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.

Previous Showing 31-39 of 39 results.

A212686 Number of (w,x,y,z) with all terms in {1,...,n} and 2|w-x|=n+|y-z|.

Original entry on oeis.org

0, 0, 4, 8, 24, 40, 76, 112, 176, 240, 340, 440, 584, 728, 924, 1120, 1376, 1632, 1956, 2280, 2680, 3080, 3564, 4048, 4624, 5200, 5876, 6552, 7336, 8120, 9020, 9920, 10944, 11968, 13124, 14280, 15576, 16872, 18316, 19760, 21360, 22960
Offset: 0

Views

Author

Clark Kimberling, May 25 2012

Keywords

Comments

a(n)=4*A005993(n-2) for n>=2.
For a guide to related sequences, see A211795.

Crossrefs

Cf. A211795.

Programs

  • Mathematica
    t = Compile[{{n, _Integer}}, Module[{s = 0},
    (Do[If[2 Abs[w - x] == n + Abs[y - z], s = s + 1],
    {w, 1, #}, {x, 1, #}, {y, 1, #}, {z, 1, #}] &[n]; s)]];
    Map[t[#] &, Range[0, 40]]   (* A212686 *)
    %/4  (* A005993 *)
    LinearRecurrence[{2, 1, -4, 1, 2, -1}, {0, 0, 4, 8, 24, 40}, 40]

Formula

a(n)=2*a(n-1)+a(n-2)-4*a(n-3)+a(n-4)+2*a(n-5)-a(n-6).
G.f.: (4*x^2 + 4*x^4)/(1 - 2*x - x^2 + 4*x^3 - x^4 - 2*x^5 + x^6).

A301739 The number of trees with 4 nodes labeled by positive integers, where each tree's label sum is n.

Original entry on oeis.org

2, 4, 10, 17, 30, 44, 67, 91, 126, 163, 213, 265, 333, 403, 491, 582, 693, 807, 944, 1084, 1249, 1418, 1614, 1814, 2044, 2278, 2544, 2815, 3120, 3430, 3777, 4129, 4520, 4917, 5355, 5799, 6287, 6781, 7321, 7868, 8463, 9065, 9718, 10378, 11091, 11812, 12588, 13372, 14214, 15064
Offset: 4

Views

Author

R. J. Mathar, Mar 26 2018

Keywords

Comments

Computed by the sum over the A000055(4)=2 shapes of the trees: the linear graph of the n-Butane, and the star graph of (1)-Methyl-Propane.

Examples

			a(4)=2 because there is a linear tree with all labels equal 1 and the star tree with all labels equal to 1.
		

Crossrefs

4th column of A303841.

Programs

  • Maple
    x^4*(2+2*x+2*x^2+x^3+x^4)/(1+x)^2/(x-1)^4/(1+x+x^2) ;
    taylor(%,x=0,80) ;
    gfun[seriestolist](%) ;

Formula

a(n) = A005993(n-4)+A000601(n-4).
G.f.: x^4*(2+2*x+2*x^2+x^3+x^4)/((1+x)^2*(x-1)^4*(1+x+x^2) ).

A317714 Chessboard rectangles sequence (see Comments), also A037270 interleaved with A163102.

Original entry on oeis.org

0, 0, 1, 2, 10, 18, 45, 72, 136, 200, 325, 450, 666, 882, 1225, 1568, 2080, 2592, 3321, 4050, 5050, 6050, 7381, 8712, 10440, 12168, 14365, 16562, 19306, 22050, 25425, 28800, 32896, 36992, 41905, 46818, 52650, 58482, 65341, 72200, 80200, 88200, 97461, 106722, 117370
Offset: 1

Views

Author

Ivan N. Ianakiev, Aug 05 2018

Keywords

Comments

Take a chessboard of n X n unit squares in which the a1 square is black. a(n) is the number of composite rectangles of p X q unit squares whose vertices are covered by black unit squares (1 < p <= n, 1 < q <= n).

Examples

			In a 4 X 4 chessboard there are two such rectangles (for both p = q = 3) and the coordinates of their lower left vertices are a1 and b2. Therefore, a(4) = 2.
		

Crossrefs

Programs

  • Magma
    [(5-5*(-1)^n-12*n+12*(-1)^n*n+14*n^2-6*(-1)^n*n^2-8*n^3+2*n^4)/64: n in [1..50]]; // Vincenzo Librandi, Aug 05 2018
    
  • Mathematica
    CoefficientList[Series[-((x^2 (1+4 x^2+x^4))/((-1+x)^5 (1+x)^3)),{x,0,44}],x]
    LinearRecurrence[{2, 2, -6, 0, 6, -2, -2, 1}, {0, 0, 1, 2, 10, 18, 45, 72}, 80] (* Vincenzo Librandi, Aug 06 2018 *)
  • PARI
    a(n) = sum(i = 1, n-1, floor(i/2)^3); \\ Jinyuan Wang, Aug 12 2019
  • Python
    n, a = 0, 0
    while n < 10:
        print(n,a)
    n, a = n+1, a+((n+1)//2)**3 # A.H.M. Smeets, Aug 09 2019
    

Formula

a(n) = 2*a(n-1) + 2*a(n-2) - 6*a(n-3) + 6*a(n-5) - 2*a(n-6) - 2*a(n-7) + a(n-8), with a(1)=0, a(2)=0, a(3)=1, a(4)=2, a(5)=10, a(6)=18, a(7)=45, a(8)=72.
G.f.: -(x^3*(1+ 4*x^2 + x^4))/((-1+x)^5*(1+x)^3).
a(n) = (5 - 5*(-1)^n - 12*n + 12*(-1)^n*n + 14*n^2 - 6*(-1)^n*n^2 - 8*n^3 + 2*n^4)/64.
a(n) = Sum_{i=1..n-1} floor(i/2)^3. - Ridouane Oudra, Jul 24 2019
E.g.f.: (1/64)*exp(-x)*(-5-6*x-6*x^2+exp(2*x)*(5-4*x+4*x^2+4*x^3+2*x^4)). - Stefano Spezia, Aug 14 2019
a(2*n) = A163102(n-1) and a(2*n+1) = A037270(n). - Ridouane Oudra, Mar 24 2024
Sum_{n>=3} 1/a(n) = Pi^2 - Pi*coth(Pi) - 5. - Amiram Eldar, Jul 04 2025

A089351 Number of planar partitions of n with trace 4.

Original entry on oeis.org

1, 2, 6, 14, 33, 64, 127, 228, 404, 672, 1100, 1724, 2661, 3974, 5849, 8402, 11911, 16556, 22751, 30772, 41198, 54436, 71283, 92316, 118609, 150950, 190753, 239090, 297783, 368236, 452782, 553240, 672532, 812980, 978211, 1171144, 1396235
Offset: 4

Views

Author

Wouter Meeussen and Vladeta Jovovic, Dec 26 2003

Keywords

Comments

Also number of partitions of n objects of 2 colors into 4 parts, each part containing at least one black object.

References

  • G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976 (Ch. XI, exercise 5 and Ch. XII, exercise 5).

Crossrefs

Column 4 of A089353. Cf. A000219, A005380, A005993 (trace 2), A050531 (trace 3).

Formula

G.f.: q^4*(q^12+q^10+2*q^9+4*q^8+2*q^7+4*q^6+2*q^5+4*q^4+2*q^3+q^2+1) / ((-1+q^4)^2*(-1+q^3)^2*(-1+q^2)^2*(-1+q)^2).

Extensions

Edited and extended by Christian G. Bower, Jan 08 2004

A143901 Rectangular array R by antidiagonals: R(m,n) = floor((m*n+1)/2).

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Sep 04 2008

Keywords

Comments

Old name was: R(m,n) = number of white squares.

Examples

			Northwest corner:
1 1 2 2 3 3 4 4
1 2 3 4 5 6 7 8
2 3 5 6 8 9 11 12
2 4 6 8 10 12 14 16
		

Crossrefs

Cf. A143902.
Antidiagonal sums: (1,2,6,10,19,...)=A005993.
Rows and columns: A004526, A000027, A007494, A005843, A047218 et al.

Extensions

Entry revised by N. J. A. Sloane, Jun 12 2015

A280186 Number of 3-element subsets of S = {1..n} whose sum is odd.

Original entry on oeis.org

0, 0, 0, 0, 2, 4, 10, 16, 28, 40, 60, 80, 110, 140, 182, 224, 280, 336, 408, 480, 570, 660, 770, 880, 1012, 1144, 1300, 1456, 1638, 1820, 2030, 2240, 2480, 2720, 2992, 3264, 3570, 3876, 4218, 4560, 4940, 5320, 5740, 6160, 6622, 7084, 7590, 8096, 8648, 9200
Offset: 0

Views

Author

Necip Fazil Patat, Dec 28 2016

Keywords

Comments

The same as A006584 (apart from the offset). - R. J. Mathar, Jan 15 2017
There are two cases: n is odd and n is even.
Let n be an odd integer and n > 3, the sum of 3 integers is odd when all of them are odd or one is odd and the others are even. Number of ways to choose 3 odd numbers: C((n+1)/2, 3). Number of ways to choose 2 even numbers and 1 odd: C((n-1)/2, 2)*C((n+1)/2, 1). Total number of ways: C((n+1)/2, 3) + C((n-1)/2, 2)*C((n+1)/2,1).
Let n be an even integer and n > 3. Number of ways to choose 3 odd numbers: C(n/2, 3). Number of ways to choose 2 even numbers and 1 odd: C(n/2, 2)*C(n/2, 1). Total number of ways: C(n/2, 3) + C(n/2, 2)*C(n/2, 1).
Take a chessboard of n X n unit squares in which the a1 square is black. a(n) is the number of composite squares having white unit squares on their vertices. For the number of composite squares having black unit squares on their vertices see A005993. - Ivan N. Ianakiev, Aug 19 2018

Examples

			For n = 5 then a(5) = 4. The subsets are: {1, 2, 4}, {1, 3, 5}, {2, 3, 4}, {2, 4, 5}.
		

Crossrefs

Essentially 2*A006918.
Same as A006584.

Programs

  • Mathematica
    Table[Binomial[(n + #)/2, 3] + Binomial[(n - #)/2, 2] Binomial[(n + #)/2, 1] &@ Boole@ OddQ@ n, {n, 0, 49}] (* or *)
    CoefficientList[Series[2 x^4/((1 - x)^4*(1 + x)^2), {x, 0, 49}], x] (* Michael De Vlieger, Jan 07 2017 *)
  • PARI
    concat(vector(4), Vec(2*x^4 / ((1-x)^4*(1+x)^2) + O(x^60))) \\ Colin Barker, Dec 28 2016

Formula

a(n) = C((n+1)/2, 3) + C((n-1)/2, 2)*C((n+1)/2,1) when n is odd.
a(n) = C(n/2, 3) + C(n/2, 2)*C(n/2, 1) when n is even.
From Colin Barker, Dec 28 2016: (Start)
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>5.
a(n) = n*(n - 1)*(n - 2)/12 for n even.
a(n) = (n - 1)*(n + 1)*(n - 3)/12 for n odd.
G.f.: 2*x^4 / ((1-x)^4*(1+x)^2). (End)
a(n) = ((-1)^n)*(-1+n)*(3 - 3*(-1)^n - 4*((-1)^n)*n + 2*((-1)^n)*n^2)/24. - Ivan N. Ianakiev, Aug 19 2018

Extensions

More terms from Colin Barker, Dec 28 2016

A361275 Number of 1423-avoiding even Grassmannian permutations of size n.

Original entry on oeis.org

1, 1, 1, 3, 5, 11, 17, 29, 41, 61, 81, 111, 141, 183, 225, 281, 337, 409, 481, 571, 661, 771, 881, 1013, 1145, 1301, 1457, 1639, 1821, 2031, 2241, 2481, 2721, 2993, 3265, 3571, 3877, 4219, 4561, 4941, 5321, 5741, 6161, 6623, 7085, 7591, 8097, 8649, 9201, 9801, 10401
Offset: 0

Views

Author

Juan B. Gil, Mar 10 2023

Keywords

Comments

A permutation is said to be Grassmannian if it has at most one descent. A permutation is even if it has an even number of inversions.
Avoiding any of the patterns 2314 or 3412 gives the same sequence.

Examples

			For n=4 the a(4) = 5 permutations are 1234, 1342, 2314, 3124, 3412.
		

Crossrefs

For the corresponding odd permutations, cf. A005993.

Programs

  • Maple
    seq(1 - 5*n/24 + n^3/12 - (-1)^n * n/8, n = 0 .. 100); # Robert Israel, Mar 10 2023

Formula

G.f.: -(x^5-x^4-4*x^3+2*x^2+x-1)/((x+1)^2*(x-1)^4).
a(n) = 1 - 5*n/24 + n^3/12 - (-1)^n * n/8. - Robert Israel, Mar 10 2023

A320657 a(n) is the number of non-unimodal sequences with n nonzero terms that arise as a convolution of sequences of binomial coefficients preceded by a finite number of ones.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 7, 12, 16, 24, 30, 41, 50, 65, 77, 96, 112, 136, 156, 185, 210, 245, 275, 316, 352, 400, 442, 497, 546, 609, 665, 736, 800, 880, 952, 1041, 1122, 1221, 1311, 1420, 1520, 1640, 1750, 1881, 2002, 2145, 2277, 2432, 2576, 2744, 2900, 3081, 3250, 3445, 3627, 3836, 4032
Offset: 1

Views

Author

Tricia Muldoon Brown, Oct 17 2018

Keywords

Comments

For integers x,y,p,q >= 0, set (s_i){i>=1} to be the sequence of p ones followed by the binomial coefficients C(x,j) for 0 <= j <= x followed by an infinite string of zeros, and set (t_i){i>=1} to be the sequence of q ones followed by the binomial coefficients C(y,j) for 0 <= j <= y followed by an infinite string of zeros. Then a(n) is the number of non-unimodal sequences (r_i){i>=1} where r_i = Sum{j=1..i} s_j*t_{i-j} for some(s_i) and (t_i) such that x + y + p + q + 1 = n.
Let T be a rooted tree created by identifying the root vertices of two broom graphs. a(n) is the number of trees T on n vertices whose poset of connected, vertex-induced subgraphs is not rank unimodal.

Crossrefs

Cf. A005993, A024206. Equals A005581 for n even.

Programs

  • Mathematica
    Table[If[EvenQ[n], 2*(Sum[Floor[i(i+4)/4], {i,0,(n/2)}]) - Floor[n^2/16], 2*(Sum[Floor[i(i+4)/4], {i,0,(n-1)/2}]) - Floor[(n-1)^2/16] + Floor[(n+1)(n+9)/16]], {n,0,40}]

Formula

a(n+10) = 2*(Sum_{i=1..n/2} floor(i*(i+4)/4)) - floor(n^2/16) for n even.
a(n+10) = 2*(Sum_{i=1..(n-1)/2} floor(i(i+4)/4)) - floor((n-1)^2/16) + floor((n+1)*(n+9)/16) for n odd.

A362711 a(n) is the hafnian of the 2n X 2n symmetric matrix defined by M[i, j] = min(i, j)*(2*n + 1) - i*j.

Original entry on oeis.org

1, 1, 17, 1177, 210249, 76961257, 50203153993, 53127675356625, 85252003916011889, 197131843368693693937, 631233222450168374457057
Offset: 0

Views

Author

Stefano Spezia, Apr 30 2023

Keywords

Comments

M(n-1)/n is the inverse of the Cartan matrix for SU(n): the special unitary group of degree n.
The elements sum of the matrix M(n) is A002415(n+1).
The antidiagonal sum of the matrix M(n) is A005993(n-1).
The n-th row of A107985 gives the row or column sums of the matrix M(n+1).

Examples

			a(2) = 17:
    [4, 3, 2, 1]
    [3, 6, 4, 2]
    [2, 4, 6, 3]
    [1, 2, 3, 4]
		

References

  • E. B. Dynkin, Semisimple subalgebras of semisimple Lie algebras, Am. Math. Soc. Translations, Series 2, Vol. 6, 1957.

Crossrefs

Cf. A000272, A000292 (trace), A002415, A003983, A003991, A005993, A106314 (antidiagonals), A107985, A362679 (permanent).

Programs

  • Mathematica
    M[i_, j_, n_]:=Part[Part[Table[Min[r,c](n+1)-r c, {r, n}, {c, n}], i], j]; a[n_]:=Sum[Product[M[Part[PermutationList[s, 2n], 2i-1], Part[PermutationList[s, 2n], 2i], 2n], {i, n}], {s, SymmetricGroup[2n]//GroupElements}]/(n!*2^n); Array[a, 6, 0]
  • PARI
    tm(n) = matrix(n, n, i, j, min(i, j)*(n + 1) - i*j);
    a(n) = my(m = tm(2*n), s=0); forperm([1..2*n], p, s += prod(j=1, n, m[p[2*j-1], p[2*j]]); ); s/(n!*2^n); \\ Michel Marcus, May 02 2023

Formula

Conjecture: det(M(n)) = A000272(n+1).
The conjecture is true (see proof in Links). - Stefano Spezia, May 24 2023

Extensions

a(6) from Michel Marcus, May 02 2023
a(7)-a(10) from Pontus von Brömssen, Oct 15 2023
Previous Showing 31-39 of 39 results.