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

A070080 Smallest side of integer triangles [a(n) <= A070081(n) <= A070082(n)], sorted by perimeter, lexicographically ordered.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 1, 2, 3, 2, 3, 1, 2, 3, 3, 2, 3, 4, 1, 2, 3, 3, 4, 2, 3, 4, 4, 1, 2, 3, 3, 4, 4, 5, 2, 3, 4, 4, 5, 1, 2, 3, 3, 4, 4, 5, 5, 2, 3, 4, 4, 5, 5, 6, 1, 2, 3, 3, 4, 4, 5, 5, 5, 6, 2, 3, 4, 4, 5, 5, 6, 6, 1, 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 2, 3, 4, 4, 5, 5
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Crossrefs

Cf. A316841, A316843, A316844, A316845 (sides (i,j,k) with j + k > i >= j >= k >= 1).
Cf. A331244, A331245, A331246 (similar, but triangles sorted by radius of enclosing circle), A331251, A331252, A331253 (triangles sorted by area), A331254, A331255, A331256 (triangles sorted by radius of circumcircle).

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]]&];
    triangles[[All, 1]] (* Jean-François Alcover, Jun 12 2012, updated Jul 09 2017 *)

Formula

a(n) = A070083(n) - A070082(n) - A070081(n).

A070081 Middle side of integer triangles [A070080(n) <= a(n) <= A070082(n)], sorted by perimeter, sides lexicographically ordered.

Original entry on oeis.org

1, 2, 2, 3, 2, 3, 4, 3, 3, 4, 3, 5, 4, 3, 4, 5, 4, 4, 6, 5, 4, 5, 4, 6, 5, 4, 5, 7, 6, 5, 6, 4, 5, 5, 7, 6, 5, 6, 5, 8, 7, 6, 7, 5, 6, 5, 6, 8, 7, 6, 7, 5, 6, 6, 9, 8, 7, 8, 6, 7, 5, 6, 7, 6, 9, 8, 7, 8, 6, 7, 6, 7, 10, 9, 8, 9, 7, 8, 6, 7, 8, 6, 7, 7, 10, 9, 8, 9, 7
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

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]]&];
    triangles[[All, 2]] (* Jean-François Alcover, Jul 09 2017 *)

Formula

a(n) = A070083(n) - A070080(n) - A070082(n).

A070082 Largest side of integer triangles [A070080(n) <= A070081(n) <= a(n)], sorted by perimeter, sides lexicographically ordered.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 4, 4, 3, 4, 4, 5, 5, 5, 4, 5, 5, 4, 6, 6, 6, 5, 5, 6, 6, 6, 5, 7, 7, 7, 6, 7, 6, 5, 7, 7, 7, 6, 6, 8, 8, 8, 7, 8, 7, 7, 6, 8, 8, 8, 7, 8, 7, 6, 9, 9, 9, 8, 9, 8, 9, 8, 7, 7, 9, 9, 9, 8, 9, 8, 8, 7, 10, 10, 10, 9, 10, 9, 10, 9, 8, 9, 8, 7, 10, 10
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

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]]&];
    triangles[[All, 3]] (* Jean-François Alcover, Jul 09 2017 *)

Formula

a(n) = A070083(n) - A070080(n) - A070081(n).

A001400 Number of partitions of n into at most 4 parts.

Original entry on oeis.org

1, 1, 2, 3, 5, 6, 9, 11, 15, 18, 23, 27, 34, 39, 47, 54, 64, 72, 84, 94, 108, 120, 136, 150, 169, 185, 206, 225, 249, 270, 297, 321, 351, 378, 411, 441, 478, 511, 551, 588, 632, 672, 720, 764, 816, 864, 920, 972, 1033, 1089, 1154, 1215, 1285, 1350, 1425, 1495
Offset: 0

Views

Author

Keywords

Comments

Molien series for 4-dimensional representation of S_4 [Nebe, Rains, Sloane, Chap. 7].
Also number of pure 2-complexes on 4 nodes with n multiple 2-simplexes. - Vladeta Jovovic, Dec 27 1999
Also number of different integer triangles with perimeter <= n+3. Also number of different scalene integer triangles with perimeter <= n+9. - Reinhard Zumkeller, May 12 2002
a(n) is the coefficient of q^n in the expansion of (m choose 4)_q as m goes to infinity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
Also number of partitions of n into parts <= 4. a(n) = A026820(n,4), for n > 3. - Reinhard Zumkeller, Jan 21 2010
Number of different distributions of n+10 identical balls in 4 boxes as x,y,z,p where 0 < x < y < z < p. - Ece Uslu and Esin Becenen, Jan 11 2016
Number of partitions of 5n+8 or 5n+12 into 4 parts (+-) 3 mod 5. a(4) = 5 partitions of 28: [7,7,7,7], [12,7,7,2], [12,12,2,2], [17,7,2,2], [22,2,2,2]. a(3) = 3 partitions of 27: [8,8,8,3], [13,8,3,3], [18,3,3,3]. - Richard Turk, Feb 24 2016
a(n) is the total number of non-isomorphic geodetic graphs of diameter n homeomorphic to a complete graph K4. - Carlos Enrique Frasser, May 24 2018

Examples

			(4 choose 4)_q = 1, (5 choose 4)_q = q^4 + q^3 + q^2 + q + 1, (6 choose 4)_q = q^8 + q^7 + 2*q^6 + 2*q^5 + 3*q^4 + 2*q^3 + 2*q^2 + q + 1, (7 choose 4) = q^12 + q^11 + 2*q^10 + 3*q^9 + 4*q^8 + 4*q^7 + 5*q^6 + 4*q^5 + 4*q^4 + 3*q^3 + 2*q^2 + q + 1 so the coefficient of q^0 converges to 1, q^1 to 1, q^2 to 2 and so on.
G.f. = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 6*x^5 + 9*x^6 + 11*x^7 + ...
a(4) = 5, i.e., {1,2,3,8}, {1,2,4,7}, {1,2,5,6}, {2,3,4,5}, {1,3,4,6}. Number of different distributions of 14 identical balls in 4 boxes as x,y,z,p where 0 < x < y < z < p. - _Ece Uslu_, Esin Becenen, Jan 11 2016
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 115, row m=4 of Q(m,n) table; p. 120, P(n,4).
  • H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 2.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 275.
  • D. E. Knuth, The Art of Computer Programming, vol. 4, Fascicle 3, Generating All Combinations and Partitions, Addison-Wesley, 2005, Section 7.2.1.4., p. 56, exercise 31.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Essentially same as A026810. Partial sums of A005044.
a(n) = A008284(n+4, 4), n >= 0.
First differences of A002621.

Programs

  • Haskell
    a001400 n = a001400_list !! n
    a001400_list = scanl1 (+) a005044_list -- Reinhard Zumkeller, Feb 28 2013
  • Magma
    K:=Rationals(); M:=MatrixAlgebra(K,4); q1:=DiagonalMatrix(M,[1,-1,1,-1]); p1:=DiagonalMatrix(M,[1,1,-1,-1]); q2:=DiagonalMatrix(M,[1,1,1,-1]); h:=M![1,1,1,1, 1,1,-1,-1, 1,-1,1,-1, 1,-1,-1,1]/2; G:=MatrixGroup<4,K|q1,q2,h>; MolienSeries(G);
    
  • Maple
    A001400 := n->if n mod 2 = 0 then round(n^2*(n+3)/144); else round((n-1)^2*(n+5)/144); fi;
    with(combstruct):ZL5:=[S,{S=Set(Cycle(Z,card<5))}, unlabeled]:seq(count(ZL5,size=n),n=0..55); # Zerinvary Lajos, Sep 24 2007
    A001400:=-(-z**8+z**9+2*z**4-z**7-1-z)/(z**2+1)/(z**2+z+1)/(z+1)**2/(z-1)**4; # [conjectured by Simon Plouffe in his 1992 dissertation; gives sequence except for an initial 1]
    B:=[S,{S = Set(Sequence(Z,1 <= card),card <=4)},unlabelled]: seq(combstruct[count](B, size=n), n=0..55); # Zerinvary Lajos, Mar 21 2009
  • Mathematica
    CoefficientList[ Series[ 1/((1 - x)*(1 - x^2)*(1 - x^3)*(1 - x^4)), {x, 0, 65} ], x ]
    LinearRecurrence[{1, 1, 0, 0, -2, 0, 0, 1, 1, -1}, {1, 1, 2, 3, 5, 6, 9, 11, 15, 18}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2012 *)
    a[n_] := Sum[Floor[(n - j - 3*k + 2)/2], {j, 0, Floor[n/4]}, {k, j, Floor[(n - j)/3]}]; Table[a[n], {n, 0, 55}] (* L. Edson Jeffery, Jul 31 2014 *)
    a[ n_] := With[{m = n + 5}, Round[ (2 m^3 - 3 m (5 + 3 (-1)^m)) / 288]]; (* Michael Somos, Dec 29 2014 *)
    a[ n_] := With[{m = Abs[n + 5] - 5}, Sign[n + 5] Length[ IntegerPartitions[ m, 4]]]; (* Michael Somos, Dec 29 2014 *)
    a[ n_] := With[{m = Abs[n + 5] - 5}, Sign[n + 5] SeriesCoefficient[ 1 / ((1 - x) (1 - x^2) (1 - x^3) (1 - x^4)), {x, 0, m}]]; (* Michael Somos, Dec 29 2014 *)
    Table[Length@IntegerPartitions[n, 4], {n, 0, 55}] (* Robert Price, Aug 18 2020 *)
  • PARI
    a(n) = round(((n+4)^3 + 3*(n+4)^2 -9*(n+4)*((n+4)% 2))/144) \\ Washington Bomfim, Jul 03 2012
    
  • PARI
    {a(n) = n+=5; round( (2*n^3 - 3*n*(5 + 3*(-1)^n)) / 288)}; \\ Michael Somos, Dec 29 2014
    
  • PARI
    a(n) = #partitions(n,,4); \\ Ruud H.G. van Tol, Jun 02 2024
    

Formula

G.f.: 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)).
a(n) = 1 + (a(n-2) + a(n-3) + a(n-4)) - (a(n-5) + a(n-6) + a(n-7)) + a(n-9). - Norman J. Meluch (norm(AT)iss.gm.com), Mar 09 2000
P(n, 4) = (1/288)*(2*n^3 + 6*n^2 - 9*n - 13 + (9*n+9)*pcr{1, -1}(2, n) - 32*pcr{1, -1, 0}(3, n) - 36*pcr{1, 0, -1, 0}(4, n)) (see Comtet).
Let c(n) = Sum_{i=0..floor(n/3)} (1 + ceiling((n-3*i-1)/2)), then a(n) = Sum_{i=0..floor(n/4)} (1 + ceiling((n-4*i-1)/2) + c(n-4*i-3)). - Jon Perry, Jun 27 2003
Euler transform of finite sequence [1, 1, 1, 1].
(n choose 4)_q = (q^n-1)*(q^(n-1)-1)*(q^(n-2)-1)*(q^(n-3)-1)/((q^4-1)*(q^3-1)*(q^2-1)*(q-1)).
a(n) = round(((n+4)^3 + 3*(n+4)^2 - 9*(n+4)*((n+4) mod 2))/144). - Washington Bomfim, Jul 03 2012
a(n) = a(n-1) + a(n-2) - 2*a(n-5) + a(n-8) + a(n-9) - a(n-10). - David Neil McGrath, Sep 12 2014
a(n) = -a(-10-n) for all n in Z. - Michael Somos, Dec 29 2014
a(n) - a(n+1) - a(n+3) + a(n+4) = 0 if n is odd, else floor(n/4) + 2 for all n in Z. - Michael Somos, Dec 29 2014
a(n) = n^3/144 + n^2/24 - 7*n/144 + 1 + floor(n/4)/4 + floor(n/3)/3 + (n+5)*floor(n/2)/8 + floor((n+1)/4)/4. - Vaclav Kotesovec, Aug 18 2015
a(n) = a(n-4) + A001399(n). - Ece Uslu, Esin Becenen, Jan 11 2016, corrected Sep 25 2020
a(6*n) - a(6*n+1) - a(6*n+4) + a(6*n+5) = n+1. - Richard Turk, Apr 19 2016
a(n) = a(n-1) + A005044(n+3) for n>0, i.e., first differences is A005044. - Yuchun Ji, Oct 12 2020
From Vladimír Modrák and Zuzana Soltysova, Dec 09 2020: (Start)
a(n) = round((n + 3)^2/12) + Sum_{i=0..floor(n/4)} round((n - 4*i - 1)^2/12).
a(n) = floor(((n + 3)^2 + 4)/12) + Sum_{i=0..floor(n/4)} floor(((n - 4*i - 1)^2 + 4)/12). (End)
a(n) - a(n-3) = A008642(n). - R. J. Mathar, Jun 23 2021
a(n) - a(n-2) = A025767(n). - R. J. Mathar, Jun 23 2021
a(n) = round((2*n^3 + 30*n^2 + 135*n + 175)/288 + (-1)^n*(n+5)/32). - Dave Neary, Oct 28 2021
From Vladimír Modrák, Jul 13 2022: (Start)
a(n) = Sum_{j=0..floor(n/4)} Sum_{i=0..floor(n/3)} ceiling((max(0,n + 1 - 3*i - 4*j))/2).
a(n) = Sum_{i=0..floor(n/4)} floor(((n + 3 - 4*i)^2 + 4)/12). (End)
a(n) = floor(((n+4)^2*(n+7) - 9*(n+4)*(n mod 2) + 32)/144). - Vladimír Modrák, Mar 23 2025

A070086 Areas of integer triangles [A070080(n), A070081(n), A070082(n)], rounded values.

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 2, 3, 4, 4, 4, 2, 4, 4, 6, 5, 6, 7, 3, 5, 5, 7, 8, 6, 7, 8, 9, 3, 6, 6, 9, 7, 10, 11, 7, 9, 10, 11, 12, 4, 6, 8, 10, 8, 12, 12, 14, 8, 10, 12, 13, 12, 15, 16, 4, 7, 9, 12, 10, 14, 10, 15, 16, 17, 9, 12, 13, 15, 14, 17, 18, 19, 5, 8, 10
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

Triangles [A070080(A070142(n)), A070081(A070142(n)), A070082(A070142(n))] have integer areas = a(A070142(k)) = A070149(k).

Examples

			[A070080(25), A070081(25), A070082(25)] = [3,5,6] and s = A070083(25)/2 = (3+5+6)/2 = 7: a(25) = sqrt(s*(s-3)*(s-5)*(s-6)) = sqrt(7*(7-3)*(7-5)*(7-6)) = sqrt(7*4*2*1) = sqrt(56) = 7.48331, rounded = 7.
		

Crossrefs

The sides are given by A070080, A070081, A070082.
See A135622 for values signifying the precise area and further crossrefs.

Programs

  • Mathematica
    m = 50; (* 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]]&];
    area[{a_, b_, c_}] := With[{p = (a+b+c)/2}, Sqrt[p(p-a)(p-b)(p-c)] // Round];
    area /@ triangles (* Jean-François Alcover, Oct 03 2021 *)

Formula

a(n) = sqrt(s*(s-u)*(s-v)*(s-w)), where u=A070080(n), v=A070081(n), w=A070082(n) and s = A070083(n)/2 = (u+v+w)/2.

A070142 Numbers n such that [A070080(n), A070081(n), A070082(n)] is an integer triangle with integer area.

Original entry on oeis.org

17, 39, 52, 116, 212, 252, 269, 368, 370, 372, 375, 493, 561, 587, 659, 839, 850, 862, 957, 972, 1156, 1186, 1196, 1204, 1297, 1582, 1599, 1629, 1912, 1920, 1955, 1971, 1988, 2115, 2352, 2555, 2574, 2713, 2774, 2778, 2790
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Examples

			a(2)=39: [A070080(39), A070081(39), A070082(39)] = [5,5,6], area^2 = s*(s-5)*(s-5)*(s-6) with s=A070083(39)/2=(5+5+6)/2=8, area^2=8*3*3*2=16*9 is an integer square, therefore A070086(39)=area=4*3=12.
		

Crossrefs

Programs

  • Mathematica
    maxPerim = 100; maxSide = Floor[(maxPerim - 1)/2]; order[{a_, b_, c_}] := (a + b + c)*maxPerim^3 + a*maxPerim^2 + b*maxPerim + c; triangles = Reap[ Do[ If[ a + b + c <= maxPerim && c - b < a < c + b && b - a < c < b + a && c - a < b < c + a, Sow[{a, b, c}]], {a, 1, maxSide}, {b, a, maxSide}, {c, b, maxSide}]][[2, 1]]; stri = Sort[ triangles, order[#1] < order[#2]&]; area[{a_, b_, c_}] := With[{p = (a + b + c)/2}, Sqrt[p*(p - a)*(p - b)*(p - c)]]; Position[ stri, tri_ /; IntegerQ[area[tri]]] // Flatten (* Jean-François Alcover, Feb 22 2013 *)

A070201 Number of integer triangles with perimeter n having integral inradius.

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, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 3, 0, 0, 0, 2, 0, 1, 0, 1, 0, 2, 0, 2, 0, 0, 0, 1, 0, 1, 0, 2, 0, 0, 0, 8, 0, 0, 0, 1, 0, 3
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

a(n) = #{k | A070083(k) = n and A070200(k) = exact inradius};
a(n) = A070203(n) + A070204(n);
a(n) = A070205(n) + A070206(n) + A024155(n);
a(odd) = 0.

Examples

			a(36)=2, as there are two integer triangles with integer inradius having perimeter=32:
First: [A070080(368), A070081(368), A070082(368)] = [9,10,17], for s = A070083(368)/2 = (9+10+17)/2 = 18: inradius = sqrt((s-9)*(s-10)*(s-17)/s) = sqrt(9*8*1/18) = sqrt(4) = 2; therefore A070200(368) = 2.
2nd: [A070080(370), A070081(370), A070082(370)] = [9,12,15], for s = A070083(370)/2 = (9+12+15)/2 = 18: inradius = sqrt((s-9)*(s-12)*(s-15)/s) = sqrt(9*6*3/18) = sqrt(9) = 3; therefore A070200(370) = 3.
		

Crossrefs

Programs

  • Ruby
    def A(n)
      cnt = 0
      (1..n / 3).each{|a|
        (a..(n - a) / 2).each{|b|
          c = n - a - b
          if a + b > c
            s = n / 2r
            t = (s - a) * (s - b) * (s - c) / s
            if t.denominator == 1
              t = t.to_i
              cnt += 1 if Math.sqrt(t).to_i ** 2 == t
            end
          end
        }
      }
      cnt
    end
    def A070201(n)
      (1..n).map{|i| A(i)}
    end
    p A070201(100) # Seiichi Manyama, Oct 06 2017

A070209 Numbers n such that [A070080(n), A070081(n), A070082(n)] is an integer triangle with integer inradius.

Original entry on oeis.org

17, 116, 212, 269, 368, 370, 493, 561, 587, 659, 850, 1204, 1297, 1582, 1599, 1629, 1920, 1988, 2115, 2352, 2555, 2574, 2774, 2778, 3251, 3473, 3746, 3751, 4286, 4298, 4307, 4313, 4319, 4330, 4370, 4406, 5008, 5251
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Examples

			a(3)=212: [A070080(212), A070081(212), A070082(212)] = [5,12,13], for s = A070083(212)/2 = (5+12+13)/2 = 15: inradius = sqrt((s-5)*(s-12)*(s-13)/s) = sqrt(10*3*2/15) = sqrt(4) = 2; therefore A070200(212)=2. [Corrected by _Rick L. Shepherd_, May 15 2008]
		

References

  • Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.

Crossrefs

A070149 Areas of integer Heronian triangles [A070080(A070142(n)), A070081(A070142(n)), A070082(A070142(n))].

Original entry on oeis.org

6, 12, 12, 24, 30, 24, 48, 36, 54, 48, 60, 60, 42, 84, 66, 84, 96, 108, 60, 120, 36, 90, 126, 108, 84, 60, 120, 150, 72, 96, 168, 120, 192, 132, 204, 210, 210, 84, 144, 216, 192, 240, 114, 156, 180, 120, 240, 300, 168, 210, 168
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Examples

			A070142(2)=39: [A070080(39), A070081(39), A070082(39)] = [5,5,6], area^2 = s*(s-5)*(s-5)*(s-6) with s=A070083(39)/2=(5+5+6)/2=8, area^2=8*3*3*2=16*9 is an integer square, therefore a(2)=A070086(39)=area=4*3=12.
		

Crossrefs

Programs

  • Mathematica
    m = 500 (* 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]] &];
    area[{a_, b_, c_}] := With[{p = (a+b+c)/2}, Sqrt[p(p-a)(p-b)(p-c)]];
    Select[area /@ triangles, IntegerQ] (* Jean-François Alcover, Oct 12 2021 *)

Formula

a(n) = A070086(A070142(n)).
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