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.

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.

Original entry on oeis.org

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

Views

Author

Ali Sada and Daniel Mondot, Mar 19 2025

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

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

Programs

  • PARI
    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

A385736 a(n) is the number of distinct nondegenerate triangles with perimeter n whose side lengths are triangular numbers.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 2, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 2, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 2, 1, 2, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 0, 0, 1, 0, 1, 1, 0, 1
Offset: 0

Views

Author

Felix Huber, Jul 16 2025

Keywords

Comments

0, 1, 6, 10, 28, 55 are the only triangular numbers <= 10^6 that are not perimeters of triangles whose side lengths are triangular numbers. Conjecture: There are no other triangular numbers that have this property.

Examples

			The a(31) = 2 distinct nondegenerate triangles with perimeter 31 and whose side lengths are triangular numbers are [1, 15, 15] and [6, 10, 15].

Links

Crossrefs

Cf. A000217, A005044, A051516, A070088, A366398, A380875, A385737, A385860, A385872.

Programs

  • Maple
    A385736:=proc(N) # To get the first N + 1 terms.
    local p,x,y,z,i;
    p:=[];
    for z to floor((sqrt(24*N+9)-3)/6) do
        for x from z to floor((sqrt(4*N-3)-1)/2) do
            for y from max(z,floor((sqrt(1+4*(x^2+x-z^2-z))-1)/2)+1) to min(x,floor((sqrt(1+4*(2*N-x^2-x-z^2-z))-1)/2)) do
                p:=[op(p),z*(z+1)/2+y*(y+1)/2+x*(x+1)/2]
            od
        od
    od;
    return seq(numboccur(p,i),i=0..N)
end proc;
A385736(87);

Formula

Trivial upper bound: a(n) <= A005044(n).
a(A385737(n)) >= 1.

A385737 Perimeters of nondegenerate triangles with integer areas, whose side lengths are triangular numbers.

Original entry on oeis.org

176, 224, 264, 336, 504, 644, 756, 950, 1196, 1232, 1280, 1500, 1566, 1650, 1700, 2100, 2112, 2250, 2366, 2754, 3036, 3306, 5676, 5796, 7296, 8064, 8316, 8526, 9576, 10206, 10260, 12474, 13200, 15872, 16236, 16896, 17094, 17150, 20172, 21714, 21726, 22382, 22644
Offset: 1

Views

Author

Felix Huber, Jul 16 2025

Keywords

Comments

224 and 1280 are the only perimeters <= 10^6 of nondegenerate triangles whose side lengths (28, 91, 105 or 325, 325, 630, respectively) and areas (1176 or 25200, respectively) are triangular numbers.

Examples

			176 is a term because it is the perimeter of the triangle [55, 55, 66], where 55 and 66 are triangular numbers, which has an integer area of sqrt(88*(88 - 55)*(88 - 55)*(88 - 66)) = 1452.
224 is a term because it is the perimeter of the triangle [28, 91, 105], where 28, 91 and 105 are triangular numbers, which has an integer area of sqrt(112*(112 - 28)*(112 - 91)*(112 - 105)) = 1176 (which is also a triangular number).

Links

Crossrefs

Subsequence of A380875.
Cf. A000217, A051516, A070083, A385736, A385872.

Programs

  • Maple
    A385737:=proc(P) # To get all perimeters <= P.
    local p,x,y,z,u,v,w,s;
    p:=[];
    for z to floor((sqrt(24*P+9)-3)/6) do
        for x from z to floor((sqrt(4*P-3)-1)/2) do
            for y from max(z,floor((sqrt(1+4*(x^2+x-z^2-z))-1)/2)+1) to min(x,floor((sqrt(1+4*(2*P-x^2-x-z^2-z))-1)/2)) do
            	u:=z*(z+1)/2;
            	v:=y*(y+1)/2;
            	w:=x*(x+1)/2;
            	s:=(u+v+w)/2;
            	if issqr(s*(s-u)*(s-v)*(s-w)) then
               	    p:=[op(p),u+v+w]
               	fi
            od
        od
    od;
    return op(sort(p))
end proc;
A385737(22644);
Showing 1-3 of 3 results.

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