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 1-5 of 5 results.

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

Views

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

A008747 Expansion of (1+x^4)/((1-x)*(1-x^2)*(1-x^3)).

Original entry on oeis.org

1, 1, 2, 3, 5, 6, 9, 11, 14, 17, 21, 24, 29, 33, 38, 43, 49, 54, 61, 67, 74, 81, 89, 96, 105, 113, 122, 131, 141, 150, 161, 171, 182, 193, 205, 216, 229, 241, 254, 267, 281, 294, 309, 323, 338, 353, 369, 384, 401, 417, 434, 451, 469, 486, 505, 523, 542, 561
Offset: 0

Views

Author

Keywords

Comments

For n>=1, the set {A008747(6n+-1)} is the set of numbers of the form a^2 + 5*(a+1)^2 for -inf < a < inf. Furthermore the set A008747(6n) is A033581(n). - Kieren MacMillan, Dec 19 2007
For n>1, a(n-1) is the number of aperiodic necklaces (Lyndon words) with k<=3 black beads and n-k white beads. For n=4 we have for example a(3)=3 aperiodic necklaces: BWWW, BBWW and BBBW. BWBW is periodic and is not counted. - Herbert Kociemba, Oct 23 2016

Examples

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 6*x^5 + 9*x^6 + 11*x^7 + 14*x^8 + ...
		

Crossrefs

Programs

  • GAP
    a:=[1,1,2,3,5,6];; for n in [7..60] do a[n]:=a[n-1]+a[n-2]-a[n-4] -a[n-5]+a[n-6]; od; a; # G. C. Greubel, Aug 03 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 60); Coefficients(R!( (1+x^4)/((1-x)*(1-x^2)*(1-x^3)) )); // G. C. Greubel, Aug 03 2019
    
  • Maple
    A008747:=n->ceil((n+1)^2/6): seq(A008747(n), n=0..100); # Wesley Ivan Hurt, Oct 25 2016
  • Mathematica
    CoefficientList[Series[(1+x^4)/((1-x)(1-x^2)(1-x^3)),{x,0,60}],x] (* or *) LinearRecurrence[{1,1,0,-1,-1,1},{1,1,2,3,5,6},60] (* Harvey P. Dale, Sep 05 2012 *)
  • PARI
    Vec((1+x^4)/((1-x)*(1-x^2)*(1-x^3))+O(x^60)) \\ Charles R Greathouse IV, Sep 25 2012
    
  • Sage
    ((1+x^4)/((1-x)*(1-x^2)*(1-x^3))).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Aug 03 2019
    

Formula

G.f.: (1+x^4)/((1-x)*(1-x^2)*(1-x^3)).
a(n) = ceiling((n+1)^2/6).
a(n) = (12*n + 23 + 6*n^2 + 9*(-1)^n + 4*A061347(n))/36. - R. J. Mathar, Mar 15 2011
a(0)=1, a(1)=1, a(2)=2, a(3)=3, a(4)=5, a(5)=6, a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6) for n > 5. - Harvey P. Dale, Sep 05 2012
From Michael Somos, Oct 25 2016: (Start)
Euler transform of length 8 sequence [ 1, 1, 1, 1, 0, 0, 0, -1].
a(n) = a(-2-n) for all n in Z.
a(2*n-1) = A071619(n).
a(3*n-1) = 2*A077043(n).
a(n) - a(n-1) = A051274(n). (End)

A051275 Expansion of (1+x^2)/((1-x^2)*(1-x^3)).

Original entry on oeis.org

1, 0, 2, 1, 2, 2, 3, 2, 4, 3, 4, 4, 5, 4, 6, 5, 6, 6, 7, 6, 8, 7, 8, 8, 9, 8, 10, 9, 10, 10, 11, 10, 12, 11, 12, 12, 13, 12, 14, 13, 14, 14, 15, 14, 16, 15, 16, 16, 17, 16, 18, 17, 18, 18, 19, 18, 20, 19, 20, 20, 21, 20, 22, 21, 22, 22, 23, 22, 24, 23, 24, 24
Offset: 0

Views

Author

Keywords

Comments

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 12 ).
Diagonal sums of A117567. - Paul Barry, Mar 29 2006
First differences of A156040. - Bob Selcoe, Feb 07 2014
Also first difference of diagonal sums of the triangle formed by rows T(2,k) k=0,1...,2m of ascending m-nomial triangles (see A004737). - Bob Selcoe, Feb 07 2014

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1+x^2)/((1-x^2)(1-x^3)),{x,0,100}],x] (* or *) LinearRecurrence[{0,1,1,0,-1},{1,0,2,1,2},100] (* Harvey P. Dale, Dec 10 2024 *)
  • PARI
    Vec((1+x^2)/((1-x^2)*(1-x^3))+ O(x^80)) \\ Michel Marcus, Nov 26 2019

Formula

From Paul Barry, Mar 29 2006: (Start)
a(n) = a(n-2) + a(n-3) - a(n-5);
a(n) = cos(2*Pi*n/3 + Pi/3)/3 - sqrt(3)*sin(2*Pi*n/3 + Pi/3)/9 + (-1)^n/2 + (2n+3)/6;
a(n) = Sum_{k=0..floor(n/2)} F(L((n-2k+2)/3)) where L(j/p) is the Legendre symbol of j and p. (End)
a(n) = 2*floor(n/2) + floor((n+4)/3) - n. - Ridouane Oudra, Nov 26 2019

A267806 a(0) = a(1) = 1; for n>1, a(n) = (a(n-1) mod 2) + a(n-2).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 3, 3, 4, 3, 5, 4, 5, 5, 6, 5, 7, 6, 7, 7, 8, 7, 9, 8, 9, 9, 10, 9, 11, 10, 11, 11, 12, 11, 13, 12, 13, 13, 14, 13, 15, 14, 15, 15, 16, 15, 17, 16, 17, 17, 18, 17, 19, 18, 19, 19, 20, 19, 21, 20, 21, 21, 22, 21, 23, 22, 23, 23, 24, 23, 25, 24, 25, 25
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[0] == a[1] == 1, a[n] == Mod[a[n - 1], 2] + a[n - 2]}, a, {n, 80}]
    Table[Floor[(n + 2)/3] + (1 + (-1)^n)/2, {n, 0, 80}] (* or *) LinearRecurrence[{0, 1, 1, 0, -1}, {1, 1, 2, 1, 3}, 80] (* Bruno Berselli, Jan 21 2016 *)
  • PARI
    a=vector(100); for(n=1, #a, if(n<3, a[n]=1, a[n]=a[n-1]%2+a[n-2])); a \\ Colin Barker, Jan 22 2016

Formula

From Bruno Berselli, Jan 21 2016: (Start)
G.f.: (1 + x + x^2 - x^3)/((1 + x)*(1 - x)^2*(1 + x + x^2)).
a(n) = a(n-2) + a(n-3) - a(n-5) for n>4.
a(n) = floor((n + 2)/3) + (1 + (-1)^n)/2. (End)
a(n) = A051274(n+2). - R. J. Mathar, May 02 2023

Extensions

Edited by Bruno Berselli, Jan 21 2016.

A063276 Dimension of the space of weight n cuspidal newforms for Gamma_1( 3 ).

Original entry on oeis.org

-1, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 1, 3, 3, 4, 2, 4, 3, 5, 3, 6, 4, 6, 3, 7, 5, 8, 4, 8, 5, 9, 5, 10, 6, 10, 5, 11, 7, 12, 6, 12, 7, 13, 7, 14, 8, 14, 7, 15, 9, 16, 8, 16, 9, 17, 9, 18, 10, 18, 9, 19, 11, 20, 10, 20, 11, 21, 11, 22, 12, 22, 11, 23, 13, 24, 12, 24
Offset: 2

Views

Author

N. J. A. Sloane, Jul 14 2001

Keywords

Crossrefs

Cf. A004523 (bisection), A051274 (bisection).

Formula

g.f.: -x^2*(x^9-x^8-x^7-2*x^4+x^3-x+1) / ((x-1)^2*(x+1)^2*(x^2-x+1)*(x^2+1)). - Colin Barker, Feb 24 2015
Showing 1-5 of 5 results.