A009000 -id:A009000 - cp's OEIS frontend

A088896 Length of longest integral ladder that can be moved horizontally around the right angled corner where two hallway corridors of integral widths meet.

125, 1000, 2197, 3375, 4913, 8000, 15625, 17576, 24389, 27000, 39304, 42875, 50653, 59319, 64000, 68921, 91125, 125000, 132651, 140608, 148877, 166375, 195112, 216000, 226981, 274625, 314432, 343000, 389017, 405224, 421875, 474552, 512000

Offset: 1

Lekraj Beedassy, Nov 28 2003

nonn

The set of values for the integral-widths corridors and longest ladder are merely the cubes of Pythagorean triples, viz. (A046083, A046084, A009000).
The corridors' widths may be parametrically expressed as d*(sin x)^3 and d*(cos x)^3, for a longest ladder length d making an angle x with one of the corridors.
A given ladder, however, is maximum-corner-bending for a family of infinite pairs of perpendicular corridor widths and that the envelope of the maximum bending positions is that of a sliding rod against the outer wall, which is a branch of an astroid or four-cusped hypocycloid.

E. Mendelson, 3000 Solved Problems in Calculus, Chapter 16 Problem 16.56 pp. 131, Mc Graw-Hill 1988.
M. Spiegel, Theory and Problems of Advanced Calculus, Chapter 4 Problem 40 pp. 75, Mc Graw-Hill 1974.

C. Azeredo, The Ladder Problem
L. Husch and M. Szapiel, The Longest Ladder
M. Kantor, Knox College, Puzzle of the Week
J. J. O'Connor and E. R. Robertson, Astroid
T. Sillke, longest ladder
D. Sjerve, Solution to problem No.3
W. H. Steeb, Solved Problem
Eric Weisstein's World of Mathematics, Astroid

a(n)=d^3, where d=A009003(n).

A096472 Numbers containing squares of Pythagorean triples in their divisor set.

Original entry on oeis.org

3600, 7200, 10800, 14400, 18000, 21600, 25200, 28800, 32400, 36000, 39600, 43200, 46800, 50400, 54000, 57600, 61200, 64800, 68400, 72000, 75600, 79200, 82800, 86400, 90000, 93600, 97200, 100800, 104400, 108000, 111600, 115200, 118800, 122400, 126000, 129600, 133200

Offset: 1

Reinhard Zumkeller, Aug 13 2004

a(n) = m * (A046083(k)*A046084(k)*A009000(k))^2 for appropriate, not necessarily unique m and k.

			5^2 + 12^2 = 13^2: 5^2, 12^2 and 13^2 are divisors of 608400 = (13*5*3*2^2)^2, therefore 608400 is a term.

Paolo Xausa, Table of n, a(n) for n = 1..10000
Tanya Khovanova, Recursive Sequences.
Eric Weisstein's World of Mathematics, Pythagorean Triple.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. Pythagorean triples: A046083, A046084, A009000.

Cf. A044102, A094519, A169823.

Range[50]*3600 (* Paolo Xausa, Jul 01 2025 *)

my(x='x+O('x^38)); Vec(3600*x/(1-x)^2) \\ Elmo R. Oliveira, Jun 30 2025

a(n) = n*60^2.
From Elmo R. Oliveira, Jun 30 2025: (Start)
G.f.: 3600*x/(1-x)^2.
E.g.f.: 3600*x*exp(x).
a(n) = 60*A169823(n) = 100*A044102(n).
a(n) = 2*a(n-1) - a(n-2) for n > 2. (End)

Name clarified by Tanya Khovanova, Jul 05 2021

More terms from Elmo R. Oliveira, Jun 30 2025

A130611 Tribonacci numbers A000073 which can be the hypotenuse of a Pythagorean triple.

Original entry on oeis.org

13, 149, 274, 1705, 19513, 35890, 66012, 121415, 755476, 1389537, 4700770

Offset: 1

Jonathan Vos Post, Jun 17 2007

nonn

The first 2 values are hypotenuses of primitive Pythagorean triples, A000073 INTERSECTION A020882: (5^2 + 12^2 = 13^2), (51^2 + 140^2 = 149^2). The other values listed have one or more nonprimitive solution: a(6) = 35890 has 13 solutions; a(8), a(9), a(10), a(11) have 4 solutions each.

			a(4) = 1705 because 1023^2 + 1364^2 = 1705^2, which is a nonprimitive Pythagorean triple 341*(3,4,5).

Cf. A000073, A020882, A009000.

A000073 INTERSECTION A009000. {c in A000073 such that there exist integers a, b with a^2 + b^2 = c^2}.

A309501 a(n) is the minimal number of steps to return to the origin for a self-avoiding walk with step-length n confined to one quadrant of a 2D plane where at each step the walk must go to an unvisited point with integer coordinates as close as possible to the origin.

Original entry on oeis.org

4, 4, 4, 4, 20, 4, 4, 4, 4, 20, 4, 4, 6, 4, 20, 4, 120, 4, 4, 20, 4, 4, 4, 4, 6, 6, 4, 4, 32, 20, 4, 4, 4, 120, 20, 4, 6, 4, 6, 20, 6, 4, 4, 4, 20, 4, 4, 4, 4, 6, 120, 6, 2452, 4, 20, 4, 4, 32, 4, 20, 6, 4, 4, 4, 62, 4, 4, 120, 4, 20, 4, 4, 222, 6, 6

Offset: 1

Scott R. Shannon, Aug 05 2019

Consider a walk on a 2D plane starting at the (0,0) origin which is confined to the positive x and y quadrant i.e. x>=0, y>=0. We introduce the following rules for the walk: 1. Each step is of length n. 2. The walk can only step to grid points with integer x and y coordinates. 3. The walk cannot go to a grid point already visited, nor can it backtrack on its very first step from the origin. 4. At each step the walk must always go to a point which is as close as possible to the origin. Given these rules, what is the minimum number of steps required for the walk to return to the origin? For step length n the sequence a(n) is the minimum number of required steps.
For step lengths n which are not hypotenuses of any Pythagorean triple the solution is 4 as the walk will simply trace out a square e.g. step up, then right, then down, then left back to the origin.  This is true for n=1,2,3,4. But for step length 5 if the initially step is to (0,5) then by the rules the closest point to the origin for the next step now becomes (3,1) - the step must take the hypotenuse of a (3,4,5) triangle as clearly (3,1) is closer to the origin than (5,5). From here the walk must continue to pick the next point as close as possible to the origin which is 5 units away from its current coordinate while remaining inside the positive x-y quadrant. This can be achieved by stepping directly left-right-up-down or by moving along the hypotenuse of a (3,4,5) triangle. Also note for n=5, and any step length which is the hypotenuse of a Pythagorean triple, the walk can also take a first step to (4,3) - this can result in a totally different path that may, or may not, lead back to the origin in fewer steps.
If the walk visits a point along the line y=x, then it is likely that for the next step two points will be available which are equidistant from the origin. As we do not know which will lead to the minimum length walk we are searching for, the walker must explore both paths. Although the values of a(n) for the primitive Pythagorean triples are not particularly large, the total number of paths that must be recursively searched to find these minimum values increases rapidly. For example to find a(193) required searching at least 230,000 different paths, these being cut off by the walk either returning to the origin in fewer steps than the current minimum path, or by their step count surpassing the current minimum path.
For many n which are Pythagorean triple hypotenuses it is found that there are multiple paths corresponding to the minimum path value e.g. for n=29 there are 2 different paths which return to the origin after 32 steps. For n=73 there are 64 different paths.
Integer multiples of the hypotenuses of the primitive Pythagorean triples result in the same values for a(n) - simply scaling up the step length does not change the minimum path. However this is not necessarily true if the resulting hypotenuse has more than one triple e.g. for n=25 the step (7,24) is available as well as the primitive multiple (15,20). The former allows the walk to return to the origin in just 6 steps as opposed to 20 steps for a walk with step length of 5.
It seems plausible that the path could be trapped by surrounding visited points during a very long walk and thus never be able to return to the origin. It is unknown if this can occur.
For step length n equal to the hypotenuse of a Pythagorean triple (a,b,n) where 2b(a,b)->(0,2b)->(n,2b)->(n-a,b)->(n,0)->(0,0). However for n values with multiple Pythagorean triples the above path may be broken if another point closer to the origin is available via a step from an alternative triple.

			a(1) = 4. Path: (0,0)->(0,1)->(1,1)->(1,0)->(0,0).
a(5) = 20. Path: (0,0)->(0,5)->(3,1)->(3,6)->(0,2)->(5,2)->(1,5)->(1,0)->(4,4)->(0,1)->(5,1)->(1,4)->(4,0)->(0,3)->(5,3)->(1,6)->(1,1)->(6,1)->(2,4)->(5,0)->(0,0).
a(13) = 6. Path: (0,0)->(12,5)->(0,10)->(13,10)->(1,5)->(13,0)->(0,0).

Scott R. Shannon, Table of n, a(n) for n = 1..336
Wikipedia, Pythagorean Triples.

Cf. A020882, A009000, A307448.

A342583 Numbers k such that prime(k) is the hypotenuse of a Pythagorean triple where one leg is also prime.

Original entry on oeis.org

3, 6, 18, 42, 82, 271, 284, 369, 445, 682, 1069, 1193, 1900, 2241, 3894, 6137, 7108, 8164, 9658, 10126, 12645, 14842, 14936, 17913, 18420, 19480, 23893, 24605, 28959, 32913, 36279, 40847, 43936, 44559, 45500

Offset: 1

Ivan N. Ianakiev, Mar 16 2021

nonn

In such a triangle, the leg that is not prime is always the largest one and is equal to prime(k)-1; these even legs are in A067755. E.g. for a(2) = 6, prime(6) = 13 and the corresponding Pythagorean triple is (5, 12, 13). - Bernard Schott, Apr 03 2021

			a(1) = 3, since prime(3) = 5 is the hypotenuse of the triple (3,4,5).

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A067756 (the hypotenuses).

Cf. A009000, A046083, A046084, A067755.

R:= NULL: count:= 0:
p:= 2:
while count < 100 do
  p:= nextprime(p); n:= (p-1)/2; q:= 2*n^2+2*n+1;
  if isprime(q) then
    count:= count+1; r:= numtheory:-pi(q); R:= R, r;
  fi
od:
R; # Robert Israel, Mar 22 2021

PrimePi[Take[Cases[Import["https://oeis.org/A067756/b067756.txt","Table"],{,}][[All,2]],100]]

A342858 a(n) is the least integer h such that there exists a Pythagorean triple (x, y, h) that satisfies f(x)+f(y)+f(h)=n where f(m)=A176774(m) is the smallest polygonality of m; a(n) = 0 if no such h exists.

Original entry on oeis.org

13530, 136, 35, 5, 4510, 10, 100, 45, 51, 1404

Offset: 9

Michel Marcus, Mar 26 2021

a(19) > 10^9 if it exists.

It appears that the triples whose sum is 10 (as in the 2nd example below) have legs n^6 = A001014(n), (n^8 - n^4)/2 = A218131(n+1)/2 and (n^8 + n^4)/2 = A071231(n) for n >= 2; they consist of 2 triangular numbers and 1 square number. - Michel Marcus, Apr 12 2021

			a(9)  = 13530 with A176774([8778, 10296, 13530]) = [3,3,3].
a(10) = 136   with A176774([64, 120, 136])       = [4,3,3].
a(11) = 35    with A176774([21, 28, 35])         = [3,3,5].
a(12) = 5     with A176774([3, 4, 5])            = [3,4,5].
a(13) = 4510  with A176774([2926, 3432, 4510])   = [3,5,5].
a(14) = 10    with A176774([6, 8, 10])           = [3,8,3].
a(15) = 100   with A176774([28, 96, 100])        = [3,8,4].
a(16) = 45    with A176774([27, 36, 45])         = [10,3,3].
a(17) = 51    with A176774([45, 24, 51])         = [3,9,5].
a(18) = 1404  with A176774([540, 1296, 1404])    = [7,4,7].

Michel Marcus, Table of n, a(n) for n = 9..10000, with 0 for a(19)

Cf. A009000, A046083, A046084, A176774.
Cf. A213188 (see 2nd comment).
Cf. A245646, A245647, A245648, A342491.
Cf. A001014, A071231, A218131.

tp(n) = if (n<3, [n], my(v=List()); fordiv(2*n, k, if(k<2, next); if(k==n, break); my(s=(2*n/k-4+2*k)/(k-1)); if(denominator(s)==1, listput(v, s))); v = Vec(v); v[#v]); \\ A176774
vsum(v) = vecsum(apply(tp, v));
lista(limp, lim) = {my(vr = vector(limp)); for(u = 2, sqrtint(lim), for(v = 1, u, if (u*u+v*v > lim, break); if ((gcd(u,v) == 1) && (0 != (u-v)%2), for (i = 1, lim, if (i*(u*u+v*v) > lim, break); my(w = [i*(u*u - v*v), i*2*u*v, i*(u*u+v*v)]); my(h = i*(u*u+v*v)); my(sw = vsum(w)); if (sw <= limp, if (vr[sw] == 0, vr[sw] = h, if (h < vr[sw], vr[sw] = h))););););); vector(#vr - 8, k, vr[k+8]);}
lista(80, 15000) \\ Michel Marcus, Apr 16 2021

A349078 List of positive integer triples a, b, c with the property a^2 + b^2 = c^4 and a

Original entry on oeis.org

7, 24, 5, 15, 20, 5, 28, 96, 10, 60, 80, 10, 65, 156, 13, 119, 120, 13, 63, 216, 15, 135, 180, 15, 136, 255, 17, 161, 240, 17, 112, 384, 20, 240, 320, 20, 175, 600, 25, 220, 585, 25, 336, 527, 25, 375, 500, 25, 260, 624, 26, 476, 480, 26, 41, 840, 29, 580, 609, 29

Offset: 1

Szymon Lukaszyk, Dec 30 2021

For a given c, the triples are sorted in order of increasing a.

Triples are listed in order of nondecreasing c. - Chai Wah Wu, Jan 02 2022

Chai Wah Wu, Table of n, a(n) for n = 1..10062

Cf. A009000.

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

A009070 Ordered sides of Pythagorean triangles.

Original entry on oeis.org

3, 4, 5, 5, 6, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 12, 12, 13, 13, 14, 15, 15, 15, 15, 15, 16, 16, 16, 17, 17, 18, 18, 19, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 23, 24, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25, 26, 26, 27, 27, 27, 28, 28, 28, 28, 29, 29, 30, 30, 30, 30, 30, 31, 32, 32, 32

Offset: 1

David W. Wilson

nonn

Sorted union of A009000, A009004, and A009012, retaining duplicates. - Sean A. Irvine, Apr 19 2018

Ron Knott, Pythagorean Triples and Online Calculators

Cf. A009000, A009004, A009012, A009041.

A235990 Consider all Pythagorean triples (X,Y,Z) consisting of refactorable numbers (A033950) only, ordered by increasing Z. This sequence gives the values of Z.

Original entry on oeis.org

104, 204, 348, 480, 488, 492, 600, 732, 936, 1248, 1356, 1360, 1440, 1448, 1644, 1788, 2000, 2172, 2196, 2700, 2784, 2824, 3060, 3084, 3228, 3360, 3368, 3372, 3552, 3712, 3744, 3816, 3924, 4080, 4240, 4392, 4500, 4500, 4812, 5052, 5088, 5220, 5280, 5856, 6000

Offset: 1

Ivan N. Ianakiev, Jan 18 2014

nonn

			104^2 = 40^2 + 96^2, i.e. A033950(17)^2 = A033950(9)^2 + A033950(16)^2.

Cf. A033950, A009000, A046083, A046084.

More terms from Michel Marcus, Jan 21 2014

cp's OEIS Frontend

A088896 Length of longest integral ladder that can be moved horizontally around the right angled corner where two hallway corridors of integral widths meet.

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A096472 Numbers containing squares of Pythagorean triples in their divisor set.

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A130611 Tribonacci numbers A000073 which can be the hypotenuse of a Pythagorean triple.

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A309501 a(n) is the minimal number of steps to return to the origin for a self-avoiding walk with step-length n confined to one quadrant of a 2D plane where at each step the walk must go to an unvisited point with integer coordinates as close as possible to the origin.

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A342583 Numbers k such that prime(k) is the hypotenuse of a Pythagorean triple where one leg is also prime.

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Maple

Mathematica

A342858 a(n) is the least integer h such that there exists a Pythagorean triple (x, y, h) that satisfies f(x)+f(y)+f(h)=n where f(m)=A176774(m) is the smallest polygonality of m; a(n) = 0 if no such h exists.

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PARI

A349078 List of positive integer triples a, b, c with the property a^2 + b^2 = c^4 and a

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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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Maple

Mathematica

Extensions

A009070 Ordered sides of Pythagorean triangles.

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