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 1-7 of 7 results.

A070093 Number of acute integer triangles with perimeter n.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 2, 4, 3, 5, 4, 5, 5, 5, 6, 6, 6, 7, 7, 9, 8, 10, 9, 10, 10, 11, 12, 12, 12, 14, 13, 16, 14, 17, 16, 17, 18, 18, 20, 20, 20, 22, 22, 24, 23, 25, 26, 26, 27, 28, 30, 30, 29, 32, 31, 35, 33, 36, 36, 38, 39, 40, 40
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

An integer triangle [A070080(k) <= A070081(k) <= A070082(k)] is acute iff A070085(k) > 0.

Examples

			For n=9 there are A005044(9)=3 integer triangles: [1,4,4], [2,3,4] and [3,3,3]; two of them are acute, as 2^2+3^2<16=4^2, therefore a(9)=2.

Links

Crossrefs

Cf. A070094, A070095, A070118.
Cf. A005044, A024155, A070101.

Programs

  • Mathematica
    Table[Sum[Sum[(1 - Sign[Floor[(n - i - k)^2/(i^2 + k^2)]]) Sign[Floor[(i + k)/(n - i - k + 1)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}] (* Wesley Ivan Hurt, May 12 2019 *)

Formula

a(n) = A005044(n) - A070101(n) - A024155(n);
a(n) = A042154(n) + A070098(n).
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} (1-sign(floor((n-i-k)^2/(i^2+k^2)))) * sign(floor((i+k)/(n-i-k+1))). - Wesley Ivan Hurt, May 12 2019

A070103 Number of obtuse integer triangles with perimeter n and prime side lengths.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Examples

			For n=11 there are A005044(11)=4 integer triangles: [1,5,5], [2,4,5], [3,3,5] and [3,4,4]; only one of the two obtuses ([2,4,5] and [3,3,5]) consists of primes, therefore a(11)=1.

Links

Crossrefs

Cf. A070080, A070081, A070082, A070093, A070098, A070101, A070088, A070105, A070108, A070129.

Programs

  • Mathematica
    Table[Sum[Sum[(PrimePi[i] - PrimePi[i - 1]) (PrimePi[k] - PrimePi[k - 1]) (PrimePi[n - i - k] - PrimePi[n - i - k - 1]) (1 - Sign[Floor[(i^2 + k^2)/(n - i - k)^2]]) Sign[Floor[(i + k)/(n - i - k + 1)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}] (* Wesley Ivan Hurt, May 13 2019 *)

Formula

a(n) = A070093(n) - A070098(n).
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} (1 - sign(floor((i^2 + k^2)/(n-i-k)^2))) * sign(floor((i + k)/(n-i-k+1))) * A010051(i) * A010051(k) * A010051(n-i-k). - Wesley Ivan Hurt, May 13 2019

A070100 Number of integer triangles with perimeter n and prime side lengths which are acute and isosceles.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 2, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 2, 0, 1, 0, 2, 0, 1, 1, 2, 0, 2, 1, 2, 0, 2, 0, 2, 0, 1, 1, 2, 0, 3, 0, 2, 0, 1, 0, 3, 0, 1, 1, 2, 0, 2, 1, 3, 0, 1, 0, 3, 0, 1, 0, 1, 0, 3, 1, 3, 0, 2, 0, 3, 0, 0, 1, 3, 0, 3, 1, 3, 0
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Links

Crossrefs

Cf. A070080, A070081, A070082, A070093, A059169, A070088, A070092, A070095, A070098, A070126.

A070106 Number of integer triangles with perimeter n which are obtuse and isosceles.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 4, 4, 3, 3, 4, 4, 3, 4, 4
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

a(n)=A070101(n)-A024156(n); a(n)=A059169(n)-A070098(n).

Examples

			For n=11 there are A005044(11)=4 integer triangles: [1,5,5], [2,4,5], [3,3,5] and [3,4,4]; only one of the two obtuses ([2,4,5] and [3,3,5]) is also isosceles; therefore a(11)=1.

Links

Crossrefs

Cf. A070080, A070081, A070082, A059169, A070107, A070108, A070133.

A070099 Number of integer triangles with perimeter n and relatively prime side lengths which are acute and isosceles.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 2, 1, 3, 1, 1, 2, 4, 1, 4, 2, 2, 2, 5, 1, 4, 2, 4, 3, 6, 2, 6, 3, 4, 3, 5, 3, 8, 3, 4, 3, 8, 3, 9, 5, 5, 4, 10, 3, 9, 4, 6, 5, 11, 4, 8, 5, 7, 6, 12, 3, 13, 6, 8, 7, 9, 4, 14, 7, 8, 5, 15, 5, 15, 7, 9, 8, 13, 6, 16, 6, 11, 8, 17, 5, 13
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Links

Crossrefs

Cf. A070080, A070081, A070082, A070093, A059169, A051493, A070091, A070094, A070098, A070084, A070125.

A070124 Numbers n such that [A070080(n), A070081(n), A070082(n)] is an acute isosceles integer triangle.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 10, 11, 12, 15, 16, 18, 19, 22, 23, 24, 27, 28, 31, 34, 35, 38, 39, 40, 43, 46, 47, 48, 51, 54, 55, 58, 63, 64, 65, 68, 71, 72, 73, 76, 81, 84, 85, 88, 93, 94, 95, 98, 103, 107, 108, 109, 112, 117, 120, 121, 124, 129
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Examples

			a(13)=18: [A070080(18), A070081(18), A070082(18)]=[4=4=4], A070085(18)=4^2+4^2-4^2=16>0.

Links

Crossrefs

Cf. A070098, A070126, A070125, A070115, A070118.

A247589 Number of integer-sided obtuse triangles with largest side n.

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 5, 7, 10, 12, 15, 17, 21, 25, 29, 33, 37, 42, 48, 53, 58, 65, 71, 76, 83, 91, 100, 106, 113, 122, 130, 140, 149, 158, 169, 177, 188, 197, 210, 221, 230, 243, 255, 269, 281, 292, 306, 318, 333, 346
Offset: 1

Views

Author

Vladimir Letsko, Sep 20 2014

Keywords

Examples

			a(5) = 2 because there are 2 integer-sided acute triangles with largest side 5: (2,4,5); (3,3,5).

Links

Crossrefs

Cf. A046080, A002623, A224921, A247586, A247587, A247588.

Programs

  • Maple
    tr_o:=proc(n) local a,b,t,d;t:=0:
for a to n do
for b from max(a,n+1-a) to n do
d:=a^2+b^2-n^2:
if d<0 then t:=t+1 fi
od od;
t; end;

Formula

a(n) = k*(k + (1+(-1)^n)/2) + Sum_{j=1..floor(n*(1-sqrt(2)/2))} floor(sqrt(2*j*n - j^2 - 1) - j), where k = floor((2*n*(sqrt(2) - 1) + 1 - (-1)^n)/4) (it appears that k(n) is A070098(n)). - Anton Nikonov, Sep 29 2014
Showing 1-7 of 7 results.

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