A009112 -id:A009112 - cp's OEIS frontend

A177063 Number of Pythagorean triangles with area n.

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

M. F. Hasler, Dec 09 2010

nonn

The first term > 1 is a(210) = 2, cf. A009127, A055193 and A024407. Up to there the sequence coincides with the characteristic function of A009112. The triangles are not necessarily primitive.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries related to Pythagorean triples.

Cf. A009111, A009127, A177021.

a(n)=my(N=1+#n=divisors(2*n));sum(i=1,N\2,issquare(n[i]^2+n[N-i]^2));

Secondary offset added by Antti Karttunen, Nov 24 2017

A228874 a(n) = L(n) * L(n+1) * L(n+2) * L(n+3), the product of four consecutive Lucas numbers, A000032.

Original entry on oeis.org

24, 84, 924, 5544, 40194, 269874, 1864584, 12741324, 87431844, 599001144, 4106310474, 28143249834, 192901471224, 1322153872644, 9062210132844, 62113226746824, 425730613530834, 2918000448971874, 20000274149827944, 137083914357154044, 939587137457703924

Offset: 0

T. D. Noe, Sep 24 2013

Mohanty and Mohanty prove in Corollary 2.6 that these numbers are Pythagorean. The number a(n) is primitive Pythagorean if Lucas(n) and Lucas(n+1) have opposite parity. Every third number, starting at a(1) = 84, is not primitive Pythagorean.

Since a(n) = L(n+1)*L(n+2)*(L(n+2)^2-L(n+1)^2), these numbers are in A073120, - Robert Israel, Apr 06 2015

Robert Israel, Table of n, a(n) for n = 0..1193
Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, Fibonacci Quarterly 28 (1990), 31-42.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000032 (Lucas numbers), A228873 (similar sequence for Fibonacci numbers).

Cf. A009112 (Pythagorean numbers), A024365, A073120.

L:= n -> 2*combinat:-fibonacci(n+1)-combinat:-fibonacci(n):
seq(mul(L(n+i),i=0..3),n=0..30); # Robert Israel, Apr 06 2015

Table[LucasL[n] LucasL[n+1] LucasL[n+2] LucasL[n+3], {n, 0, 25}]
Times@@@Partition[LucasL[Range[0,30]],4,1] (* Harvey P. Dale, Jul 11 2017 *)

Vec(6*(x^4-4*x^3-24*x^2+6*x-4)/((x-1)*(x^2-7*x+1)*(x^2+3*x+1)) + O(x^100)) \\ Colin Barker, Oct 29 2013

G.f.: 6*(x^4-4*x^3-24*x^2+6*x-4) / ((x-1)*(x^2-7*x+1)*(x^2+3*x+1)). - Colin Barker, Oct 29 2013
From Robert Israel, Apr 06 2015: (Start)
a(n+5) = 5*a(n+4) + 15*a(n+3) - 15*a(n+2) - 5*a(n+1) + a(n).
a(n) = -A228873(n+3) + 4*A228873(n+2) + 24*A228873(n+1) - 6*A228873(n) + 4*A228873(n-1) for n >= 2. (End)
Sum_{n>=0} 1/a(n) = (10 - 3*sqrt(5))/60. - Diego Rattaggi, Aug 16 2021

A364169 Smallest integer m = bc which satisfies (b + c)n = m - 1.

Original entry on oeis.org

6, 21, 40, 105, 126, 301, 204, 273, 550, 1221, 936, 697, 690, 3165, 2176, 4641, 1242, 1333, 4200, 8841, 1786, 3213, 2508, 15025, 9126, 18981, 3700, 6105, 13950, 3901, 3876, 4161, 6106, 5781, 23976, 49321, 8178, 6765, 32800, 67281, 6930, 18565, 7440, 11001, 49726, 8925, 9072, 26977

Offset: 1

Jose Aranda, Jul 12 2023

All terms have b,c > 1.
a(n) is the smallest of a certain n-class. The 1-class would be related to the numbers denoted as "pqrs" in A009112.
From David A. Corneth, Jul 19 2023: (Start)
n < min(b, c) <= 2*n.
Proof of n < min(b, c):
As m = b*c and (b + c)*n = m - 1 we have m - (m - 1) = 1 = b*c - (b + c)*n.
Solving 1 = b*c = (b + c)*n for c gives c = (n*b + 1) / (b - n) > 0.
As 0 < b, c, n we have b - n > 0 so b > n.
Similarily as b = (n*c + 1)/(c - n) we have c > n.
Proof of min(b, c) <= 2*n by contradiction.
Suppose 2*n < min(b, c). Then 2*n + 1 <= min(b, c) as both b and c are integers.
Let b = 2n + b' and c = 2n + c' where 1 <= b', c', n. Then
1 = b*c - (b + c)*n = (2n + c') * (2n + b') - (2n + c' + 2n + b') * n = (b' + c')*n + c'*b' > 1. A contradiction. Q.e.d. (End)

			a(1) = 6 as 6 = 2*3, with (2 + 3)*1 = 6 - 1.
a(2) = 21 as 21 = 3*7, with (3 + 7)*2 = 21 - 1.
a(6) = 301 as 301 = 7*43, with (7 + 43)*6 = 301 - 1.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A009112 ("1-pqrs" numbers), A364171 (increasing m).

f:= proc(n) local t,d,b,c,m;
   d:= max(select(`<=`,numtheory:-divisors(n^2+1),n));
   b:= d+n;
   c:= (n^2+1)/d + n;
   b*c
end proc:
map(f, [$1..100]); # Robert Israel, Jul 19 2023

seq[max_] := Module[{len = Floor[Sqrt[max]/2], s, r}, s = Table[max + 1, {len}]; Do[r = (b*c - 1)/(b + c); If[IntegerQ[r] && r <= len && b*c < s[[r]], s[[r]] = b*c], {b, 2, max}, {c, 2, max/b}]; TakeWhile[s, # <= max &]]; seq[70000] (* Amiram Eldar, Jul 12 2023 *)

a(n) = for (x=1, oo, my(d=divisors(x)); for (i=1, #d\2, b = d[i]; c = x/d[i]; if ((b+c)*n == (x-1), return(x)););); \\ Michel Marcus, Jul 12 2023

a(n) = {forstep(i = 1, oo, n, if(iscan(i, n), return(i)))}
iscan(c, n) = {D = (1 - c)^2 - 4*n^2*c; if(!issquare(D), return(0)); b = ((c - 1) + sqrtint((1-c)^2 - 4*n^2*c)) / (2*n); if(denominator(b) == 1, return(1))} \\ David A. Corneth, Jul 12 2023

a(n) = {res = oo; for(b = n+1, 2*n, c = (n*b + 1)/(b - n); if(denominator(c) == 1, res = min(res, b*c))); res} \\ David A. Corneth, Jul 19 2023

a(n) = my(d = divisors(n^2 + 1), t = d[#d \ 2], b = t+n, c = (n^2 + 1)/t + n); return(b*c) \\ David A. Corneth, Jul 20 2023, adapted from Robert Israel, Jul 19 2023

More terms from Michel Marcus, Jul 12 2023

A057228 a(n) = u * v = x * y with (u - v) = (x + y) = A009000(n) (u>v, y>0, v>0, x>0, y>0).

Original entry on oeis.org

6, 24, 30, 54, 60, 96, 84, 150, 120, 210, 216, 240, 294, 210, 270, 384, 180, 486, 336, 600, 540, 480, 630, 726, 840, 864, 330, 504, 750, 924, 1014, 960, 1176, 1320, 840, 756, 1350, 1080, 1536, 720, 546, 1386, 1500, 1734, 1890, 1560, 1944, 1470, 2166

Offset: 1

Naohiro Nomoto, Sep 19 2000

nonn

Areas of Pythagorean triangles.

			a(1) = 6 = 6 * 1 = 3 * 2, (6 - 1)=(3 + 2) = 5 = A009000(1).

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A009000, A009003, A009111, A009112.

list(lim) = {my(lh = List()); for(u = 2, sqrtint(lim), for(v = 1, u, if (u^2+v^2 > lim, break); if ((gcd(u, v) == 1) && (0 != (u-v)%2), for (i = 1, lim, if (i*(u^2+v^2) > lim, break); /* if (u^2 - v^2 < 2*u*v, w = [i*(u^2 - v^2), i*2*u*v, i*(u^2+v^2)], w = [i*2*u*v, i*(u^2 - v^2), i*(u^2+v^2)]); */ listput(lh, [i*(u^2+v^2), i^2*(u^2 - v^2)*u*v]); ); ); ); ); lh = vecsort(Vec(lh)); vector(#lh, i, lh[i][2])} \\ David A. Corneth, Apr 10 2021, adapted from A009000

Offset changed to 1 by David A. Corneth, Apr 10 2021

A057101 Area of a Pythagorean triangle (ordered by the product of the sides).

Original entry on oeis.org

6, 24, 30, 54, 60, 96, 84, 120, 150, 210, 216, 180, 210, 240, 294, 270, 384, 336, 330, 486, 480, 540, 600, 504, 630, 726, 546, 840, 750, 864, 756, 720, 924, 840, 960, 1014, 1176, 1080, 840, 1320, 990, 1350, 1386, 1536, 1500, 1470, 1344, 1560, 1734, 1320

Offset: 1

Henry Bottomley, Aug 01 2000

nonn

			a(1)=3*4/2=6 since 3*4*5=60 is smallest possible positive product

Cf. A009111, A009112, A057096.

a(n) =A057096(n)/(2*A057100(n)) =A057098(n)*A057099(n)/2

A177887 The areas of Pythagorean triangles divided by 6.

Original entry on oeis.org

1, 4, 5, 9, 10, 14, 16, 20, 25, 30, 35, 36, 40, 45, 49, 55, 56, 64, 80, 81, 84, 90, 91, 100, 105, 120, 121, 125, 126, 140, 144, 154, 160, 165, 169, 180, 196, 204, 220, 224, 225, 231, 245, 250, 256, 260, 270, 285, 286, 289, 315, 320, 324, 336, 350, 360, 361, 364, 385, 390

Offset: 1

Robert G. Wilson v, Dec 14 2010

nonn

Number of terms less than 10^k: 4, 23, 109, 462, 1861, 7255, 27042, 95703, 314534, ..., .
This is A009112/6. - T. D. Noe, Oct 08 2013
Same as A256418/24, i.e., congrua/24. - Jonathan Sondow, Nov 07 2017

Albert H. Beiler, Recreations in the Theory of Numbers, The Queen of Mathematics Entertains, 2nd Ed., Chpt. XIV, "The Eternal Triangle", pp. 104-134, Dover Publ., NY, 1964.

Cf. A009112, A256418.

(* First run the program given for A009112 *) Union[t/6]

A009112(n)/6.

A231328 Integer areas of the reflection triangles of integer-sided triangles.

Original entry on oeis.org

18, 72, 90, 162, 180, 252, 288, 360, 450, 540, 630, 648, 720, 810, 882, 990, 1008, 1152, 1440, 1458, 1512, 1620, 1638, 1800, 1890, 2160, 2178, 2250, 2268, 2520, 2592, 2772, 2880, 2970, 3042, 3240, 3528, 3672, 3960, 4032, 4050, 4158, 4410, 4500, 4608, 4680, 4860

Offset: 1

Michel Lagneau, Nov 07 2013

nonn

The triangle A'B'C' obtained by reflecting the vertices of a reference triangle ABC about the opposite sides is called the reflection triangle (Grinberg 2003).
The area of the reflection triangle is given by
A' = A*t/(a^2*b^2*c^2) where A is the area of the reference triangle of sides (a, b, c) and
t=-(a^6-b^2*a^4-c^2*a^4-b^4*a^2-c^4*a^2-b^2*c^2*a^2+b^6+c^6-b^2*c^4-b^4*c^2)/(a^2*b^2*c^2).
See the link for the side lengths of the reflection triangles.
Properties of this sequence:
The areas corresponding to the primitive reflection triangles are 18, 90, 180, 252, 540,...
The non-primitive triangles of areas 4*a(n),9*a(n),...,p^2*a(n),... are in the sequence.
It appears that one of the side of the reflection triangles equals the greatest side of the initial triangle (see the table below), and the initial triangles are Pythagorean triangles => a(n) = 3*A009112(n).
The following table gives the first values (A, A', a, b, c, a', b', c') where A' is the area of the reflection triangles, A is the area of the initial triangles, a, b, c are the integer sides of the initial triangles, and a', b', c' are the sides of the reflection triangles.
-------------------------------------------------------------------------
|  A'  |   A |  a |  b |   c |       a'         |       b'         |  c'|
-------------------------------------------------------------------------
|  18  |   6 |  3 |  4 |   5 |  9*sqrt(17)/5    |  4*sqrt(97)/5    |  5 |
|  72  |  24 |  6 |  8 |  10 | 18*sqrt(17)/5    |  8*sqrt(97)/5    | 10 |
|  90  |  30 |  5 | 12 |  13 |  5*sqrt(1321)/13 | 36*sqrt(41)/13   | 13 |
| 162  |  54 |  9 | 12 |  15 | 27*sqrt(17)/5    | 12*sqrt(97)/5    | 15 |
| 180  |  60 |  8 | 15 |  17 |  8*sqrt(2089)/17 | 45*sqrt(89)/17   | 17 |
| 252  |  84 |  7 | 24 |  25 |  7*sqrt(5233)/25 | 72*sqrt(113)/25  | 25 |
| 288  |  96 | 12 | 16 |  20 | 36*sqrt(17)/5    | 16*sqrt(97)/5    | 20 |
| 360  | 120 | 10 | 24 |  26 | 10*sqrt(1321)/13 | 72*sqrt(41)/13   | 26 |
| 450  | 150 | 15 | 20 |  25 |  9*sqrt(17)      |  4*sqrt(97)      | 25 |
| 540  | 180 |  9 | 40 |  41 | 27*sqrt(1609)/41 | 40*sqrt(2329)/41 | 41 |
| 630  | 210 | 12 | 35 |  37 | 36*sqrt(1241)/37 | 35*sqrt(2521)/37 | 37 |
| 648  | 216 | 18 | 24 |  30 | 54*sqrt(17)/5    | 24*sqrt(97)/5    | 30 |
.......................................................................

			18 is in the sequence. We use two ways:
First way: with the triangle (3, 4, 5) the formula A' = A*t/(a^2*b^2*c^2) gives directly the result: A'= 18 where the area A = 6 is obtained by Heron's formula A =sqrt(s*(s-a)*(s-b)*(s-c))= sqrt(6*(6-3)*(6-4)*(6-5)) = 6, where s is the semiperimeter.
Second way: by calculation of the sides a', b', c' and by using Heron's formula. We obtain from the formulas given in the link:
a' = 9*sqrt(17)/5;
b' = 4*sqrt(97/5);
c' = 5.
Now, we use Heron's formula with (a',b',c'). We find A'=sqrt(s1*(s1-a')*(s1-b')*(s1-c')) with:
s1 =(a'+b'+c')/2 = (9*sqrt(17)/5+ 4*sqrt(97/5)+ 5)/2. We find A'= 18.

D. Grinberg, On the Kosnita Point and the Reflection Triangle, Forum Geom. 3, 105-111, 2003.

Eric W. Weisstein, MathWorld: Reflection Triangle

Cf. A188158, A009111, A009112.

nn = 300 ; lst = {}; Do[s = (a + b + c)/2 ; If[IntegerQ[s],area2 = s (s-a)(s-b) (s-c); If[area2 > 0 && IntegerQ[Sqrt[area2] + (a^2 + b^2 + c^2)/8], AppendTo[lst, Sqrt[area2] + (a^2 + b^2 + c^2)/8]]],{a,nn},{b,a},{c,b}] ; Union[lst]

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A177063 Number of Pythagorean triangles with area n.

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A228874 a(n) = L(n) * L(n+1) * L(n+2) * L(n+3), the product of four consecutive Lucas numbers, A000032.

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A364169 Smallest integer m = b*c which satisfies (b + c)*n = m - 1.

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A057228 a(n) = u * v = x * y with (u - v) = (x + y) = A009000(n) (u>v, y>0, v>0, x>0, y>0).

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A057101 Area of a Pythagorean triangle (ordered by the product of the sides).

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A177887 The areas of Pythagorean triangles divided by 6.

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A231328 Integer areas of the reflection triangles of integer-sided triangles.

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A364169 Smallest integer m = bc which satisfies (b + c)n = m - 1.