A057100 -id:A057100 - cp's OEIS frontend

A057096 Saint-Exupéry numbers: ordered products of the three sides of Pythagorean triangles.

60, 480, 780, 1620, 2040, 3840, 4200, 6240, 7500, 12180, 12960, 14760, 15540, 16320, 20580, 21060, 30720, 33600, 40260, 43740, 49920, 55080, 60000, 65520, 66780, 79860, 92820, 97440, 97500, 103680, 113400, 118080, 120120, 124320, 130560, 131820, 164640

Offset: 1

Henry Bottomley, Aug 01 2000

nonn

It is an open question whether any two distinct Pythagorean Triples can have the same product of their sides.
From Amiram Eldar, Nov 22 2020: (Start)
Named after the French writer Antoine de Saint-Exupéry (1900-1944).
The problem of finding two distinct Pythagorean triples with the same product was proposed by Eckert (1984). It is equivalent of finding a nontrivial solution of the Diophantine equation x*y*(x^4-y^4) = z*w*(z^4-w^4) (Bremner and Guy, 1988). (End)

			a(1) = 3*4*5 = 60.

Richard K. Guy, "Triangles with Integer Sides, Medians and Area." D21 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 188-190, 1994.
Antoine de Saint-Exupéry, Problème du Pharaon, Liège : Editions Dynamo, 1957.

T. D. Noe, Table of n, a(n) for n = 1..10000
Andrew Bremner and Richard K. Guy, A Dozen Difficult Diophantine Dilemmas, The American Mathematical Monthly, Vol. 95, No. 1 (1988), pp. 31-36.
Ernest J. Eckert, Problem 994, Crux Mathematicorum, Vol. 10, No. 10 (1984), p. 318, entire issue.
Richard K. Guy, Comment to Problem 994, Crux Mathematicorum, Vol. 12, No. 5 (1986), p. 109, entire issue.
Henry Plane, Calcule-moi un parallélépipède..., AMPEP, PLOT No. 22 (2002), pp. 22-23.
Giovanni Resta, Saint-Exupery numbers.
Antoine de Saint Exupéry, Le Problème du Pharaon, Succession Saint Exupéry - d'Agay, 2018.
Eric Weisstein's World of Mathematics, Pythagorean Triple.

Cf. A009004, A009012, A009111, A020886, A057097, A057098, A057099, A057100.

k=5000000; lst={}; Do[Do[If[IntegerQ[a=Sqrt[c^2-b^2]], If[a>=b, Break[]]; x=a*b*c; If[x<=k, AppendTo[lst,x]]], {b,c-1,4,-1}], {c,5,400,1}]; Union@lst (* Vladimir Joseph Stephan Orlovsky, Sep 05 2009 *)

a(n) = 60*A057097(n) = A057098(n)*A057099(n)*A057100(n).

A089982 Triangular numbers that can be expressed as the sum of 2 positive triangular numbers.

Original entry on oeis.org

6, 21, 36, 55, 66, 91, 120, 136, 171, 210, 231, 276, 351, 378, 406, 496, 561, 666, 703, 741, 820, 861, 946, 990, 1035, 1081, 1176, 1225, 1326, 1378, 1431, 1485, 1540, 1596, 1653, 1711, 1770, 1891, 1953, 2016, 2080, 2211, 2278, 2346, 2556, 2701, 2775, 2850

Offset: 1

Jon Perry, Jan 13 2004

Intersection of triangular numbers with sumset of triangular numbers. Triangular number analog of what for squares is {A057100(n)^2} = {A009000(n)^2}. {A000217} INTERSECT {A000217 + A000217}. - Jonathan Vos Post, Mar 09 2007

A subsequence of A051533. - Wolfdieter Lang, Jan 11 2017

			Generally, A000217(A000217(n)) = A000217(A000217(n)-1) + A000217(n) and so is automatically included. These are 6=T(3), 21=T(6), 55=T(10), etc. Other solutions occur when a partial sum from x to y is triangular, e.g., 15 + 16 + 17 + 18 = 66 = T(11), so T(14) + T(11) = T(18). This particular example arises since 10+4k is triangular (at k=14, 10 + 4k = 66), and we therefore have a solution.
All other solutions occur when 3+2k, 6+3k, 10+4k, etc. -- in general, T(j) + j*k -- is triangular.

Michael S. Branicky, Table of n, a(n) for n = 1..11915

Cf. A000217, A012132, A051533, A057100.

trn[i_]:=Module[{trnos=Accumulate[Range[i]],t2s},t2s=Union[Total/@ Tuples[ trnos,2]];Intersection[trnos,t2s]] (* Harvey P. Dale, Nov 08 2011 *)
Select[Range[75], ! PrimeQ[#^2 + (# + 1)^2] &] /. Integer_ -> (Integer^2 + Integer)/2 (* Arkadiusz Wesolowski, Dec 03 2015 *)

t(i) = i*(i+1)/2;
{ v=vector(100,i,t(i)); y=vector(100); c=0; for (i=1,30, for (j=i,30, x=t(i)+t(j); f=0; for (k=1,100,if (x==v[k],f=1;break)); if (f==1,y[c++ ]=x))); select(x->(x>0), vecsort(y,,8)) } \\ slightly edited by Michel Marcus, Apr 15 2021

lista(nn) = {for (n=1, nn, my(t = n*(n+1)/2); for (k=1, n-1, if (ispolygonal(t - k*(k+1)/2, 3), print1(t, ", "); break;)););} \\ Michel Marcus, Apr 15 2021

from itertools import count, takewhile
def aupto(lim):
    t = list(takewhile(lambda x: x<=lim, (i*(i+1)//2 for i in count(1))))
    s = set(a+b for i, a in enumerate(t) for b in t[i:])
    return sorted(s & set(t))
print(aupto(3000)) # Michael S. Branicky, Jun 21 2021

Triangular number m is in this sequence iff A000161(4*m+1)>1 or, alternatively, A083025(4*m+1)>1. - Max Alekseyev, Oct 24 2008

a(n) = A000217(A012132(n)). - Ivan N. Ianakiev, Jan 17 2013

More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net) and David Wasserman, Sep 23 2005

A057098 Shortest side of a Pythagorean triangle (ordered by the product of the sides).

Original entry on oeis.org

3, 6, 5, 9, 8, 12, 7, 10, 15, 20, 18, 9, 12, 16, 21, 15, 24, 14, 11, 27, 20, 24, 30, 16, 28, 33, 13, 40, 25, 36, 21, 18, 33, 24, 32, 39, 42, 30, 15, 48, 20, 45, 36, 48, 40, 35, 28, 39, 51, 22, 60, 54, 17, 27, 40, 57, 36, 48, 65, 60, 24, 32, 35, 56, 63, 45, 60, 19, 66, 44, 56

Offset: 1

Henry Bottomley, Aug 01 2000

nonn

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

Cf. A009004, A009005, A046083, A057096.

maxShortLeg = 66; terms = 71;
r[a_] := {a, b, c} /. {ToRules[Reduce[a <= b < c && a^2+b^2 == c^2, {b, c}, Integers]]};
abc = r /@ Complement[Range[maxShortLeg], {1, 2, 4}] // Flatten[#, 1]&;
SortBy[abc, Times @@ # &][[;; terms, 1]] (* Jean-François Alcover, Nov 21 2019 *)

a(n) =A057096(n)/(A057099(n)*A057100(n)) =sqrt(A057100(n)^2-A057099(n)^2)

A057099 Middle side of a Pythagorean triangle (ordered by the product of the sides).

Original entry on oeis.org

4, 8, 12, 12, 15, 16, 24, 24, 20, 21, 24, 40, 35, 30, 28, 36, 32, 48, 60, 36, 48, 45, 40, 63, 45, 44, 84, 42, 60, 48, 72, 80, 56, 70, 60, 52, 56, 72, 112, 55, 99, 60, 77, 64, 75, 84, 96, 80, 68, 120, 63, 72, 144, 120, 96, 76, 105, 90, 72, 80, 143, 126, 120, 90, 84, 108, 91

Offset: 1

Henry Bottomley, Aug 01 2000

nonn

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

Cf. A009012, A009023, A046084, A057096.

maxShortLeg = 66; terms = 67;
r[a_] := {a, b, c} /. {ToRules[Reduce[a <= b < c && a^2+b^2 == c^2, {b, c}, Integers]]};
abc = r /@ Complement[Range[maxShortLeg], {1, 2, 4}] // Flatten[#, 1]&;
SortBy[abc, Times @@ # &][[;; terms, 2]] (* Jean-François Alcover, Nov 21 2019 *)

a(n) =A057096(n)/(A057098(n)*A057100(n)) =sqrt(A057100(n)^2-A057098(n)^2)

A057097 Products of the three sides of Pythagorean triangles divided by 60.

Original entry on oeis.org

1, 8, 13, 27, 34, 64, 70, 104, 125, 203, 216, 246, 259, 272, 343, 351, 512, 560, 671, 729, 832, 918, 1000, 1092, 1113, 1331, 1547, 1624, 1625, 1728, 1890, 1968, 2002, 2072, 2176, 2197, 2744, 2808, 3164, 3212, 3333, 3375, 3927, 4096, 4250, 4459, 4480

Offset: 1

Henry Bottomley, Aug 01 2000

nonn

Note that if m appears in the sequence then k^3*m will also appear for all k and so in particular all cubes appear; the reverse is not always true (for example, 32*255*257/60 = 34952 = 2^3*4369 eventually appears, but 4369 does not).

By considering the Pythagorean triangle (3k, 4k, 5k) we see that all numbers k^3 are in the sequence. - Sergey Pavlov, Mar 29 2017

			a(1) = 3*4*5/60 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..386

Cf. A000578 (cubes).

(k=600000; lst={}; Do[Do[If[IntegerQ[a=Sqrt[c^2 - b^2]], If[a>=b, Break[]]; x=a b c; If[x<=k, AppendTo[lst, x]]], {b, c - 1, 4, -1}], {c, 5, 400, 1}]; Union@lst)/60 (* Vincenzo Librandi Mar 30 2017 *)

a(n) = A057096(n)/60 = A057098(n)*A057099(n)*A057100(n)/60.

A088511 Hypotenuses of primitive Pythagorean triangles sorted on product of sides.

Original entry on oeis.org

5, 13, 17, 25, 29, 41, 37, 61, 65, 53, 85, 65, 113, 73, 101, 85, 89, 145, 97, 145, 109, 181, 125, 221, 149, 197, 137, 173, 265, 157, 185, 257, 169, 313, 185, 229, 193, 205, 365, 325, 205, 269, 233, 421, 221, 293, 241, 265, 401, 481, 317, 277, 545, 365, 485, 281

Offset: 1

Lekraj Beedassy, Nov 14 2003

nonn

Cf. A063011.

Cf. A057100.

More terms from Ray Chandler, Nov 16 2003

A379985 Numbers k such that k^2 is of the form b^2 + (4c)^2 where bc is squarefree.

Original entry on oeis.org

5, 13, 17, 25, 29, 37, 61, 65, 85, 109, 137, 145, 149, 157, 169, 173, 193, 197, 205, 221, 229, 241, 265, 269, 293, 305, 325, 365, 377, 401, 409, 421, 433, 445, 485, 505, 533, 541, 557, 565, 569, 629, 673, 685, 689, 701, 709, 725, 761, 773, 797

Offset: 1

Lei Zhou, Jan 07 2025

nonn

It is known that the sum of squares of two odd numbers cannot be a square number, and when the sum of square of two numbers is the square of an odd number, the even one among the two numbers has to be multiple of 4. Thus the Mathematica program will not miss any entries.
a(n) == 1 (mod 4).
Numbers 4x^2 + y^2 where x, y are coprime numbers such that y is odd and x, y, 2x+y, 2x-y are squarefree. - Yifan Xie, Jan 09 2025, corrected by Robert Israel, Feb 03 2025

			5 is a term since 5^2 = 3^2 + (4*1)^2 and 3*1 is squarefree.
149 is a term since 149^2 = 51^2 + (4*35)^2 and 51*35 = 3*5*7*17 is squarefree.

Robert Israel, Table of n, a(n) for n = 1..10000

Subsequence of A009003.

Cf. A005117, A084645, A009000, A057100.

N:= 1000: # for terms <= N
Res:= {}:
for x from 1 while 4*x^2 < N do
  if not numtheory:-issqrfree(x) then next fi;
  for y from 1 by 2  while 4*x^2 + y^2 <= N do
    if igcd(x,y) = 1 and andmap(numtheory:-issqrfree,[y,2*x+y,2*x-y]) then Res:= Res union  {4*x^2 + y^2} fi
od od:
sort(convert(Res,list)); # Robert Israel, Feb 03 2025

a = {}; Do[m = n^2; b = n; While[b = b - 2; b > 1, k = m - b^2; If[c = Sqrt[k]/4; IntegerQ[c] && SquareFreeQ[b*c], AppendTo[a, n]]], {n, 5, 800, 2}]; a

Edited by Robert Israel, Feb 03 2025

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

cp's OEIS Frontend

A057096 Saint-Exupéry numbers: ordered products of the three sides of Pythagorean triangles.

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A089982 Triangular numbers that can be expressed as the sum of 2 positive triangular numbers.

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A057098 Shortest side of a Pythagorean triangle (ordered by the product of the sides).

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A057099 Middle side of a Pythagorean triangle (ordered by the product of the sides).

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A057097 Products of the three sides of Pythagorean triangles divided by 60.

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A088511 Hypotenuses of primitive Pythagorean triangles sorted on product of sides.

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A379985 Numbers k such that k^2 is of the form b^2 + (4*c)^2 where b*c is squarefree.

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

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A379985 Numbers k such that k^2 is of the form b^2 + (4c)^2 where bc is squarefree.