5017588, 10506008, 29244646, 61233502, 170450288, 356895004, 993457082, 2080136522, 5790292204, 12123924128, 33748296142, 70663408246, 196699484648

Offset: 1

Sture Sjöstedt, Oct 16 2012

a(n) gives y-values solving the Diophantine equation x^2 + 7 = 2*y^2. A077446(n) gives the x-values. - Sture Sjöstedt, Oct 16 2012

Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 28 = 0. - Colin Barker, Feb 08 2014

			Since 2(4^2) - 7 = 25 = 5^2, and 4 is the second number with this property, a(2) = 4.

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

Cf. A077442 (2*n^2 + 7 is a square).

I:=[2, 4, 8, 22]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..31]]; // Vincenzo Librandi, Oct 16 2012

LinearRecurrence[{0, 6, 0, -1}, {2, 4, 8, 22}, 50] (* Sture Sjöstedt, Oct 16 2012 *)

Vec(2*x*(1-x)*(x^2+3*x+1)/(x^2-2*x-1)/(x^2+2*x-1)+O(x^99)) \\ Charles R Greathouse IV, Oct 24 2012

a(n) = 6*a(n - 2) - a(n - 4) with a(1)=2, a(2)=4, a(3)=8, a(4)=22. - Sture Sjöstedt, Oct 16 2012
a(n)*a(n+3)-a(n+1)*a(n+2) = 10-2*(-1)^n. - Bruno Berselli, Oct 25 2012
a(n) = 2*A006452(n). - R. J. Mathar, Oct 17 2012
G.f.: -2*x*(x - 1)*(x^2 + 3*x + 1)/((x^2 - 2*x - 1)*(x^2 + 2*x - 1)). - Colin Barker, Oct 24 2012
a(n) = a(-n+1) = ((4+sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))+(4-sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2)))/4. - Bruno Berselli, Oct 25 2012
a(2n-1) = A078343(2n-1), a(2n) = A100525(n-1). - Bruno Berselli, Oct 25 2012

A201157 y-values in the solution to 5*x^2 - 20 = y^2.

Original entry on oeis.org

0, 5, 15, 40, 105, 275, 720, 1885, 4935, 12920, 33825, 88555, 231840, 606965, 1589055, 4160200, 10891545, 28514435, 74651760, 195440845, 511670775, 1339571480, 3507043665, 9181559515, 24037634880, 62931345125, 164756400495, 431337856360, 1129257168585

Offset: 1

Sture Sjöstedt, Nov 27 2011

Except a(1), the same as A054888. - R. J. Mathar, Nov 28 2011

			15 is in the sequence because 15^2 = 5*7^2 - 20.

Michael De Vlieger, Table of n, a(n) for n = 1..2392
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A005248.

Mathematica
```
LinearRecurrence[{3, -1}, {0, 5}, 50]
```

a(n) = 3*a(n-1) - a(n-2), n>2.
G.f.: 5*x^2 / (x^2 - 3*x + 1). - Colin Barker, Apr 08 2013
a(n) = 5*Fibonacci(2*n-2) = Lucas(2*n-1) + Lucas(2*n-3) with Lucas(-1) = -1. - Bruno Berselli, Feb 15 2017
a(n) = Lucas(n)^2 - Lucas(n-2)^2. - Greg Dresden, Apr 15 2022

More terms from Colin Barker, Apr 08 2013

A200407 The x-values in the solution to 19*x^2 - 18 = y^2.

Original entry on oeis.org

1, 9, 131, 209, 3051, 44539, 71059, 1037331, 15143129, 24159851, 352689489, 5148619321, 8214278281, 119913388929, 1750515426011, 2792830455689, 40770199546371, 595170096224419, 949554140655979, 13861747932377211, 202356082200876449, 322845614992577171

Offset: 1

Sture Sjöstedt, Nov 17 2011

When are both n+1 and 19*n+1 perfect squares? This gives the equation 19*x^2-18=y^2.

			a(7)=340*209-1=71059.

Index entries for linear recurrences with constant coefficients, signature (0,0,340,0,0,-1).

Cf. A199773, A199774, A199798, A200409.

LinearRecurrence[{0, 0, 340, 0, 0, -1}, {1, 9, 131, 209, 3051, 44539}, 50]

Vec(-x*(x-1)*(x^4+10*x^3+141*x^2+10*x+1)/(x^6-340*x^3+1) + O(x^100)) \\ Colin Barker, Sep 01 2013

a(n) = 340*a(n-3)+a(n-6), a(1)=1, a(2)=9, a(3)=131, a(4)=209, a(5)=3051, a(6)=44539.

G.f.: -x*(x-1)*(x^4+10*x^3+141*x^2+10*x+1) / (x^6-340*x^3+1). - Colin Barker, Sep 01 2013

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User: Sture Sjöstedt

A293846 Numbers such that k is the altitude of a Heronian triangle with sides m-13, m, m+13.

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A293817 Numbers k such that m=2*k is the middle side in a Heronian triangle with sides m-13, m , m+13.

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A296796 Numbers k such that k is the altitude of a Heronian triangle with sides m - 11, m, m + 11.

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A296795 Numbers k such that m = 2*k is the middle side in a Heronian triangle with sides m - 11, m, m + 11.

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A291591 Numbers k such that there exist exactly five distinct Pythagorean triangles, at least one of them primitive, with area k.

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A291420 Numbers n such that there exist exactly four distinct Pythagorean triangles, at least one of them primitive, with area n.

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A266698 x-values of solutions to the Diophantine equation x^2 - 7*y^2 = 2.

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