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 1-10 of 17 results. Next

A298942 Where records occur in A070138.

Original entry on oeis.org

12, 36, 54, 84, 98, 162, 242, 338, 484, 578, 722, 1058, 1682, 1922, 2738, 3362, 3698, 4418, 5618, 6962, 8978, 10082, 12482, 13778, 15842, 18818, 20402, 21218, 22898, 25538, 29282, 32258, 34322, 37538, 38642, 44402, 45602, 49298, 53138, 55778, 59858, 64082, 72962
Offset: 1

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Author

Peter Kagey, Jan 30 2018

Keywords

Crossrefs

Cf. A070138.

Extensions

a(27)-a(43) from Giovanni Resta, Jan 31 2018

A096468 Perimeters of primitive Heronian triangles.

Original entry on oeis.org

12, 16, 18, 30, 32, 36, 40, 42, 44, 48, 50, 54, 56, 60, 64, 66, 68, 70, 72, 76, 78, 80, 84, 90, 96, 98, 100, 104, 108, 110, 112, 114, 120, 126, 128, 130, 132, 136, 140, 144, 150, 152, 154, 156, 160, 162, 164, 168, 170, 172, 174, 176, 180, 182, 186, 190, 192, 196
Offset: 1

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Author

T. D. Noe, Jun 22 2004

Keywords

Comments

Here a primitive Heronian triangle has integer sides a,b,c with GCD(a,b,c) = 1 and integral area. The perimeter is always even. Cheney's article contains many theorems about these triangles.

Examples

			12 is on this list because the triangle with sides 3, 4, 5 has integral area and perimeter 12.
		

Crossrefs

Cf. A070138 (number of primitive Heronian triangles having perimeter n), A083875 (area/6 of primitive Heronian triangles), A096467 (longest side of primitive Heronian triangles).

Programs

  • Mathematica
    nn=150; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s] && GCD[a, b, c]==1, area2=s(s-a)(s-b)(s-c); If[area2>0 && IntegerQ[Sqrt[area2]], AppendTo[lst, 2s]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

Extensions

Name changed by Wesley Ivan Hurt, May 17 2020

A051493 Triangles with perimeter n and relatively prime integer side lengths.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 2, 1, 2, 1, 4, 2, 5, 2, 5, 4, 8, 4, 10, 6, 9, 6, 14, 8, 15, 9, 16, 12, 21, 11, 24, 16, 22, 16, 27, 18, 33, 20, 31, 24, 40, 23, 44, 30, 39, 30, 52, 32, 54, 35, 52, 42, 65, 38, 65, 48, 64, 49, 80, 48, 85, 56, 77, 64, 90, 58, 102, 72, 93, 69, 114, 72, 120, 81
Offset: 1

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Author

Keywords

Comments

From Peter Munn, Jul 26 2017: (Start)
The triangles that meet the conditions are listed by nondecreasing n in A070110.
Without the requirement for relatively prime side lengths, this sequence becomes A005044.
Counting the triangles by longest side instead of perimeter, this sequence becomes A123323.
a(n) = A070094(n) + A070102(n) + A070109(n).
(End)

Examples

			There are 3 triangles with integer-length sides and perimeter 9: 1-4-4, 2-3-4, 3-3-3. 3-3-3 is omitted because isomorphic to 1-1-1, so a(9)=2.
		

Crossrefs

Equivalent sequences, restricted to subsets: A070091 (isosceles), A070094 (acute), A070102 (obtuse), A070109 (right-angled), A070138 (with integer area), A070202 (with integer inradius).

Programs

  • Mathematica
    nmax = 100;
    A005044[n_] := Quotient[n^2 + 6n Mod[n, 2] + 24, 48];
    A = Array[A005044, nmax];
    mob[m_, n_] := If[ Mod[m, n] == 0, MoebiusMu[m/n], 0];
    Reap[Do[Sow[Sum[mob[n, d] A[[d]], {d, 1, n}]], {n, 1, nmax}]][[2, 1]] (* Jean-François Alcover, Oct 05 2021 *)

Formula

Moebius transform of A005044.

Extensions

Corrected and extended with formula by Christian G. Bower, Nov 15 1999
Formula updated due to change to referenced sequence, and definition clarified by Peter Munn, Jul 26 2017

A070109 Number of right integer triangles with perimeter n and relatively prime side lengths.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

Right integer triangles have integer areas: see A070142, A051516.
a(n) is nonzero iff n is in A024364.

Examples

			For n=30 there are A005044(30) = 19 integer triangles; only one is right: 5+12+13 = 30, 5^2+12^2 = 13^2; therefore a(30) = 1.
		

Crossrefs

Programs

  • Mathematica
    unitaryDivisors[n_] := Cases[Divisors[n], d_ /; GCD[d, n/d] == 1];
    A078926[n_] := Count[unitaryDivisors[n], d_ /; OddQ[d] && Sqrt[n] < d < Sqrt[2n]];
    a[n_] := If[EvenQ[n], A078926[n/2], 0];
    Table[a[n], {n, 1, 1716}] (* Jean-François Alcover, Oct 04 2021 *)

Formula

a(n) = A078926(n/2) if n is even; a(n)=0 if n is odd.
a(n) = A051493(n) - A070094(n) - A070102(n).
a(n) <= A024155(n).

Extensions

Secondary offset added by Antti Karttunen, Oct 07 2017

A298614 Number of primitive Heronian triangles with perimeter A096468(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 1, 1, 1, 2, 5, 3, 1, 6, 3, 2, 5, 1, 4, 1, 1, 1, 4, 1, 6, 1, 3, 4, 5, 1, 3, 3, 2, 8, 1, 1, 1, 1, 1, 5, 1, 3, 2, 2, 7, 8, 3, 4, 3, 2, 6, 3, 3, 1, 5, 4, 3, 1, 5, 1, 2, 6, 16, 1, 9, 4, 9, 1, 3, 2, 3, 1, 4
Offset: 1

Views

Author

Peter Kagey, Jan 26 2018

Keywords

Comments

A Heronian triangle is a triangle with integer sides and area.

Crossrefs

This is A070138 with the zeros omitted.
Cf. A096468.

Formula

a(n) = A070138(A096468(n)).

A305717 Number of distinct Heronian triangles with perimeter A051518(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 4, 1, 2, 1, 3, 2, 4, 1, 3, 5, 1, 1, 2, 5, 1, 1, 4, 8, 1, 5, 5, 6, 5, 2, 12, 1, 6, 1, 5, 4, 9, 1, 9, 2, 5, 11, 8, 2, 3, 5, 7, 12, 1, 10, 1, 1, 1, 7, 10, 3, 2, 2, 15, 14, 5, 10, 5, 5, 11, 16, 5, 1, 12, 7, 3, 1, 7, 1, 2, 15, 16, 1, 11, 16, 18
Offset: 1

Views

Author

Peter Kagey, Jun 08 2018

Keywords

Examples

			For n = 7, the a(7) = 4 Heronian triangles with perimeter A051518(7) = 36 are:
the 10-13-13 isosceles triangle (area 60),
the 10-10-16 isosceles triangle (area 48),
the  9-12-15 scalene   triangle (area 54), and
the  9-10-17 scalene   triangle (area 36).
		

Crossrefs

Programs

  • Mathematica
    htc[p_] := Block[{t=0, c, q=p/2}, Do[c=p-a-b; If[c >= b && a+c > b && a+b > c && IntegerQ[Sqrt[q (q-a) (q-b) (q-c)]], t++], {a, p/3}, {b, a, p-a-1}]; t]; Select[htc /@ (Range[128] 2), # > 0 &] (* Giovanni Resta, Jun 14 2018 *)

Formula

a(n) = A051516(A051518(n)).

A330912 Sum of the smallest side lengths of all Heronian triangles with perimeter A051518(n).

Original entry on oeis.org

3, 5, 5, 6, 5, 14, 38, 8, 20, 11, 37, 29, 43, 7, 31, 64, 11, 17, 37, 84, 19, 15, 70, 130, 22, 87, 101, 133, 122, 38, 241, 25, 149, 25, 111, 123, 225, 39, 220, 54, 120, 327, 254, 57, 103, 162, 227, 371, 41, 321, 34, 43, 29, 278, 373, 76, 70, 95, 577, 567, 157, 476, 221
Offset: 1

Views

Author

Wesley Ivan Hurt, May 02 2020

Keywords

Examples

			a(1) = 3; there is one Heronian triangle with perimeter A051518(1) = 12, which is [3,4,5] and its smallest side length is 3.
a(6) = 14; there are two Heronian triangles with perimeter A051518(6) = 32, [4,13,15] and [10,10,12]. The sum is 4 + 10 = 14.
		

Crossrefs

Formula

a(n) = Sum_{k=1..floor(c(n)/3)} Sum_{i=k..floor((c(n)-k)/2)} sign(floor((i+k)/(c(n)-i-k+1))) * chi(sqrt((c(n)/2)*(c(n)/2-i)*(c(n)/2-k)*(c(n)/2-(c(n)-i-k)))) * k, where chi(n) = 1 - ceiling(n) + floor(n) and c(n) = A051518(n). - Wesley Ivan Hurt, May 12 2020

A330915 Sum of the "middle" side lengths (b such that a <= b <= c) of all Heronian triangles with perimeter A051518(n).

Original entry on oeis.org

4, 5, 5, 8, 12, 23, 45, 15, 29, 13, 48, 30, 77, 24, 69, 117, 25, 25, 46, 119, 20, 26, 110, 246, 26, 167, 172, 205, 169, 79, 468, 33, 229, 38, 222, 167, 429, 41, 429, 101, 270, 560, 416, 100, 153, 276, 390, 717, 50, 615, 61, 61, 60, 404, 634, 214, 130, 130, 1033, 975, 382
Offset: 1

Views

Author

Wesley Ivan Hurt, May 02 2020

Keywords

Examples

			a(1) = 4; there is one Heronian triangle with perimeter A051518(1) = 12, which is [3,4,5] and its "middle" side length is 4.
a(6) = 23; there are two Heronian triangles with perimeter A051518(6) = 32, [4,13,15] and [10,10,12]. The sum is 13 + 10 = 23.
		

Crossrefs

Formula

a(n) = Sum_{k=1..floor(c(n)/3)} Sum_{i=k..floor((c(n)-k)/2)} sign(floor((i+k)/(c(n)-i-k+1))) * chi(sqrt((c(n)/2)*(c(n)/2-i)*(c(n)/2-k)*(c(n)/2-(c(n)-i-k)))) * i, where chi(n) = 1 - ceiling(n) + floor(n) and c(n) = A051518(n). - Wesley Ivan Hurt, May 12 2020

A330916 Sum of the largest side lengths of all Heronian triangles with perimeter A051518(n).

Original entry on oeis.org

5, 6, 8, 10, 13, 27, 61, 17, 35, 20, 59, 41, 96, 25, 80, 139, 30, 26, 57, 157, 37, 37, 140, 296, 40, 196, 207, 250, 209, 91, 587, 52, 294, 51, 267, 214, 498, 50, 539, 117, 310, 697, 530, 147, 206, 342, 503, 856, 73, 744, 75, 68, 85, 550, 793, 256, 172, 155, 1270, 1202
Offset: 1

Views

Author

Wesley Ivan Hurt, May 02 2020

Keywords

Examples

			a(1) = 5; there is one Heronian triangle with perimeter A051518(1) = 12, which is [3,4,5] and its largest side length is 5.
a(6) = 27; there are two Heronian triangles with perimeter A051518(6) = 32, [4,13,15] and [10,10,12]. The sum is 15 + 12 = 27.
		

Crossrefs

Formula

a(n) = Sum_{k=1..floor(c(n)/3)} Sum_{i=k..floor((c(n)-k)/2)} sign(floor((i+k)/(c(n)-i-k+1))) * chi(sqrt((c(n)/2)*(c(n)/2-i)*(c(n)/2-k)*(c(n)/2-(c(n)-i-k)))) * (c(n)-i-k), where chi(n) = 1 - ceiling(n) + floor(n) and c(n) = A051518(n). - Wesley Ivan Hurt, May 12 2020

A239246 Number of primitive Heronian triangles with n as greatest side length.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 2, 0, 2, 0, 3, 0, 0, 2, 2, 0, 0, 1, 3, 2, 0, 1, 2, 3, 0, 0, 0, 0, 2, 1, 5, 0, 4, 3, 4, 1, 0, 2, 1, 0, 0, 2, 0, 2, 4, 6, 4, 0, 2, 2, 0, 2, 0, 1, 3, 0, 1, 0, 8, 2, 0, 5, 1, 2, 0, 0, 6, 2, 7, 0, 3, 0, 0, 3, 0, 2, 0, 0, 9
Offset: 1

Views

Author

Frank M Jackson, Mar 13 2014

Keywords

Examples

			a(17)=3 as there are 3 primitive Heronian triangles with greatest side length of 17. They are (9, 10, 17), (8, 15, 17) and (16, 17, 17).
		

Crossrefs

Programs

  • Mathematica
    nn=200; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s]&&GCD[a, b, c]==1, area2=s(s-a)(s-b)(s-c); If[area2>0&&IntegerQ[Sqrt[area2]], AppendTo[lst, c]]], {c, 3, nn}, {b, c}, {a, b}]; Table[Length@Select[lst, #==n &], {n, 1, nn}] (* using T. D. Noe's program at A083875 *)
Showing 1-10 of 17 results. Next