A380868 -id:A380868 - cp's OEIS frontend

A382268 Numbers k such that a right triangle can be formed from a chain of linked rods of lengths 1, 2, 3, ..., k, with the perimeter equal to the total length.

15, 20, 24, 35, 39, 44, 48, 55, 56, 63, 75, 76, 80, 84, 91, 95, 99, 104, 111, 119, 120, 132, 135, 140, 143, 144, 152, 155, 168, 175, 176, 187, 188, 195, 203, 207, 215, 216, 219, 224, 252, 259, 260, 264, 272, 275, 279, 287, 288, 296, 299, 308, 315, 320, 324, 335, 351, 360

Offset: 1

Ali Sada and Daniel Mondot, Mar 19 2025

nonn

The corresponding perimeters T(a(n)) must be in the intersection of T = A000217 (triangular numbers) with A010814 (perimeters of integer sided right triangles). A number k is in the sequence if there exists a solution {k1, k2, k3} with k > k1 > k2 > k3 >= 0 such that for a < b < c in { T(k1) - T(k2), T(k2) - T(k3), T(k) - T(k1) + T(k3) } one has a^2 + b^2 = c^2. - M. F. Hasler, Mar 20 2025

			The first triangle is when k = 15. The segments are [6+7+8+9+10] [11+12+13+14] [15+1+2+3+4+5]. The sums of the segments are (40, 50, 30), which is 10 times the primitive right triangle (3, 4, 5).
The second term, k = 20, corresponds to 5 distinct solutions:
  S1 = {18, 16, 9}: a = 9+...+1 + 20+19 = 84, b = 18+17 = 35, c = 16+...+10 = 91,
  S2 = {17, 11, 3}: a = 20+19+18 + 3+2+1 = 63, c = 17+...+12 = 87, b = 11+...+4 = 60,
  S3 = {17, 11, 2}: a = 20+19+18 + 2+1 = 60, c = 17+...+12 = 87, b = 11+...+3 = 63,
  S4 = {16, 9, 4}: a = 20+...+17 + 4+...+1 = 84, c = 16+...+10 = 91, b = 9+...+5 = 35,
  S5 = {15, 8, 1}: c = 20+...+16 + 1 = 91, a = 15+...+9 = 84, b = 8+...+2 = 35.
We note that S2 and S3, and S1, S4 and S5, have the same side lengths, but different decompositions.

Daniel Mondot, Table of n, a(n) for n = 1..3052

Cf. A000217, A010814, A380867, A380868, A380875.

select( {is_A382268(n)=my(Tn=n*(n+1)\2,T1,T2,S); Tn%2==0 && is_A005279(Tn\2) && forstep(n1=n-1,sqrtint(Tn-1)+1,-1, T1=n1*(n1+1)\2; forstep(n2=n1-1,sqrtint(2*T1-Tn-1)+1,-1, T2=n2*(n2+1)\2; forstep(n3=n2-1,0,-1, #(S=Set([Tn-T1+S=n3*(n3+1)\2,T2-S,T1-T2]))>2 && S[3]^2 == S[1]^2+S[2]^2 && return(S))))}, [1..100])\\ M. F. Hasler, Mar 22 2025

A382605 Number of distinct solutions to the problem of folding in half a chain of linked rods of length 1, ..., n.

Original entry on oeis.org

0, 0, 1, 1, 0, 0, 1, 2, 0, 0, 1, 1, 0, 0, 2, 1, 0, 0, 1, 3, 0, 0, 1, 2, 0, 0, 3, 1, 0, 0, 1, 2, 0, 0, 4, 1, 0, 0, 4, 1, 0, 0, 1, 4, 0, 0, 1, 2, 0, 0, 3, 1, 0, 0, 3, 3, 0, 0, 1, 1, 0, 0, 2, 1, 0, 0, 1, 3, 0, 0, 1, 1, 0, 0, 4, 1, 0, 0, 1, 4, 0, 0, 1, 5, 0, 0, 3, 1, 0, 0, 1, 3, 0, 0, 2, 1, 0, 0, 6, 1

Offset: 1

Daniel Mondot, Mar 31 2025

nonn

In order to be able to fold such chain in half, the total length of the chain (A000217(n)) has to be even, which is true when n=3 (mod 4) or n=0 (mod 4).

Conjecture: Whenever the length of the chain is even, there is at least one solution. That makes A154708 the sequence that lists numbers k that have at least one solution.

			A chain of 7 rods of length 1 to 7, can be folded in half in only one way: 2+3+4+5 on one side, 6+7+1, on the other, both sides being 14 in total length. Therefore a(7) = 1.

Daniel Mondot, Table of n, a(n) for n = 1..10000
Allan Gottlieb, Puzzle Corner, Technology Review, December 2, 2003.

Cf. A380868, A382268, A380867, A382632.

A382632 Numbers k such that one can make an equilateral triangle from a chain of linked rods of length 1, 2, 3, ..., k, with perimeter equal to the total length.

Original entry on oeis.org

9, 90, 125, 153, 189, 440, 819, 989, 1295, 1394, 1484, 1701, 2079, 2448, 2925, 3024, 4004, 5453, 6174, 7865, 8910, 13509, 13689, 13923, 16235, 19683, 20294, 21824, 24804, 26649, 32760, 33488, 37169, 37925, 39024, 40733, 42704, 44225, 44289, 47915, 48734, 52325, 97335, 101870

Offset: 1

Daniel Mondot, Apr 01 2025

nonn

For all known terms (n<157), there is only one solution, except for 125 and 158949 which both have 2 solutions.

Conjecture: Out of some linked rods of length 1, 2...k, we can fold them in half (digon), or make equilateral triangles, but no other regular polygons (squares, regular pentagons, etc...) can be made.

			For k=9, the sides of the triangle are 15 in length: [4+5+6], [7+8] and [9+1+2+3]. Therefore 9 is in the sequence.
For k=90, the sides of the triangle are 1365 in length: [16+...+54], [55+...+75] and [76+...+90 + 1+...+15]. Therefore 90 is in the sequence.
The 2 solutions for 125 are:
    [3+4+...+71+72], [73+74+...+101+102] and [103+104+...+124+125 +1+2]
and [58+...+92], [93+...+117] and [118+...+125 + 1+...+58], all sides 2625 in length.

Daniel Mondot, Table of n, a(n) for n = 1..156
Allan Gottlieb, Puzzle Corner, Technology Review, December 2, 2003.

Cf A380867, A380868, A382268, A382605.

A380869 Numbers k such that one can make a rectangle from a chain of linked rods of lengths 1, 2, 3, ..., k, with perimeter equal to the total length, and with one side consisting of a single rod.

Original entry on oeis.org

8, 15, 20, 24, 27, 35, 39, 80, 84, 104, 143, 215, 220, 252, 264, 351, 363, 459, 476

Offset: 1

Ali Sada and M. F. Hasler, Mar 14 2025

Subsequence of A380867, with the additional requirement in the set {n1, n2, n3, n4} corresponding to the solutions (cf. there), there are two consecutive integers.

			The smallest such number is a(1) = 8, for which we have (n1..n4) = (2, 4, 6, 7), that is, the rectangle:
    o--+--o--o--+--+--+--+--+--+--+--o
    |  2   1             8           |
    |3                               |
    |                                |
    o                                7
    |                                |
    |4                               |
    |                                |
    |       5               6        |
    o--+--+--+--+--o--+--+--+--+--+--o
We see that one of the sides, of length 7, is made of only one single rod.

Cf. A000217 (triangular numbers), A334720 (2D cycles on square lattice).

Cf. A380867 (contains this as a subsequence), A380868 (total number of solutions for given n).

T(n)=n*(n+1)/2 \\ = A000217
select( {is_A380869(n)=my(Tn=T(n), T1, T2, T3, T4, n3, n4); Tn%2==0 && forstep(n1=n-1, 3, -1, T1=T(n1); forstep(n2=n1-1, 2, -1, (B = Tn/2 - A = T1 - T2 = T(n2)) < 3 && break; iferr((1+n3=sqrtint(2*T3 = T2-B))*n3==2*T3 && (1+n4=sqrtint(2*T4 = T3-A))*n4==2*T4 && vecmin([n1-n2,n2-n3,n3-n4])==1 && return(n), E, )))}, [1..99])

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A382268 Numbers k such that a right triangle can be formed from a chain of linked rods of lengths 1, 2, 3, ..., k, with the perimeter equal to the total length.

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A382605 Number of distinct solutions to the problem of folding in half a chain of linked rods of length 1, ..., n.

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A382632 Numbers k such that one can make an equilateral triangle from a chain of linked rods of length 1, 2, 3, ..., k, with perimeter equal to the total length.

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A380869 Numbers k such that one can make a rectangle from a chain of linked rods of lengths 1, 2, 3, ..., k, with perimeter equal to the total length, and with one side consisting of a single rod.

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PARI