A024406 -id:A024406 - cp's OEIS frontend

A249869 Triangle giving the area of primitive Pythagorean triangles, with zero entries for non-primitive triangles.

6, 0, 30, 60, 0, 84, 0, 210, 0, 180, 210, 0, 0, 0, 330, 0, 630, 0, 924, 0, 546, 504, 0, 1320, 0, 1560, 0, 840, 0, 1386, 0, 2340, 0, 0, 0, 1224, 990, 0, 2730, 0, 0, 0, 3570, 0, 1710, 0, 2574, 0, 4620, 0, 5610, 0, 5016, 0, 2310, 1716, 0, 0, 0, 7140, 0, 7980, 0, 0, 0, 3036

Offset: 2

Wolfdieter Lang, Dec 03 2014

See A249866 for comments and references.
For the sorted areas of all primitive Pythagorean triangles (x, y, z) with, say y even, see A024406.
Note that in a row > N there may appear smaller numbers than the maximal number up to row N. Therefore the sorted nonvanishing numbers up to a given row N will in general not produce a subsequence of A024406. The minimal areas in rows n = 2..20 are 6, 30, 60, 180, 210, 546, 504, 1224, 990, 2310, 1716, 3900, 2730, 6090, 4080, 8976, 5814, 12654, 7980. For example, one has to go up to row n = 16 to cover all areas <= 4080.
See the link for more details on a safe row number n = N to cover all areas not exceeding a given one, and also for all areas <= 10^6 with their squarefree parts. - Wolfdieter Lang, Nov 25 2016

			The triangle T(n, m) begins:
n\m    1    2    3     4     5     6    7     8     9   10    11
2:     6
3:     0   30
4:    60    0   84
5:     0  210    0   180
6:   210    0    0     0   330
7:     0  630    0   924     0   546
8:   504    0 1320     0  1560     0  840
9:     0 1386    0  2340     0     0    0 1224
10:  990    0 2730     0     0     0 3570    0   1710
11:    0 2574    0  4620     0  5610    0 5016      0 2310
12: 1716    0    0     0  7140     0 7980     0     0    0  3036
...
For more rows see the link.
T(5, 2) = 210 for the primitive triangle (21, 20, 29).
T(6, 1) = 210 for the primitive triangle (35, 12, 37).

Wolfdieter Lang, First rows of the triangle.
Wolfdieter Lang, A Note on the Area table A249869 for Primitive Pythagorean Triangles.

Cf. A024406, A249866, A258150 (one sixth of this triangle), A225949 (leg sums), A225951 (perimeters), A222946 (hypotenuses), A208854 (odd catheti), A208855 (even catheti), A278711.

T(n, m) = n*m*(n+m)(n-m) if n > m >= 1, (-1)^(n+m) = -1 and gcd(n,m) = 1, else 0.

A155176 Perimeter s/6 (divided by 6) of primitive Pythagorean triangles such that perimeters are Averages of twin prime pairs, q=p+1, a=q^2-p^2, c=q^2+p^2, b=2pq, s=a+b+c, s-+1 are primes.

Original entry on oeis.org

2, 5, 40, 77, 287, 590, 1335, 1717, 2882, 3337, 3927, 4030, 6902, 7315, 7740, 8932, 15965, 20592, 26070, 27405, 34277, 34580, 40920, 50692, 92132, 96647, 113575, 139690, 160557, 167167, 220225, 237407, 279720, 300832, 310765, 336777, 389895

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 21 2009

nonn

p=1,q=2,a=3,b=4,c=5,s=12-+1primes, ...

Cf. A020882, A020886, A020884, A020883, A024364, A024406, A155171, A155173, A155174, A155175

lst={};Do[p=n;q=p+1;a=q^2-p^2;c=q^2+p^2;b=2*p*q;s=a+b+c;If[PrimeQ[s-1]&&PrimeQ[s+1],AppendTo[lst,s/6]],{n,8!}];lst

A258150 Triangle of Fibonacci's congruum (congruous) numbers divided by 24 based on primitive Pythagorean triangles. Areas divided by 6 of these triangles.

Original entry on oeis.org

1, 0, 5, 10, 0, 14, 0, 35, 0, 30, 35, 0, 0, 0, 55, 0, 105, 0, 154, 0, 91, 84, 0, 220, 0, 260, 0, 140, 0, 231, 0, 390, 0, 0, 0, 204, 165, 0, 455, 0, 0, 0, 595, 0, 285, 0, 429, 0, 770, 0, 935, 0, 836, 0, 385, 286, 0, 0, 0, 1190, 0, 1330, 0, 0, 0, 506

Offset: 2

Wolfdieter Lang, Jun 11 2015

The problem is: given a square, find a positive integer that, whether added to or subtracted from that square, yields a square. That is, both x^2 + C = y^2 and x^2 - C = z^2. Equivalently: z^2 + C = x^2 and x^2 + C = y^2 (squares in arithmetic progression). This is treated in Fibonacci's 'The book of squares' (Liber Quadratorum (1225) but for rational x,y,z). See the Sigler reference, Proposition 14, pp. 53-74 (note that the formulation of this problem on p. 53 is not correct, 'from a square' should read 'from the same square'). See also van der Waerden, pp. 40-42, and A. Weil, pp. 13-14. The desired number C is called a congruum by Fibonacci (a congruous number in Sigler's translation).
For the history of this problem, see Dickson, pp. 459-472 (he uses the (misleading) term congruent number).
The following solution is based on primitive Pythagorean triangles. (Fibonacci's solution is based on sums of odd squares.) The triangle T(n, m) = 24*C(n, m) will have 0's for those (n, m) not leading to primitive Pythagorean triples.
Addition of the two equations, substitution of y = u + v > 0 and z = |u - v| and division by 2 leads to x^2 = u^2 + v^2. Consider primitive Pythagorean triples (u, v, x) with even v which are pairwise relatively prime. Then also GCD(u,v,x) = 1. A common factor f for u, v and x would lead to a multiplication by f^2 on both sides of the two equations. For primitive Pythagorean triples see A249866. One has u = n^2 - m^2, v = 2*n*m and x = n^2 + m^2 with GCD(n, m) = 1 , n > m >= 1, n + m odd. Then C = C(n, m) = 4*n*m*(n^2 - m^2) = 2*v(n, m)*u(n, m). This is four times the area of the Pythagorean triangle. C is divisible by 4! = 24 (see A020885). Define T(n, m) = C(n, m)/4!, for 2 <= m + 1 <= n. This is the area of the corresponding primitive Pythagorean triangle divided by 6.
The corresponding x = x(n, m), y = y(n, m) and z(n, m) number triangles are given in A222946, A225949 and A258149 respectively.
T(n, m) = n*m*(n^2 - m^2)/6, for m = 1, 2, ..., n-1, has for n >= 2 the minimum value at m = 1, and the next largest value appears for n >= 3 at m = n-1. Note all (n, m) pairs are considered here. The proof of the first part is easy. The proof of T(n, m) - T(n, n-1) > 0, for m = 2, 3, ..., n-2, and n >= 3, is equivalent to n^2*(m-2) + 3*n > m^3 +1 and this is easy to prove with n >= m+2 and m >= 2. Therefore the triangle T(n, m) with 0's attains for even n the smallest nonzero row entry at m = 1, and for odd n the smallest nonzero row entry appears at m = n-1 (last entry).
This allows us to find (after solution of two cubic equations for even and odd n, named ne = ne(N) and no = no(N)) a row number nmin(N) = max(ne(n), no(N)) such that N will not appear in any row n > nmin(N).
The original problem posed to Fibonacci by Giovanni di Palermo (Master John of Palermo) was to find a [rational] square that when increased or decreased by 5 gives a square. Fibonacci gave the solution in his Liber Quadratorum in Proposition 17 (see Sigler, pp. 77-81) as x^2 = (41/12)^2 = 1681/144, y^2 = (49/12)^2 = 2401/144 and z^2 = (31/12)^2 = 961/144. This corresponds to the integer quartet (C; x, y, z) = (720; 41, 49, 31) corresponding to the primitive Pythagorean triple [9, 40, 41]. See the examples for (n, m) = (5, 4).
The numbers without zeros, in nondecreasing order, are given in A020885 = A024406/6.
Comments from Eric Snyder, Feb 07 2023: (Start)
If m+n > 3 and not divisible by 3, then m+n | T(n,m).
Additionally, if 2n-1 > 3 and not divisible by 3, then 2n-1 = 6k+-1, and T(n,n-1) = (2n-1)*P(-+k), where P(-+k) is a generalized pentagonal number (A001318). For example, T(6,5) = 11*P(-2) = 11*5.
T(n,n-1) = A000330(n-1) for n>=2. (End)

			The triangle T(n, m) begins:
n\m   1   2   3   4    5   6    7   8   9  10    11
2:    1
3:    0   5
4:   10   0  14
5:    0  35   0  30
6:   35   0   0   0   55
7:    0 105   0 154    0  91
8:   84   0 220   0  260   0  140
9:    0 231   0 390    0   0    0 204
10: 165   0 455   0    0   0  595   0 285
11:   0 429   0 770    0 935    0 836   0 385
12: 286   0   0   0 1190   0 1330   0   0   0   506
...
The smallest nonzero number for each row with even n is T(n, 1), and for odd n it is T(n, n-1).
The above mentioned nmin(N) will for N = 300 be 12.
Therefore, no number > 300 will appear for rows with n > 12.
-----------------------------------------------------
The corresponding quartets (C; x, y, z) are:
n=2:  (24; 5, 7, 1),
n=3:  (120; 13, 17, 7),
n=4:  (240; 17, 23, 7), (336; 25, 31, 17),
n=5:  (840; 29, 41, 1), (720; 41, 49, 31),
n=6:  (840; 37, 47, 23), (1320; 61, 71, 49),
n=7:  (2520; 53, 73, 17), (3696; 65, 89, 23),
      (2184; 85, 97, 71),
n=8:  (2016; 65, 79, 47), (5280; 73, 103, 7),
      (6240; 89, 119, 41), (3360; 113, 127, 97),
n=9:  (5544; 85, 113, 41), (9360; 97, 137, 7),
      (4896; 145, 161, 127),
n=10: (3960; 101, 119, 79), (10920; 109, 151, 31),
      (14280; 149, 191, 89), (6840; 181, 199, 161),
n=11: (10296; 125, 161, 73), (18480; 137, 193, 17),
      (22440; 157, 217, 47), (20064; 185, 233, 119),
      (9240; 221, 241, 199),
n=12: (6864; 145, 167, 119), (28560; 169, 239, 1),
      (31920; 193, 263, 73), (12144; 265, 287, 241),
...
-----------------------------------------------------
The corresponding primitive Pythagorean triples
(u, v, x) are:
n=2:  [3, 4, 5],
n=3:  [5, 12, 13],
n=4:  [15, 8, 17], [7, 24, 25],
n=5:  [21, 20, 29],[9, 40, 41],
n=6:  [35, 12, 37], [11, 60, 61],
n=7:  [45, 28, 53], [33, 56, 65],
      [13, 84, 85],
n=8:  [63, 16, 65], [55, 48, 73],
      [39, 80, 89], [15, 112, 113],
n=9:  [77, 36, 85], [65, 72, 97],
      [17, 144, 145],
n=10: [99, 20, 101], [91, 60, 109],
      [51, 140, 149], [19, 180, 181],
n=11: [117, 44, 125], [105, 88, 137],
      [85, 132, 157], [57, 176, 185],
      [21, 220, 221],
n=12: [143, 24, 145], [119, 120, 169],
      [95, 168, 193], [23, 264, 265],
...

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 2, 1920, pp. 459-472.
L. E. Sigler, Leonardo Pisano, Fibonacci, The book of squares, Academic Press, 1987.
B. L. van der Waerden, A History of Algebra, Springer, 1985, pp. 40-42.
André Weil, Number Theory, An approach through history, From Hammurapi to Legendre, Birkhäuser, 1984, pp. 13-14.

Cf. A020885, A024365, A024406, A249866, A222946, A225949, A258149.

T[n_, m_] /; 2 <= m+1 <= n && OddQ[n+m] && CoprimeQ[n, m] := n*m*(n^2 - m^2)/6; T[, ] = 0; Table[T[n, m], {n, 2, 12}, {m, 1, n-1}] // Flatten (* Jean-François Alcover, Jun 16 2015, after given formula *)

T(n, m) = n*m*(n^2 - m^2)/6 if 2 <= m+1 <= n, n+m odd, GCD(n, m) = 1 and 0 otherwise.

A155177 Area ar/6 (divided by 6) of primitive Pythagorean triangles such that perimeters are Averages of twin prime pairs, q=p+1, a=q^2-p^2, c=q^2+p^2, b=2pq, ar=a*b/2; s=a+b+c, s-+1 are primes.

Original entry on oeis.org

1, 5, 140, 385, 2870, 8555, 29370, 42925, 93665, 116795, 149226, 155155, 348551, 380380, 414090, 513590, 1229305, 1801800, 2567895, 2767905, 3873301, 3924830, 5053620, 6970150, 17090486, 18362930, 23396450, 31919165, 39336465, 41791750

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 21 2009

nonn

p=1,q=2,a=3,b=4,c=5, ar=3*4/2=6, s=12-+1primes, ...

Cf. A020882, A020886, A020884, A020883, A024364, A024406, A155171, A155173, A155174, A155175, A155176

lst={};Do[p=n;q=p+1;a=q^2-p^2;c=q^2+p^2;b=2*p*q;s=a+b+c;ar=a*b/2;If[PrimeQ[s-1]&&PrimeQ[s+1],AppendTo[lst,ar/6]],{n,8!}];lst

A381009 Ordered areas of the Pythagorean triangles defined by a = 2^(4n) + 2^(2n+1), b = 2^(4n) - 2^(4n-2) - 2^(2n) - 1, c = 2^(4n) + 2^(4n-2) + 2^(2n) + 1.

Original entry on oeis.org

84, 25200, 6350784, 1614708480, 412583721984, 105570270965760, 27022696873181184, 6917599389942743040, 1770891934572664848384, 453347470584212823736320, 116056897129722086198083584, 29710562123440325102508441600, 7605903676927233379495034486784, 1947111326786263531071061496954880

Offset: 1

Robert C. Lyons, Feb 12 2025

Proper subset of A024406.

Paolo Xausa, Table of n, a(n) for n = 1..400
John D. Cook, Sparse binary Pythagorean triples (2025).
H. S. Uhler, A Colossal Primitive Pythagorean Triangle, The American Mathematical Monthly, Vol. 57, No. 5 (May, 1950), pp. 331-332.
Wikipedia, Pythagorean triple.
Index entries for linear recurrences with constant coefficients, signature (340,-22848,348160,-1048576).

Cf. A024406.

Cf. A381005 (short legs), A381006 (long legs), A381007 (hypotenuses), A381008 (perimeters).

[(2^(4*n) + 2^(2*n+1)) * (2^(4*n) - 2^(4*n-2) - 2^(2*n) - 1) / 2: n in [1..20]];

A381009[n_] := (3*# + 2)*(# + 2)*(# - 2)*2^(2*n - 3) & [4^n]; Array[A381009, 20] (* or *)
LinearRecurrence[{340, -22848, 348160, -1048576}, {84, 25200, 6350784, 1614708480}, 20] (* Paolo Xausa, Feb 26 2025 *)

a(n) = (2^(4*n) + 2^(2*n+1)) * (2^(4*n) - 2^(4*n-2) - 2^(2*n) - 1) / 2

def A381009(n): return (m:=1<<(n<<1)-1)*(m-1)*(m+1)*(3*m+1)<<1 # Chai Wah Wu, Feb 13 2025

a(n) = A381005(n) * A381006(n) / 2.
a(n) = (2^(4n) + 2^(2n+1)) * (2^(4n) - 2^(4n-2) - 2^(2n) - 1) / 2.
G.f.: 12*(7 - 280*x - 24832*x^2 + 163840*x^3)/((1 - 4*x)*(1 - 16*x)*(1 - 64*x)*(1 - 256*x)). - Stefano Spezia, Feb 13 2025

A155178 Numbers p of primitive Pythagorean triangles such that perimeters and products of 3 sides are Averages of twin prime pairs, q=p+1, a=q^2-p^2, c=q^2+p^2, b=2pq, ar=ab/2; s=a+b+c, s-+1 are primes, pr=ab*c, pr-+1 are primes.

Original entry on oeis.org

1, 7916, 35882, 37816, 47491, 128429, 131830, 146471, 154799, 157579, 170219, 174964, 187544, 207829, 208039, 222887, 223142, 262502, 291544, 319825, 327602, 331627, 353857, 476681, 477659, 494207, 522025, 537454, 540682, 558161, 571670

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 21 2009

nonn

p=1,q=2,a=3,b=4,c=5,s=12-+1 primes,pr=3*4*5=60-+1 primes, ...

Cf. A020882, A020886, A020884, A020883, A024364, A024406, A155171, A155173, A155174, A155175, A155176, A155177

lst={};Do[p=n;q=p+1;a=q^2-p^2;c=q^2+p^2;b=2*p*q;ar=a*b/2;s=a+b+c;pr=a*b*c;If[PrimeQ[s-1]&&PrimeQ[s+1]&&PrimeQ[pr-1]&&PrimeQ[pr+1],AppendTo[lst,n]],{n,3*9!}];lst

A277557 The ordered image of the 1-to-1 mapping of an integer ordered pair (x,y) into an integer using Cantor's pairing function, where 0 < x < y, gcd(x,y)=1 and x+y odd.

Original entry on oeis.org

8, 18, 19, 32, 33, 34, 50, 52, 53, 72, 73, 74, 75, 76, 98, 99, 100, 101, 102, 103, 128, 131, 133, 134, 162, 163, 164, 165, 166, 167, 168, 169, 200, 201, 202, 203, 204, 205, 206, 207, 208, 242, 244, 247, 248, 250, 251, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 338

Offset: 1

Frank M Jackson, Oct 19 2016

nonn

The mapping of the ordered pair (x,y) to an integer uses Cantor's pairing function to generate the integer as (x+y)(x+y+1)/2+y. Also for every ordered pair (x,y) such that 0 < x < y, gcd(x,y)=1 and x+y odd, there exists a primitive Pythagorean triple (PPT) (a, b, c) such that a = y^2-x^2, b = 2xy, c = x^2+y^2. Therefore each term in the sequence represents a unique PPT.
Numbers n for which 0 < A025581(n) < A002262(n) and A025581(n)+A002262(n) is odd, and gcd(A025581(n), A002262(n)) = 1. [The definition expressed with A-numbers.] - Antti Karttunen, Nov 02 2016
See also the triangle T(y, x) with the values for PPTs given in A278147. - Wolfdieter Lang, Nov 24 2016

			a(5)=33 because the ordered pair (2,5) maps to 33 by Cantor's pairing function (see below) and is the 5th such occurrence. Also x=2, y=5 generates a PPT with sides (21,20,29).
Note: Cantor's pairing function is simply A001477 in its two-argument tabular form A001477(k, n) = n + (k+n)*(k+n+1)/2, thus A001477(2,5) = 5 + (2+5)*(2+5+1)/2 = 33. - _Antti Karttunen_, Nov 02 2016

Wikipedia, Cantor's pairing function, and Pythagorean triple

Cf. A001477, A002262, A025581, A243808, A277632.
Cf. A020882 (is obtained when A048147(a(n)) is sorted into ascending order), A008846 (same with duplicates removed).
Cf. A020887, A020888, A120427, A024362, A024406, A046079, A046087, A070151, A156678, A156679, A156680, A156683, A156685, A222946, A278147.

Cantor[{i_, j_}] := (i+j)(i+j+1)/2+j; getparts[n_] := Reverse@Select[Reverse[IntegerPartitions[n, {2}], 2], GCD@@#==1 &]; pairs=Flatten[Table[getparts[2n+1], {n, 1, 20}], 1]; Table[Cantor[pairs[[n]]], {n, 1, Length[pairs]}]

A378148 a(n) is the number of distinct trapezoids having integer sides and height with exactly one pair of parallel sides and area n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 2, 1, 0, 3, 0, 1, 2, 1, 0, 3, 0, 1, 2, 1, 0, 4, 0, 2, 2, 1, 1, 5, 0, 1, 2, 3, 0, 5, 0, 2, 3, 1, 0, 6, 0, 2, 2, 2, 0, 7, 1, 3, 2, 1, 0, 9, 0, 1, 3, 3, 2, 8, 0, 3, 2, 3, 0, 10, 0, 1, 5, 3, 0, 9, 0, 6, 3, 1, 0, 10, 2, 1, 2

Offset: 1

Felix Huber, Dec 02 2024

nonn

The number of trapezoids having integer sides and height, which are neither right-angled nor isosceles, is a(n) - A378149(n) - A378150(n). The first trapezoid, which is neither right-angled nor isosceles, appears at a(36).
a(p) = 0 for prime p. Proof: Suppose there is a trapezoid with integer sides and prime area p. Then in p = m*h (m is the average of the parallel sides and h is the height of the trapezoid) m = p and h = 1 or m = p/2 and h = 2. At least one nonparallel side of the trapezoid is the hypotenuse of a right triangle with leg h. Legs in integer right triangles are >= 3. This is a contradiction and therefore a(p) = 0.
A214602 is the index of the positive terms in this sequence.
There are also integer-sided trapezoids with integer area that do not have an integer height. For example, the trapezoid with sides p = 630, d = 615, q = 5, f = 40 (p and q are parallel) has an area of 12192 and a height of h = 38.4.

			a(54) = 7 because there are 7 distinct trapezoids [p, d, q, f, h] (p and q are parallel, height h) having integer sides and height with area 54:[17, 10, 1, 10, 6], [13, 6, 5, 10, 6], [22, 5, 14, 5, 3], [20, 3, 16, 5, 3], [8, 15, 1, 20, 12], [7, 12, 2, 13, 12], [15, 4, 12, 5, 4].
For a(54) = 7 and (92) = 4 see the linked illustrations.
See also the linked Maple program "Trapezoids having integer sides and height with area n".

Felix Huber, Table of n, a(n) for n = 1..10000
Felix Huber, Illustration of a(54) = 7
Felix Huber, Illustration of a(92) = 4
Felix Huber, Trapezoids having integer sides and height with area n
Eric Weisstein's World of Mathematics, Trapezoid

Cf. A024406, A027750, A103606, A214602, A340858, A340859, A340860, A365049, A374594, A378149, A378150.

A378148:=proc(n)
   local a,m,p,q,h,x,y,M;
   a:=0;
   M:=map(x->x/2,NumberTheory:-Divisors(2*n) minus {1,2});
   for m in M do
      for q from 1 to m-1/2 do
         p:=2*m-q;
         h:=n/m;
         for x from max(3,floor((p-q+1)/2)) to (h^2-1)/2 do
            y:=p-q-x;
            if issqr(x^2+h^2) and issqr(y^2+h^2) then
               a:=a+1
            fi
         od
      od
   od;
   return a
end proc;
seq(A378148(n),n=1..87);

a(p) = 0 for prime p.

A155180 Short leg A of primitive Pythagorean triangles such that perimeters and products of 3 sides are Averages of twin prime pairs, q=p+1, a=q^2-p^2, c=q^2+p^2, b=2pq, ar=ab/2; s=a+b+c, s-+1 are primes, pr=ab*c, pr-+1 are primes.

Original entry on oeis.org

3, 15833, 71765, 75633, 94983, 256859, 263661, 292943, 309599, 315159, 340439, 349929, 375089, 415659, 416079, 445775, 446285, 525005, 583089, 639651, 655205, 663255, 707715, 953363, 955319, 988415, 1044051, 1074909, 1081365, 1116323

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 21 2009

nonn

p=1,q=2,a=3,b=4,c=5,s=12-+1 primes,pr=3*4*5=60-+1 primes, ...

Cf. A020882, A020886, A020884, A020883, A024364, A024406, A155171, A155173, A155174, A155175, A155176, A155177, A155178

lst={};Do[p=n;q=p+1;a=q^2-p^2;c=q^2+p^2;b=2*p*q;ar=a*b/2;s=a+b+c;pr=a*b*c;If[PrimeQ[s-1]&&PrimeQ[s+1]&&PrimeQ[pr-1]&&PrimeQ[pr+1],AppendTo[lst,a]],{n,3*9!}];lst

A225414 Ordered counts of internal lattice points within primitive Pythagorean triangles (PPT).

Original entry on oeis.org

3, 22, 49, 69, 156, 187, 190, 295, 465, 498, 594, 777, 880, 931, 1144, 1269, 1330, 1501, 1611, 1633, 2190, 2272, 2494, 2619, 2655, 2893, 3475, 3732, 3937, 4182, 4524, 4719, 4900, 5502, 5635, 5866, 6490, 7021, 7185, 7719, 7761, 7828, 7849, 8688

Offset: 1

Frank M Jackson, May 07 2013

nonn

A PPT can be drawn as a closed lattice polygon with the hypotenuse intersecting no lattice points other than at its start and end. Consequently the PPT is subject to Pick's theorem.

			a(5)=156 as when x = 5 and n = 4, the PPT generated has area A = 180 and sides 9, 40, 41. Hence 156=180-(9+40+1)/2+1 and is the 5th such occurrence.

Eric W. Weisstein, MathWorld: Pick's Theorem
Wikipedia, Pick's theorem

Cf. A024406.

getpairs[k_] := Reverse[Select[IntegerPartitions[k, {2}], GCD[#[[1]], #[[2]]]==1 &]]; getlist[j_] := (newlist=getpairs[j]; Table[(newlist[[m]][[1]]^2-newlist[[m]][[2]]^2-1)*(2 newlist[[m]][[1]]*newlist[[m]][[2]]-1)/2, {m, 1, Length[newlist]}]); maxterms = 60; Sort[Flatten[Table[getlist[2p+1], {p, 1, 10*maxterms}]]][[1;;maxterms]] (* corrected with suggestion from Giovanni Resta, May 07 2013 *)

Let x and y be integers used to generate the set of PPT's where x > y > 0, x + y is odd and GCD(x, y) = 1. Then the PPT area A = x*y(x^2-y^2) and the lattice points lying on the PPT boundary B = x^2-y^2+2x*y+1. Applying Pick's theorem gives internal lattice points I = A - B/2 + 1. Hence I = (x^2-y^2-1)*(2x*y-1)/2.

cp's OEIS Frontend

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A155177 Area ar/6 (divided by 6) of primitive Pythagorean triangles such that perimeters are Averages of twin prime pairs, q=p+1, a=q^2-p^2, c=q^2+p^2, b=2*p*q, ar=a*b/2; s=a+b+c, s-+1 are primes.

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A155178 Numbers p of primitive Pythagorean triangles such that perimeters and products of 3 sides are Averages of twin prime pairs, q=p+1, a=q^2-p^2, c=q^2+p^2, b=2*p*q, ar=a*b/2; s=a+b+c, s-+1 are primes, pr=a*b*c, pr-+1 are primes.

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A155180 Short leg A of primitive Pythagorean triangles such that perimeters and products of 3 sides are Averages of twin prime pairs, q=p+1, a=q^2-p^2, c=q^2+p^2, b=2*p*q, ar=a*b/2; s=a+b+c, s-+1 are primes, pr=a*b*c, pr-+1 are primes.

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A155180 Short leg A of primitive Pythagorean triangles such that perimeters and products of 3 sides are Averages of twin prime pairs, q=p+1, a=q^2-p^2, c=q^2+p^2, b=2pq, ar=ab/2; s=a+b+c, s-+1 are primes, pr=ab*c, pr-+1 are primes.