cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A380875 Indices of triangular numbers (A000217) which are also perimeters of integer-sided right triangles (A010814).

Original entry on oeis.org

8, 15, 20, 23, 24, 27, 32, 35, 39, 44, 47, 48, 51, 55, 56, 59, 60, 63, 64, 68, 71, 72, 75, 76, 79, 80, 84, 87, 91, 92, 95, 96, 99, 104, 111, 112, 115, 116, 119, 120, 123, 124, 128, 132, 135, 139, 140, 143, 144, 147, 152, 155, 159, 160, 164, 167, 168, 171, 175, 176, 179, 180, 183, 184, 187, 188
Offset: 1

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Author

M. F. Hasler, Mar 20 2025

Keywords

Comments

This is relevant for considering integer-sided right triangles whose sides are made of sticks of length 1, 2, 3, ..., n, using all those.
Sequence A382268 considers the special case where these sticks must be used in order.

Links

Crossrefs

Cf. A000217 (triangular numbers), A010814 (perimeters of right triangles), A382268 (subsequence of k for which a right triangle can be made with linked rods of length 1, ..., k).

Programs

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

Views

Author

Daniel Mondot, Mar 31 2025

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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

Views

Author

Daniel Mondot, Apr 01 2025

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

Cf A380867, A380868, A382268, A382605.
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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