A003500 -id:A003500 - cp's OEIS frontend

A060964 Table by antidiagonals where T(n,k) = n*T(n,k-1) - T(n,k-2) with T(n,0) = 2 and T(n,1) = n.

2, 0, 2, -2, 1, 2, 0, -1, 2, 2, 2, -2, 2, 3, 2, 0, -1, 2, 7, 4, 2, -2, 1, 2, 18, 14, 5, 2, 0, 2, 2, 47, 52, 23, 6, 2, 2, 1, 2, 123, 194, 110, 34, 7, 2, 0, -1, 2, 322, 724, 527, 198, 47, 8, 2, -2, -2, 2, 843, 2702, 2525, 1154, 322, 62, 9, 2, 0, -1, 2, 2207, 10084, 12098, 6726, 2207, 488, 79, 10, 2

Offset: 0

Henry Bottomley, May 09 2001

			Square array begins as:
  2, 0, -2,   0,   2,    0,    -2, ...
  2, 1, -1,  -2,  -1,    1,     2, ...
  2, 2,  2,   2,   2,    2,     2, ...
  2, 3,  7,  18,  47,  123,   322, ...
  2, 4, 14,  52, 194,  724,  2702, ...
  2, 5, 23, 110, 527, 2525, 12098, ...

G. C. Greubel, Antidiagonals n = 0..100, flattened

Rows include A003499, A003500, A003501, A005248, A007395, A056854, A056918, A057079.

Columns include A001477, A007395, A008865, A058794.

T:= function(n,k)
    if k=0 then return 2;
    elif k=1 then return n;
    else return n*T(n,k-1) - T(n,k-2);
    fi; end;
Flat(List([0..12], n-> List([0..n], k-> T(k,n-k) ))); # G. C. Greubel, Jan 15 2020

function T(n,k)
  if k eq 0 then return 2;
  elif k eq 1 then return n;
  else return n*T(n, k-1) - T(n, k-2);
  end if; return T; end function;
[T(k,n-k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 15 2020

seq(seq( simplify(k*ChebyshevU(n-k-1, k/2) -2*ChebyshevU(n-k-2, k/2)), k=0..n), n=0..12); # G. C. Greubel, Jan 15 2020

Table[k*ChebyshevU[n-k-1, k/2] -2*ChebyshevU[n-k-2, k/2], {n,0,12}, {k,0,n} ]//Flatten

T(n,k) = n*polchebyshev(k-1,2,n/2) -2*polchebyshev(k-2,2,n/2);
for(n=0,12, for(k=0,n, print1(T(k,n-k), ", "))) \\ G. C. Greubel, Jan 15 2020

[[k*chebyshev_U(n-k-1, k/2) -2*chebyshev_U(n-k-2, k/2) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 15 2020

For all m, T(n, k) = T(n, |m|)*T(n, |k - m|) - T(n, |k - 2m|).
T(n, 2k) = T(n, k)^2 - 2.
T(n, 2k + 1) = T(n, k)*T(n, k + 1) - n.
T(n, 3k) = T(n, k)^3 - 3*T(n, k).
T(n, 4k) = T(n, k)^4 - 4*T(n, k)^2 + 2.
T(n, 5k) = T(n, k)^5 - 5*T(n, k)^3 + 5*T(n, k) etc.
T(n, -k) = T(n, k).
T(-n, k) = T(-n, -k) = (-1)^k * T(n, k).
T(n, k) = ( n*( ((n + sqrt(n^2 -4))/2)^k - ((n - sqrt(n^2 -4))/2)^k ) - 2*( ((n + sqrt(n^2 -4))/2)^(k-1) - ((n - sqrt(n^2 -4))/2)^(k-1) ) )/sqrt(n^2 -4).
T(n, k) = n*ChebyshevU(k-1, n/2) - 2*ChebyshevU(k-2, n/2). - G. C. Greubel, Jan 15 2020

A268281 Numbers n such that n-tau(n), phi(n) and n form a Heronian triangle, where tau=A000005 is the number of divisors and phi=A000010 the totient.

Original entry on oeis.org

5, 34, 53, 90, 120, 440, 780, 1954, 120994, 140453, 28813276834

Offset: 1

Frank M Jackson, Jan 29 2016

For all n, n > tau(n) and n > phi(n) and if n is prime then n-tau(n) = n-2 and phi(n) = n-1. So n = 5 gives the triangle {3, 4, 5} which is a primitive Pythagorean triangle and this is the only one. Other Pythagorean triangles are {30, 16, 34} and {756, 192, 780}, the remainder are only Heronian.
It is not known if this sequence is infinite. Prime numbers in the sequence are 5, 53 and 140453 and generate triangles {3, 4, 5}, {51, 52, 53} and {140451, 140452, 140453}.
If n = 2p where p is prime then n-tau(n) = n-4 and phi(n) = n/2-1. So n = 34 gives the triangle {16, 30, 34}. Similar numbers in this sequence are a(8), a(9) and a(11). See A272365 for generating Heronian triangles with sides n, n-4, n/2-1.
a(12) > 2*10^12. - Giovanni Resta, Apr 14 2016
Next prime value of a(n) after 140453 is > 2*10^5719. See A003500 for generating Heronian triangles with consecutive sides. - Frank M Jackson, Apr 19 2016
A003500(n)+1 is a member of this sequence iff it is prime. Also A272365(n) is a member of this sequence iff A272365(n)/2 is prime. - Frank M Jackson, Apr 29 2016

			a(2) = 34 because the triangle so formed has sides 30, 16, 34. It is Heronian with integer area 240 and is also Pythagorean. It is the second Heronian triangle.
The triangle corresponding to a(11) has sides n = 28813276834, n-tau(n) = 28813276830, phi(n) = 14406638416, and area 200960614753814018640.

Eric Weisstein's World of Mathematics, Divisor Function
Eric Weisstein's World of Mathematics, Totient Function
Eric Weisstein's World of Mathematics, Heronian Triangle.

Cf. A000005, A000010, A003500, A046022, A049820, A084820, A272365.

triples[n_] := ({a, b, c}={n-DivisorSigma[0, n], EulerPhi[n], n}; s=(a+b+c)/2; If[a+b>c&&IntegerQ[Sqrt[s(s-a)(s-b)(s-c)]], {a, b, c}, {}]); lst={}; Do[If[triples[n]!={}, AppendTo[lst, Last[triples[n]]]], {n, 1, 200000}]; lst

a(11) from Giovanni Resta, Apr 14 2016

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

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

Original entry on oeis.org

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

A297384 Number of Eulerian cycles in the n-antiprism graph.

Original entry on oeis.org

4, 44, 372, 2932, 22484, 170196, 1279828, 9590612, 71736660, 536055124, 4003591508, 29892900180, 223162389844, 1665861735764, 12434781197652, 92816950121812, 692805066118484, 5171207088198996, 38598573880071508, 288104312443589972, 2150442403051689300

Offset: 1

Eric W. Weisstein, Dec 29 2017

Sequence extrapolated to n=1 and n=2 using the recurrence. - Andrew Howroyd, Jan 11 2018

Andrew Howroyd, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Antiprism Graph
Eric Weisstein's World of Mathematics, Eulerian Cycle
Index entries for linear recurrences with constant coefficients, signature (13, -48, 52, -16).

Table[2^n ((2 - Sqrt[3])^n + (2 + Sqrt[3])^n) - 2/3 (2 + 4^n), {n, 20}] // Expand
Table[2^n LucasL[2 n, Sqrt[2]] - 2/3 (2 + 4^n), {n, 20}] // Round
LinearRecurrence[{13, -48, 52, -16}, {4, 44, 372, 2932}, 20]
CoefficientList[Series[-4 (-1 + 2 x + 2 x^2)/(1 - 13 x + 48 x^2 - 52 x^3 + 16 x^4), {x, 0, 20}], x]

Vec(4*(1 - 2*x - 2*x^2)/((1 - 8*x + 4*x^2)*(1 - 4*x)*(1 - x)) + O(x^30)) \\ Andrew Howroyd, Jan 11 2018

From Andrew Howroyd, Jan 11 2018: (Start)
a(n) = 13*a(n-1) - 48*a(n-2) + 52*a(n-3) - 16*a(n-4).
G.f.: 4*x*(1 - 2*x - 2*x^2)/((1 - 8*x + 4*x^2)*(1 - 4*x)*(1 - x)).
(End)
From Eric W. Weisstein, Jan 12 2018: (Start)
a(n) = 2^n*((2 - sqrt(3))^n + (2 + sqrt(3))^n) - 2/3*(2 + 4^n).
a(n) = 2^n*A003500(n) - 2/3*(2 + 4^n).
a(n) = A003500(n) - A039301(n-1).
(End)

A087601 Primes occurring as lesser side of Heronian triangle (sides are consecutive integers, area and inradius are integers).

Original entry on oeis.org

3, 13, 193, 37633, 7300801, 1416317953

Offset: 1

Zak Seidov, Aug 07 2003

Next term (if it exists) is greater than 10^10000. - Ray Chandler, Jul 04 2015

			a(1)=3 because 3, 4 and 5 are sides of Heronian triangle (area is 6, inradius is 1).

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.

Cf. A001353, A016064, A003500, A087602.

Prime elements of A016064. Alternatively, primes p such that 3*(p+1)^2-12 is a square, i.e., p+1 belongs to A003500. - Max Alekseyev, May 14 2010

A087602 Primes occurring as larger side of Heronian triangle (sides are consecutive integers, area and inradius are integers).

Original entry on oeis.org

3, 5, 53, 140453

Offset: 1

Zak Seidov, Aug 07 2003

Next term (if it exists) is greater than 10^10000. - Ray Chandler, Jul 04 2015

			a(3)=53 because 53, 52 and 51 are sides of Heronian triangle (area is 1170, inradius is 15).

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.

Cf. A001353, A016064, A003500, A087601.

p, p-1 and p-2 are sides of Heronian triangle.

Primes p such that 3*(p-1)^2-12 is a square, i.e., p-1 belongs to A003500. - Max Alekseyev, May 14 2010

A242497 Sides of (Heronian) triangles where sides are consecutive integers and area is an integer.

Original entry on oeis.org

3, 4, 5, 13, 14, 15, 51, 52, 53, 193, 194, 195, 723, 724, 725, 2701, 2702, 2703, 10083, 10084, 10085, 37633, 37634, 37635, 140451, 140452, 140453, 524173, 524174, 524175, 1956243, 1956244, 1956245, 7300801, 7300802, 7300803, 27246963, 27246964, 27246965

Offset: 1

R. K. Guy and Charles R Greathouse IV, May 16 2014

Let the edge lengths of the triangle be 2x-1, 2x, 2x+1 so that area = sqrt{3x * x * (x-1) * (x+1)} and we need x^2 - 1 to be of shape 3y^2. That is, x/y is an even rank convergent to the continued fraction of sqrt(3) and x is A001075.

The intermediate length sides are given by A003500(n), n >= 1. Note that A003500(0) = 2 corresponds to the degenerate (Heronian) triangle with sides {1, 2, 3} and area 0. - Daniel Forgues, May 28 2014

Nakane Genkei (Nakane the Elder), Shichijo Beki Yenshiki, 1691.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
David Eugene Smith and Yoshio Mikami, A history of Japanese mathematics, Dover, 2004, p. 168.
Index entries for linear recurrences with constant coefficients, signature (-1,-1,4,4,4,-1,-1,-1).

A016064 is the main entry for this sequence.

LinearRecurrence[{-1,-1,4,4,4,-1,-1,-1},{3,4,5,13,14,15,51,52},40] (* Harvey P. Dale, May 04 2021 *)

Vec((-3*x^7 - 5*x^6 - 6*x^5 + 4*x^4 + 10*x^3 + 12*x^2 + 7*x + 3)/(x^8 + x^7+ x^6 - 4*x^5 - 4*x^4 - 4*x^3 + x^2 + x + 1)+O(x^99))

G.f.: (-3*x^7 - 5*x^6 - 6*x^5 + 4*x^4 + 10*x^3 + 12*x^2 + 7*x + 3)/ ((1+x+x^2)*(1-4*x^3+x^6)). - R. J. Mathar, May 30 2023

A272365 a(n) = 9a(n-1) - 9a(n-2) + a(n-3).

Original entry on oeis.org

10, 34, 250, 1954, 15370, 120994, 952570, 7499554, 59043850, 464851234, 3659766010, 28813276834, 226846448650, 1785958312354, 14060820050170, 110700602088994, 871543996661770, 6861651371205154, 54021666972979450, 425311684412630434, 3348471808328064010, 26362462782211881634

Offset: 1

Frank M Jackson, Apr 27 2016

For n>1 this linear recurrence generates Heronian triangles whose sides are a(n), a(n)-4, a(n)/2-1 and whose area K = (a(n)-2)*sqrt(15(a(n)-10)(a(n)+6))/16. When a(n) = 2p and p is an odd prime then 2p-tau(2p) = a(n)-4 and phi(2p) = a(n)/2-1, where tau=A000005 is the number of divisors and phi=A000010 the totient. Hence when a(n) = 2p for some odd prime p, it is a member of A268281.

			a(2) = 34 because the triangle so formed has sides 34, 30, 16. It is Heronian with integer area 240 and is also Pythagorean. Because 34 = 2*17 and 17 is prime, it is also a member of A268281.

Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

A268281 is a member of this sequence iff A268281/2 is prime.

Cf. A000005, A000010, A003500, A268281

LinearRecurrence[{9, -9, 1}, {10, 34, 250}, 24]

Vec(2*x*(5-28*x+17*x^2)/(1-9*x+9*x^2-x^3) + O(x^99)) \\ Altug Alkan, Apr 28 2016

a(n) = 9a(n-1) - 9a(n-2) + a(n-3).
From Ilya Gutkovskiy, Apr 27 2016: (Start)
G.f.: -2*x*(5-28*x+17*x^2) / ( (x-1)*(x^2-8*x+1) )
a(n) = 2*(2*(4 + sqrt(15))*(4 - sqrt(15))^n - 2*(sqrt(15) - 4)*(4 + sqrt(15))^n + 1). (End)

A349948 Tetrahedral-sided isosceles Heron triangle pairs.

Original entry on oeis.org

0, 10, 48, 190, 720, 2698, 10080, 37630, 140448, 524170, 1956240, 7300798, 27246960, 101687050, 379501248, 1416317950, 5285770560, 19726764298, 73621286640, 274758382270, 1025412242448, 3826890587530, 14282150107680, 53301709843198, 198924689265120

Offset: 1

Randall L Rathbun, Mar 26 2022

Isosceles Heron triangle pairs with tetrahedral sides: [t(a(n)+1), t(a(n)+1), t(a(n))] and [t(a(n)+6), t(a(n)+5), t(a(n)+5)] where t(n) = A000292(n) is a tetrahedral number, i.e., t(n) = n*(n+1)*(n+2)/6. The Heron triangle pair areas have been checked for rationality to 100 terms of {a(n)}.
Not all isosceles Heron triangles with tetrahedral sides are generated by this sequence. For example, [t(63),t(50),t(50)] is not included. Also, scalene Heron triangles with tetrahedral sides are not included. For example, [t(111),t(104),t(62)]. - Michael Somos, Mar 27 2022
Area of triangles: T1(n) = (b(n)-2)^2*(b(n)-3)^2*(b(n)-4)*c(n)/48 and T2(n) = (b(n)+2)^2*(b(n)+3)^2*(b(n)+4)*c(n)/48, where b(n) = A003500(n) and c(n) =  A052530(n). - Randall L Rathbun, Apr 01 2022
Conjecture: for k a positive integer, the sequence {a(k^n): n >= 1} is a strong divisibility sequence; that is, for n, m >= 1,  gcd(a(k^n), a(k^m)) = a(k^gcd(n,m)). - Peter Bala, Dec 03 2022

			10 is a term, so there exists one Heron isosceles triangle whose sides are the 10th, 11th, and 11th tetrahedral numbers (220, 286, 286) and another whose sides are the 15th, 15th, and 16th tetrahedral numbers (680, 680, 816). Those two triangles have areas 29040 and 221952, respectively. (See the n=2 row of the table below.)
.
             Triangle sides               Triangle sides
     k=    ------------------          --------------------
  n a(n)   T(k) T(k+1) T(k+1)  Area    T(k+5) T(k+5) T(k+6)   Area
  - ----   ---- ------ ------ ------   ------ ------ ------  ------
  1    0      0      1      1      0*      35     35     56     588
  2   10    220    286    286  29040      680    680    816  221952
*(degenerate triangle)

Eric Weisstein's World of Mathematics, Tetrahedral Number
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A000292, A001075, A003500, A052530.

a[ n_] := 2*ChebyshevT[n, 2] - 4; (* Michael Somos, Mar 27 2022 *)

Vec(2*x^2*(5 - x)/(1 - 5*x + 5*x^2 - x^3) + O(x^42))

{a(n) = 2*polchebyshev(n,1, 2) - 4}; /* Michael Somos, Mar 27 2022 */

a(n+2) = 4*a(n+1) - a(n) + 8.
From Stefano Spezia, Mar 26 2022: (Start)
G.f.: 2*x^2*(5 - x)/((1-x)*(1 - 4*x +x^2)).
a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) for n > 3.
a(n) = (2 + sqrt(3))^n + (2 - sqrt(3))^n - 4. (End)
a(n) = 2*A001075(n) - 4. - Michael Somos, Mar 27 2022

cp's OEIS Frontend

A060964 Table by antidiagonals where T(n,k) = n*T(n,k-1) - T(n,k-2) with T(n,0) = 2 and T(n,1) = n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A268281 Numbers n such that n-tau(n), phi(n) and n form a Heronian triangle, where tau=A000005 is the number of divisors and phi=A000010 the totient.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A297384 Number of Eulerian cycles in the n-antiprism graph.

Views

Author

Keywords

Comments

Links

Programs

Mathematica

PARI

Formula

A087601 Primes occurring as lesser side of Heronian triangle (sides are consecutive integers, area and inradius are integers).

Views

Author

Keywords

Comments

Examples

References

Crossrefs

Formula

A087602 Primes occurring as larger side of Heronian triangle (sides are consecutive integers, area and inradius are integers).

Views

Author

Keywords

Comments

Examples

References

Crossrefs

Formula

A242497 Sides of (Heronian) triangles where sides are consecutive integers and area is an integer.