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.

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A005044 Alcuin's sequence: expansion of x^3/((1-x^2)*(1-x^3)*(1-x^4)).

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 1, 2, 1, 3, 2, 4, 3, 5, 4, 7, 5, 8, 7, 10, 8, 12, 10, 14, 12, 16, 14, 19, 16, 21, 19, 24, 21, 27, 24, 30, 27, 33, 30, 37, 33, 40, 37, 44, 40, 48, 44, 52, 48, 56, 52, 61, 56, 65, 61, 70, 65, 75, 70, 80, 75, 85, 80, 91, 85, 96, 91, 102, 96, 108, 102, 114, 108, 120
Offset: 0

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Author

Keywords

Comments

a(n) is the number of triangles with integer sides and perimeter n.
Also a(n) is the number of triangles with distinct integer sides and perimeter n+6, i.e., number of triples (a, b, c) such that 1 < a < b < c < a+b, a+b+c = n+6. - Roger Cuculière
With a different offset (i.e., without the three leading zeros, as in A266755), the number of ways in which n empty casks, n casks half-full of wine and n full casks can be distributed to 3 persons in such a way that each one gets the same number of casks and the same amount of wine [Alcuin]. E.g., for n=2 one can give 2 people one full and one empty and the 3rd gets two half-full. (Comment corrected by Franklin T. Adams-Watters, Oct 23 2006)
For m >= 2, the sequence {a(n) mod m} is periodic with period 12*m. - Martin J. Erickson (erickson(AT)truman.edu), Jun 06 2008
Number of partitions of n into parts 2, 3, and 4, with at least one part 3. - Joerg Arndt, Feb 03 2013
For several values of p and q the sequence (A005044(n+p) - A005044(n-q)) leads to known sequences, see the crossrefs. - Johannes W. Meijer, Oct 12 2013
For n>=3, number of partitions of n-3 into parts 2, 3, and 4. - David Neil McGrath, Aug 30 2014
Also, a(n) is the number of partitions mu of n of length 3 such that mu_1-mu_2 is even and mu_2-mu_3 is even (see below example). - John M. Campbell, Jan 29 2016
For n > 1, number of triangles with odd side lengths and perimeter 2*n-3. - Wesley Ivan Hurt, May 13 2019
Number of partitions of n+1 into 4 parts whose largest two parts are equal. - Wesley Ivan Hurt, Jan 06 2021
For n>=3, number of weak partitions of n-3 (that is, allowing parts of size 0) into three parts with no part exceeding (n-3)/2. Also, number of weak partitions of n-3 into three parts, all of the same parity as n-3. - Kevin Long, Feb 20 2021
Also, a(n) is the number of incongruent acute triangles formed from the vertices of a regular n-gon. - Frank M Jackson, Nov 04 2022

Examples

			There are 4 triangles of perimeter 11, with sides 1,5,5; 2,4,5; 3,3,5; 3,4,4. So a(11) = 4.
G.f. = x^3 + x^5 + x^6 + 2*x^7 + x^8 + 3*x^9 + 2*x^10 + 4*x^11 + 3*x^12 + ...
From _John M. Campbell_, Jan 29 2016: (Start)
Letting n = 15, there are a(n)=7 partitions mu |- 15 of length 3 such that mu_1-mu_2 is even and mu_2-mu_3 is even:
(13,1,1) |- 15
(11,3,1) |- 15
(9,5,1) |- 15
(9,3,3) |- 15
(7,7,1) |- 15
(7,5,3) |- 15
(5,5,5) |- 15
(End)
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 74, Problem 7.
  • I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. Wiley, NY, Chap.10, Section 10.2, Problems 5 and 6, pp. 451-2.
  • D. Olivastro: Ancient Puzzles. Classic Brainteasers and Other Timeless Mathematical Games of the Last 10 Centuries. New York: Bantam Books, 1993. See p. 158.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • A. M. Yaglom and I. M. Yaglom: Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 8, #30 (First published: San Francisco: Holden-Day, Inc., 1964)

Crossrefs

See A266755 for a version without the three leading zeros.
Both bisections give (essentially) A001399.
(See the comments.) Cf. A008615 (p=1, q=3, offset=0), A008624 (3, 3, 0), A008679 (3, -1, 0), A026922 (1, 5, 1), A028242 (5, 7, 0), A030451 (6, 6, 0), A051274 (3, 5, 0), A052938 (8, 4, 0), A059169 (0, 6, 1), A106466 (5, 4, 0), A130722 (2, 7, 0)
Cf. this sequence (k=3), A288165 (k=4), A288166 (k=5).
Number of k-gons that can be formed with perimeter n: this sequence (k=3), A062890 (k=4), A069906 (k=5), A069907 (k=6), A288253 (k=7), A288254 (k=8), A288255 (k=9), A288256 (k=10).

Programs

  • Haskell
    a005044 = p [2,3,4] . (subtract 3) where
    p _ 0 = 1
    p [] _ = 0
    p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m
    -- Reinhard Zumkeller, Feb 28 2013
  • Maple
    A005044 := n-> floor((1/48)*(n^2+3*n+21+(-1)^(n-1)*3*n)): seq(A005044(n), n=0..73);
    A005044 := -1/(z**2+1)/(z**2+z+1)/(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation
  • Mathematica
    a[n_] := Round[If[EvenQ[n], n^2, (n + 3)^2]/48] (* Peter Bertok, Jan 09 2002 *)
    CoefficientList[Series[x^3/((1 - x^2)*(1 - x^3)*(1 - x^4)), {x, 0, 105}], x] (* Robert G. Wilson v, Jun 02 2004 *)
    me[n_] := Module[{i, j, sum = 0}, For[i = Ceiling[(n - 3)/3], i <= Floor[(n - 3)/2], i = i + 1, For[j = Ceiling[(n - i - 3)/2], j <= i, j = j + 1, sum = sum + 1] ]; Return[sum]; ] mine = Table[me[n], {n, 1, 11}]; (* Srikanth (sriperso(AT)gmail.com), Aug 02 2008 *)
    LinearRecurrence[{0,1,1,1,-1,-1,-1,0,1},{0,0,0,1,0,1,1,2,1},80] (* Harvey P. Dale, Sep 22 2014 *)
    Table[Length@Select[IntegerPartitions[n, {3}], Max[#]*180 < 90 n &], {n, 1, 100}] (* Frank M Jackson, Nov 04 2022 *)
  • PARI
    a(n) = round(n^2 / 12) - (n\2)^2 \ 4
    
  • PARI
    a(n) = (n^2 + 6*n * (n%2) + 24) \ 48
    
  • PARI
    a(n)=if(n%2,n+3,n)^2\/48 \\ Charles R Greathouse IV, May 02 2016
    
  • PARI
    concat(vector(3), Vec((x^3)/((1-x^2)*(1-x^3)*(1-x^4)) + O(x^70))) \\ Felix Fröhlich, Jun 07 2017
    

Formula

a(n) = a(n-6) + A059169(n) = A070093(n) + A070101(n) + A024155(n).
For odd indices we have a(2*n-3) = a(2*n). For even indices, a(2*n) = nearest integer to n^2/12 = A001399(n).
For all n, a(n) = round(n^2/12) - floor(n/4)*floor((n+2)/4) = a(-3-n) = A069905(n) - A002265(n)*A002265(n+2).
For n = 0..11 (mod 12), a(n) is respectively n^2/48, (n^2 + 6*n - 7)/48, (n^2 - 4)/48, (n^2 + 6*n + 21)/48, (n^2 - 16)/48, (n^2 + 6*n - 7)/48, (n^2 + 12)/48, (n^2 + 6*n + 5)/48, (n^2 - 16)/48, (n^2 + 6*n + 9)/48, (n^2 - 4)/48, (n^2 + 6*n + 5)/48.
Euler transform of length 4 sequence [ 0, 1, 1, 1]. - Michael Somos, Sep 04 2006
a(-3 - n) = a(n). - Michael Somos, Sep 04 2006
a(n) = sum(ceiling((n-3)/3) <= i <= floor((n-3)/2), sum(ceiling((n-i-3)/2) <= j <= i, 1 ) ) for n >= 1. - Srikanth K S, Aug 02 2008
a(n) = a(n-2) + a(n-3) + a(n-4) - a(n-5) - a(n-6) - a(n-7) + a(n-9) for n >= 9. - David Neil McGrath, Aug 30 2014
a(n+3) = a(n) if n is odd; a(n+3) = a(n) + floor(n/4) + 1 if n is even. Sketch of proof: There is an obvious injective map from perimeter-n triangles to perimeter-(n+3) triangles defined by f(a,b,c) = (a+1,b+1,c+1). It is easy to show f is surjective for odd n, while for n=2k the image of f is only missing the triangles (a,k+2-a,k+1) for 1 <= a <= floor(k/2)+1. - James East, May 01 2016
a(n) = round(n^2/48) if n is even; a(n) = round((n+3)^2/48) if n is odd. - James East, May 01 2016
a(n) = (6*n^2 + 18*n - 9*(-1)^n*(2*n + 3) - 36*sin(Pi*n/2) - 36*cos(Pi*n/2) + 64*cos(2*Pi*n/3) - 1)/288. - Ilya Gutkovskiy, May 01 2016
a(n) = A325691(n-3) + A000035(n) for n>=3. The bijection between partition(n,[2,3,4]) and not-over-half partition(n,3,n/2) + partition(n,2,n/2) can be built by a Ferrers(part)[0+3,1,2] map. And the last partition(n,2,n/2) is unique [n/2,n/2] if n is even, it is given by A000035. - Yuchun Ji, Sep 24 2020
a(4n+3) = a(4n) + n+1, a(4n+4) = a(4n+1) = A000212(n+1), a(4n+5) = a(4n+2) + n+1, a(4n+6) = a(4n+3) = A007980(n). - Yuchun Ji, Oct 10 2020
a(n)-a(n-4) = A008615(n-1). - R. J. Mathar, Jun 23 2021
a(n)-a(n-2) = A008679(n-3). - R. J. Mathar, Jun 23 2021

Extensions

Additional comments from Reinhard Zumkeller, May 11 2002
Yaglom reference and mod formulas from Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), May 27 2000
The reference to Alcuin of York (735-804) was provided by Hermann Kremer (hermann.kremer(AT)onlinehome.de), Jun 18 2004

A070115 Numbers m such that [A070080(m), A070081(m), A070082(m)] is an isosceles integer triangle.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 18, 19, 22, 23, 24, 26, 27, 28, 31, 32, 34, 35, 38, 39, 40, 43, 46, 47, 48, 51, 52, 54, 55, 58, 61, 63, 64, 65, 68, 71, 72, 73, 76, 81, 82, 84, 85, 88, 91, 93, 94, 95, 98, 103, 104, 107, 108
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Examples

			26 is a term because [A070080(26), A070081(26), A070082(26)] = [4=4<6].
		

Crossrefs

Programs

  • Mathematica
    m = 55 (* max perimeter *);
    sides[per_] := Select[Reverse /@ IntegerPartitions[per, {3}, Range[ Ceiling[per/2]]], #[[1]] < per/2 && #[[2]] < per/2 && #[[3]] < per/2 &];
    triangles = DeleteCases[Table[sides[per], {per, 3, m}], {}] // Flatten[#, 1]& // SortBy[Total[#] m^3 + #[[1]] m^2 + #[[2]] m + #[[1]] &];
    Position[triangles, {a_, a_, b_} | {a_, b_, b_}] // Flatten (* Jean-François Alcover, Oct 12 2021 *)

A285869 a(n) is the number of zeros of the Chebyshev S(n, x) polynomial in the open interval (-sqrt(2), +sqrt(2)).

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 4, 3, 4, 5, 6, 5, 6, 7, 8, 7, 8, 9, 10, 9, 10, 11, 12, 11, 12, 13, 14, 13, 14, 15, 16, 15, 16, 17, 18, 17, 18, 19, 20, 19, 20, 21, 22, 21, 22, 23, 24, 23, 24, 25, 26, 25, 26, 27, 28, 27, 28, 29, 30, 29, 30, 31, 32, 31, 32, 33, 34, 33, 34
Offset: 0

Views

Author

Wolfdieter Lang, May 10 2017

Keywords

Comments

See a May 06 2017 comment on A049310 where these problems are considered which originated in a conjecture by Michel Lagneau (see A008611) on Fibonacci polynomials.

Crossrefs

Programs

  • Mathematica
    Table[2 (Floor[n/2] - Floor[(n + 1)/4]) + Boole[OddQ@ n], {n, 0, 52}] (* Michael De Vlieger, May 10 2017 *)
  • PARI
    concat(0, Vec(x*(1 + x - x^2 + x^3) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^100))) \\ Colin Barker, May 18 2017

Formula

a(n) = 2*b(n) if n is even, else a(n) = 1 + 2*b(n), with b(n) = floor(n/2) - floor((n + 1)/4) = A059169(n+1).
G.f. for {b(n)}: Sum_{n>=0} b(n)*x^n = x^2*(1 - x + x^2)/((1 - x)*(1 - x^4)) (see A059169).
From Colin Barker, May 18 2017: (Start)
G.f.: x*(1 + x - x^2 + x^3) / ((1 - x)^2*(1 + x)*(1 + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>4.
(End)
a(n) = A162330(n-1) for n >= 2. - Michel Marcus, Nov 01 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2) (A016627). - Amiram Eldar, Sep 17 2023
a(n) = A183041(n-1) for n>=2. - R. J. Mathar, May 13 2025

A070098 Number of integer triangles with perimeter n which are acute and isosceles.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

Equivalently, the number of obtuse isosceles integer triangles with base n. - Charlie Marion, Jun 18 2019

Examples

			For n=9 there are A005044(9)=3 integer triangles: [1,4,4], [2,3,4] and [3,3,3]; both isosceles are also acute.
		

Crossrefs

Programs

  • Magma
    [Floor(k/2)-Floor(k/(2 + Sqrt(2)))-((k + 1) mod 2): k in [1..76]]; // Marius A. Burtea, Jun 21 2019

Formula

a(n) = A070093(n)-A024154(n); a(n) = A059169(n)-A070106(n).
a(n) = floor(n/2) - floor(n/(2 + sqrt(2))) - ((n + 1) mod 2). - David Pasino, Jun 27 2016
a(n) = A004526(n-1) - A183138(n). - R. J. Mathar, May 22 2019

A008624 Expansion of g.f. (1 + x^3)/((1 - x^2)*(1 - x^4)) = (1 - x + x^2)/((1 + x)*(1 - x)^2*(1 + x^2)).

Original entry on oeis.org

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

Views

Author

Keywords

Comments

Molien series of 2-dimensional representation of group of order 16 over GF(3).

References

  • D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 107.

Crossrefs

Essentially the same as A059169.

Programs

  • Maple
    f := x -> (1+x^3)/((1-x^2)*(1-x^4)): seq(coeff(series(f(x), x, n+1), x, n), n=0..64);
    a := n -> floor(n/4) + ((n mod 2 + 1 - floor((n mod 4)/3)) mod 2): seq(a(n), n=0..64); # Johannes W. Meijer, Oct 08 2013
  • Mathematica
    CoefficientList[Series[(1 + x^3) / (1 - x^2) / (1 - x^4), {x, 0, 70}], x] (* Vincenzo Librandi, Aug 15 2013 *)
    LinearRecurrence[{1,0,0,1,-1},{1,0,1,1,2},70] (* Harvey P. Dale, Sep 27 2024 *)
  • PARI
    a(n) = (3 + 3*(-1)^n + (1-I)*(-I)^n + (1+I)*I^n + 2*n) / 8 \\ Colin Barker, Oct 15 2015
    
  • PARI
    my(x='x+O('x^100)); Vec((1+x^3)/((1-x^2)*(1-x^4))) \\ Altug Alkan, Dec 24 2015

Formula

From Reinhard Zumkeller, Aug 05 2005: (Start)
a(n) = floor(n/4) + ((n mod 2 + 1 - floor((n mod 4)/3)) mod 2).
a(n) = A110654(A028242(n)). (End)
a(n) = (3 + 3*(-1)^n + (1-i)*(-i)^n + (1+i)*i^n + 2*n) / 8 where i = sqrt(-1). - Colin Barker, Oct 15 2015
a(n) = (2*n+3+2*cos(n*Pi/2)+3*cos(n*Pi)-2*sin(n*Pi/2))/8. - Wesley Ivan Hurt, Oct 01 2017
E.g.f.: (cos(x) + (3 + x)*cosh(x) - sin(x) + x*sinh(x))/4. - Stefano Spezia, Jan 03 2023

Extensions

Replaced x^2 three times with x in the generating function (un-aerated). - R. J. Mathar, Oct 23 2008

A070092 Number of isosceles integer triangles with perimeter n and prime side lengths.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Examples

			For n=17 there are A005044(17)=8 integer triangles: [1,8,8], [2,7,8], [3,6,8], [3,7,7], [4,5,8], [4,6,7], [5,5,7] and [5,6,6]: four are isosceles: [1<8=8], [3<7=7], [5=5<7] and [5<6=6], but only two of them consist of primes, therefore a(17)=2.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[(PrimePi[i] - PrimePi[i - 1]) (PrimePi[k] - PrimePi[k - 1]) (PrimePi[n - i - k] - PrimePi[n - i - k - 1]) (KroneckerDelta[i, k] + KroneckerDelta[i, n - i - k] - KroneckerDelta[k, n - i - k]) Sign[Floor[(i + k)/(n - i - k + 1)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}] (* Wesley Ivan Hurt, May 14 2019 *)

Formula

a(n) = A070088(n) - A070090(n).
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} sign(floor((i+k)/(n-i-k+1))) * ([i = k] + [i = n-i-k] - [k = n-i-k]) * A010051(i) * A010051(k) * A010051(n-i-k), where [] is the Iverson bracket. - Wesley Ivan Hurt, May 14 2019

A070091 Number of isosceles integer triangles with perimeter n and relatively prime side lengths.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 2, 1, 1, 1, 3, 1, 3, 1, 2, 2, 4, 2, 5, 2, 2, 2, 6, 2, 5, 3, 5, 3, 7, 2, 8, 4, 4, 4, 6, 3, 9, 4, 6, 4, 10, 4, 11, 5, 6, 5, 12, 4, 10, 5, 8, 6, 13, 4, 10, 6, 8, 7, 15, 4, 15, 7, 10, 8, 12, 6, 17, 8, 10, 6, 18, 6, 18, 9, 10, 9, 14, 6, 20, 8, 13
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

a(n) = A051493(n) - A005044(n-6).

Examples

			For n=15 there are A005044(15)=7 integer triangles: [1,7,7], [2,6,7], [3,5,7], [3,6,6], [4,4,7], [4,5,6] and [5,5,5]: four are isosceles: [1<7=7], [3<6=6], [4=4<7] and [5=5=5], but GCD(3,6,6)>1 and GCD(5,5,5)>1, therefore a(15)=2.
		

Crossrefs

Programs

  • Mathematica
    m = 81 (* max perimeter *);
    sides[per_] := Select[Reverse /@ IntegerPartitions[per, {3}, Range[ Ceiling[per/2]]], #[[1]] < per/2 && #[[2]] < per/2 && #[[3]] < per/2 &];
    triangles = DeleteCases[Table[sides[per], {per, 3, m}], {}] // Flatten[#, 1] & // SortBy[Total[#] m^3 + #[[1]] m^2 + #[[2]] m + #[[1]] &] ;
    a[n_] := Count[triangles, t_ /; Total[t] == n && Length[Union[t]] < 3 && GCD @@ t == 1];
    Table[a[n], {n, 1, m}] (* Jean-François Alcover, Oct 05 2021 *)

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

Crossrefs

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.
		

Crossrefs

A070108 Number of integer triangles with perimeter n and prime side lengths which are obtuse and isosceles.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Examples

			a(k)<=1 until k = 140, for k = 141 there are A005044(141)=432 integer triangles, a(141)=2 as
[37=37<67]: 37+37+67 = 141 and 2*(37^2)<67^2 and 37, 67 are primes,
[41=41<59]: 41+41+59 = 141 and 2*(41^2)<59^2 and 41, 59 are primes.
		

Crossrefs

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