A031175 -id:A031175 - cp's OEIS frontend

A197040 Occurrences of edge-lengths of Euler bricks in every 100 consecutive integers.

3, 8, 9, 8, 9, 9, 6, 9, 10, 8, 7, 9, 6, 8, 7, 8, 11, 6, 7, 8, 9, 8, 7, 6, 8, 10, 6, 6, 6, 8, 8, 8, 8, 9, 6, 9, 7, 6, 7, 8, 8, 9, 7, 11, 7, 8, 5, 9, 8, 9, 9, 7, 6, 7, 9, 6, 7, 9, 7, 8, 10, 5, 9, 7, 7, 7, 7, 6, 9, 9, 6, 8, 7, 9, 8, 6, 9, 5, 9, 9, 8, 6, 6, 7, 7

Offset: 1

Christopher Monckton of Brenchley, Oct 08 2011

Distribution of edge-length occurrences for Euler bricks is remarkably near-uniform.

			For n=1 (i.e., the integers 1..100), there are only 3 possible edge-lengths for Euler bricks: 44, 85, 88.

L. E. Dickson, History of the Theory of Numbers, vol. 2, Diophantine Analysis, Dover, New York, 2005.
P. Halcke, Deliciae Mathematicae; oder, Mathematisches sinnen-confect., N. Sauer, Hamburg, Germany, 1719, page 265.

Robin Visser, Table of n, a(n) for n = 1..10000
E. W. Weisstein, MathWorld: Euler brick

cf. A195816, A196943, A031173, A031174, A031175. Edge lengths of Euler bricks are A195816; face diagonals are A196943.

def a(n):
    ans = set()
    for x in range(100*(n-1)+1, 100*n+1):
        divs = Integer(x^2).divisors()
        for d in divs:
            if (d <= x^2/d): continue
            if (d-x^2/d)%2==0:
                y = (d-x^2/d)/2
                for e in divs:
                    if (e <= x^2/e): continue
                    if (e-x^2/e)%2==0:
                        z = (e-x^2/e)/2
                        if (y^2+z^2).is_square(): ans.add(x)
    return len(ans)  # Robin Visser, Jan 02 2024

A239618 Number of primitive Euler bricks with side length a < b < c < 10^n, i.e., in a boxed parameter space with dimension 10^n.

Original entry on oeis.org

0, 0, 5, 19, 65, 242, 704, 1884, 4631

Offset: 1

Martin Renner, Mar 22 2014

An Euler brick is a cuboid of integer side dimensions a, b, c such that the face diagonals are integers. It is called primitive if gcd(a,b,c)=1.

Because the sides of a cuboid are permutable without changing its shape, the total number of primitive Euler bricks in the parameter space a, b, c < 10^n is b(n) = 6*a(n) = 0, 0, 30, 114, 390, ...

			a(3) = 5, since there are the five primitive Euler bricks [44, 117, 240], [85, 132, 720], [140, 480, 693], [160, 231, 792], [240, 252, 275] with longest side length < 1000.

Eric Weisstein's World of Mathematics, Euler Brick
Index entries for sequences related to bricks

Cf. A031173, A031174, A031175, A239620.

def a(n):
    ans = 0
    for x in range(1,10^n):
        divs = Integer(x^2).divisors()
        for d in divs:
            if (d <= x^2/d): continue
            if (d-x^2/d >= 2*x): break
            if (d-x^2/d)%2==0:
                y = (d-x^2/d)/2
                for e in divs:
                    if (e <= x^2/e): continue
                    if (e-x^2/e >= 2*y): break
                    if (e-x^2/e)%2==0:
                        z = (e-x^2/e)/2
                        if (gcd([x,y,z])==1) and (y^2+z^2).is_square():
                            ans += 1
    return ans  # Robin Visser, Jan 01 2024

a(6)-a(8) from Giovanni Resta, Mar 22 2014

a(9) from Robin Visser, Jan 01 2024

A239620 Number of Euler bricks with side length a < b < c < 10^n, i.e., in a boxed parameter space with dimension 10^n.

Original entry on oeis.org

0, 0, 10, 151, 1714, 17873, 180953, 1815841, 18174211

Offset: 1

Martin Renner, Mar 22 2014

An Euler brick is a cuboid of integer side dimensions a, b, c such that the face diagonals are integers.

Because the sides of a cuboid are permutable without changing its shape, the total number of Euler bricks in the parameter space is b(n) = 6*a(n) = 0, 0, 60, 906, 10284, ...

			a(3) = 10, since there are the ten Euler bricks [44, 117, 240], [85, 132, 720], [88, 234, 480], [132, 351, 720], [140, 480, 693], [160, 231, 792], [176, 468, 960], [240, 252, 275], [480, 504, 550], [720, 756, 825] with longest side length < 1000.

Eric Weisstein's World of Mathematics, Euler Brick
Index entries for sequences related to bricks

Cf. A031173, A031174, A031175, A239618.

def a(n):
    ans = 0
    for x in range(1,10^n):
        divs = Integer(x^2).divisors()
        for d in divs:
            if (d <= x^2/d): continue
            if (d-x^2/d >= 2*x): break
            if (d-x^2/d)%2==0:
                y = (d-x^2/d)/2
                for e in divs:
                    if (e <= x^2/e): continue
                    if (e-x^2/e >= 2*y): break
                    if (e-x^2/e)%2==0:
                        z = (e-x^2/e)/2
                        if (y^2+z^2).is_square():
                            ans += 1
    return ans  # Robin Visser, Jan 01 2024

a(6)-a(8) from Giovanni Resta, Mar 22 2014

a(9) from Robin Visser, Jan 01 2024

A306120 Lengths of largest face diagonal in primitive Euler bricks or Pythagorean cuboids: possible values of max(d, e, f) for solutions to a^2 + b^2 = d^2, a^2 + c^2 = e^2, b^2 + c^2 = f^2 in coprime positive integers a, b, c, d, e, f.

Original entry on oeis.org

267, 373, 732, 825, 843, 1595, 1884, 2500, 2775, 3725, 3883, 6380, 6409, 8140, 8579, 9188, 9272, 9512, 11764, 12125, 13123, 14547, 14681, 14701, 19572, 20503, 20652, 24695, 25121, 25724, 29307, 30032, 30695, 31080, 32595, 34484, 37104, 37895, 38201, 38965

Offset: 1

M. F. Hasler, Oct 11 2018

nonn

These are the values obtained as sqrt(A031173(n)^2 + A031174(n)^2), sorted by size (n = 3 yields 843, n = 4 yields 732) and duplicates removed: The first duplicate is 71402500^2 = A031173(1428)^2 + A031174(1428)^2 = A031173(1626)^2 + A031174(1626)^2, there is no other among the first 3500 terms.
This considers only the face diagonals, not the space diagonals.
See the main entry A031173 for links, cross-references, and further comments.

M. F. Hasler, Table of n, a(n) for n = 1..2500

Cf. A031173, A031174, A031175, A195816, A196943, A268396.

A306120=Set(vector(1000,n,sqrtint(A031173(n)^2+A031174(n)^2)))[1..-100] \\ Discard the last 100 values, which may have holes. This is empirical: better find the smallest sqrtint(A031173(n)^2+A031174(n)^2) with n > 1000 not in the set, and discard all elements larger than that.

A306120 = { sqrtint(A031173(n)^2+A031174(n)^2); n >= 1 }.

A340178 Euler brick triples, side dimensions (a,b,c) in increasing order for a.

Original entry on oeis.org

44, 117, 240, 85, 132, 720, 88, 234, 480, 132, 351, 720, 140, 480, 693, 160, 231, 792, 170, 264, 1440, 176, 468, 960, 187, 1020, 1584, 195, 748, 6336, 220, 585, 1200, 240, 252, 275, 255, 396, 2160, 264, 702, 1440, 280, 960, 1386, 308, 819, 1680, 320, 462, 1584

Offset: 1

Ctibor O. Zizka, Dec 30 2020

How are primitive Euler bricks distributed in the (a,b,c) parameter space?

Robin Visser, Table of n, a(n) for n = 1..10000
Oliver Knill, Treasure Hunting Perfect Euler bricks, Math table, February 24, 2009.
Randall L. Rathbun, The Integer Cuboid Table, arXiv:1705.05929 [math.NT], 2017-2020.

Cf. A031173, A031174, A031175, A103605.

A141029 Nearest integer to the space diagonal of the smallest (measured by the longest edge) primitive (gcd(a,b,c)=1) Euler bricks (a, b, c, sqrt(a^2 + b^2), sqrt(b^2 + c^2), sqrt(a^2 + c^2) are integers). If the space diagonal is an integer then the Euler brick is called a "perfect cuboid". There are no known perfect cuboids.

Original entry on oeis.org

271, 444, 855, 737, 840, 1887, 1893, 2537, 2897, 3961, 3816, 6596, 8595, 6383, 9260, 8327, 9525, 9405, 13454, 16525, 12122, 12167, 15336, 14721, 22943, 20988, 22444, 25844, 28443, 26336, 30382, 29714, 35079, 31094, 31700, 38989, 32965

Offset: 1

Darrell Minor, Jul 29 2008

nonn

			a(1)=271 because sqrt(240^2 + 117^2 + 44^2) = 270.60, where 240 is the longest edge, 117 the intermediate edge and 44 the smallest edge, of the smallest primitive Euler brick.

Eric Weisstein's World of Mathematics, Euler Brick.

Cf. A031173, A031174, A031175.

A297327 1/36 of the square of the basis of a primitive 3-simplex.

Original entry on oeis.org

6434041, 89002225, 865125625, 89610625, 353440516, 29160156025, 18989880481, 37434450625, 72399370000, 444515646025, 346008660625, 2003915162500, 9475360381201, 166729268761, 13110591519025, 8007417968121, 11201866562500, 3095696620900, 61956758281561

Offset: 1

Ralf Steiner, Dec 28 2017

nonn

For every primitive trirectangular tetrahedron (0, a, b, c) with coprime integer sides, (b*c)^2 + (a*b)^2 + (c*a)^2 is divisible by 144.
The square of the basis is related by De Gua's theorem on the square of the main diagonal of a (different, not necessarily primitive) Euler brick (a*b/12=A031173(k), a*c/12=A031174(k), b*c/12=A031175(k)) also having integer sides and integer face diagonals including a trirectangular tetrahedron (0, a*b/12, a*c/12, b*c/12), such as a(1) = 6434041 = A023185(8) = A031173(8)^2 + A031174(8)^2 + A031175(8)^2.
By this process a cycle of primitive trirectangular tetrahedrons is defined, such as with indices k: (1 8), (2 6), (3 5), (4 7), (9 19), ...

Eric Weisstein's World of Mathematics, Trirectangular Tetrahedron
Wikipedia, De Gua's theorem

Cf. A295507, A023185, A031173, A031174, A031175.

a(n) = (1/144)*(A031174(n)^2*A031175(n)^2 + A031173(n)^2*(A031174(n)^2 + A031175(n)^2)).

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A197040 Occurrences of edge-lengths of Euler bricks in every 100 consecutive integers.

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A239618 Number of primitive Euler bricks with side length a < b < c < 10^n, i.e., in a boxed parameter space with dimension 10^n.

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A239620 Number of Euler bricks with side length a < b < c < 10^n, i.e., in a boxed parameter space with dimension 10^n.

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A306120 Lengths of largest face diagonal in primitive Euler bricks or Pythagorean cuboids: possible values of max(d, e, f) for solutions to a^2 + b^2 = d^2, a^2 + c^2 = e^2, b^2 + c^2 = f^2 in coprime positive integers a, b, c, d, e, f.

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A340178 Euler brick triples, side dimensions (a,b,c) in increasing order for a.

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A297327 1/36 of the square of the basis of a primitive 3-simplex.

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