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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A000574 Coefficient of x^5 in expansion of (1 + x + x^2)^n.

Original entry on oeis.org

3, 16, 51, 126, 266, 504, 882, 1452, 2277, 3432, 5005, 7098, 9828, 13328, 17748, 23256, 30039, 38304, 48279, 60214, 74382, 91080, 110630, 133380, 159705, 190008, 224721, 264306, 309256, 360096, 417384, 481712, 553707, 634032, 723387, 822510
Offset: 3

Views

Author

N. J. A. Sloane

Keywords

Comments

If Y is a 3-subset of an n-set X then, for n>=7, a(n-4) is the number of 5-subsets of X having at most one element in common with Y. - Milan Janjic, Nov 23 2007

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
  • 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).

Links

Crossrefs

Cf. A000389, A005581, A005712, A005714-A005716, A111808.
Column m=5 of (1, 3) Pascal triangle A095660.
Cf. A000217, A055998.

Programs

  • Magma
    [3*Binomial(n+2,5)-2*Binomial(n+1,5): n in [3..50]]; // Vincenzo Librandi, Jun 10 2012
  • Maple
    A000574:=-(-3+2*z)/(z-1)**6; # conjectured by Simon Plouffe in his 1992 dissertation
seq(3*binomial(n+2,5)-2*binomial(n+1,5),n=3..100); # Robert Israel, Aug 04 2015
A000574 := n -> GegenbauerC(`if`(5A000574(n)), n=3..20); # Peter Luschny, May 10 2016
  • Mathematica
    CoefficientList[Series[(3-2*x)/(1-x)^6,{x,0,40}],x] (* Vincenzo Librandi, Jun 10 2012 *)
  • PARI
    x='x+O('x^50); Vec(x^3*(3-2*x)/(1-x)^6) \\ G. C. Greubel, Nov 22 2017

Formula

G.f.: x^3*(3-2*x)/(1-x)^6.
a(n) = 3*binomial(n+2,5) - 2*binomial(n+1,5).
a(n) = A111808(n,5) for n>4. - Reinhard Zumkeller, Aug 17 2005
a(n) = binomial(n+1, 4)*(n+12)/5 = 3*b(n-3)-2*b(n-4), with b(n)=binomial(n+5, 5); cf. A000389.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). - Vincenzo Librandi, Jun 10 2012
a(n) = 3*binomial(n, 3) + 4*binomial(n, 4) + binomial(n, 5). - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012
a(n) = GegenbauerC(N, -n, -1/2) where N = 5 if 5Peter Luschny, May 10 2016
a(n) = Sum_{i=1..n-1} A000217(i)*A055998(n-1-i). - Bruno Berselli, Mar 05 2018
E.g.f.: exp(x)*x^3*(60 + 20*x + x^2)/120. - Stefano Spezia, Jul 09 2023

Extensions

More terms from Vladeta Jovovic, Oct 02 2000

A241765 a(n) = n*(n + 1)*(n + 2)*(3*n + 17)/24.

Original entry on oeis.org

0, 5, 23, 65, 145, 280, 490, 798, 1230, 1815, 2585, 3575, 4823, 6370, 8260, 10540, 13260, 16473, 20235, 24605, 29645, 35420, 41998, 49450, 57850, 67275, 77805, 89523, 102515, 116870, 132680, 150040, 169048, 189805, 212415, 236985, 263625, 292448
Offset: 0

Views

Author

Bruno Berselli, Apr 28 2014

Keywords

Comments

Equivalently, Sum_{i=0..n} (i+4)*A000217(i).
Sequences of the type Sum_{i=0..n} (i+k)*A000217(i):
k = 0, A001296: 0, 1, 7, 25, 65, 140, 266, 462, ...
k = 1, A000914: 0, 2, 11, 35, 85, 175, 322, 546, ...
k = 2, A050534: 0, 3, 15, 45, 105, 210, 378, 630, ... (deleting two 0)
k = 3, A215862: 0, 4, 19, 55, 125, 245, 434, 714, ...
k = 4, a(n): 0, 5, 23, 65, 145, 280, 490, 798, ...
k = 5, A239568: 0, 6, 27, 75, 165, 315, 546, 882, ...
Antidiagonal sums (without 0) give A034263: 1, 9, 39, 119, 294, ...
Diagonal: 1, 11, 45, 125, 280, 546, ... is A051740.
Also: k = -1 gives A050534 deleting a 0; k = -2 gives 0 followed by A059302.
After 0, partial sums of A212343 and third column of A118788.
This sequence is even related to A005286 by a(n) = n*A005286(n) - Sum_{i=0..n-1} A005286(i).

Examples

			a(7) = 4*0 + 5*1 + 6*3 + 7*6 + 8*10 + 9*15 + 10*21 + 11*28 = 798.

Links

Crossrefs

Cf. A000217, A005286, A118788, A212343, A227342.
Cf. similar sequences A000914, A001296, A050534, A059302, A215862, A239568 (see table in Comments lines).

Programs

  • Magma
    /* By first comment: */ k:=4; A000217:=func; [&+[(i+k)*A000217(i): i in [0..n]]: n in [0..40]];
  • Maple
    A241765:=n->n*(n + 1)*(n + 2)*(3*n + 17)/24; seq(A241765(n), n=0..40); # Wesley Ivan Hurt, May 09 2014
  • Mathematica
    Table[n (n + 1) (n + 2) (3 n + 17)/24, {n, 0, 40}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 5, 23, 65, 145}, 40]
CoefficientList[Series[x (5 - 2 x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)
  • Maxima
    makelist(coeff(taylor(x*(5-2*x)/(1-x)^5, x, 0, n), x, n), n, 0, 40);
  • PARI
    a(n)=n*(n+1)*(n+2)*(3*n+17)/24 \\ Charles R Greathouse IV, Oct 07 2015
  • PARI
    x='x+O('x^99); concat(0, Vec(x*(5-2*x)/(1-x)^5)) \\ Altug Alkan, Apr 10 2016
  • Sage
    [n*(n+1)*(n+2)*(3*n+17)/24 for n in (0..40)]

Formula

G.f.: x*(5 - 2*x)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A227342(A055998(n+1)).
a(n) = Sum_{j=0..n+2} (-1)^(n-j)*binomial(-j,-n-2)*S1(j,n), S1 Stirling cycle numbers A132393. - Peter Luschny, Apr 10 2016

A027379 Expansion of (1+x^2-x^3)/(1-x)^3.

Original entry on oeis.org

1, 3, 7, 12, 18, 25, 33, 42, 52, 63, 75, 88, 102, 117, 133, 150, 168, 187, 207, 228, 250, 273, 297, 322, 348, 375, 403, 432, 462, 493, 525, 558, 592, 627, 663, 700, 738, 777, 817, 858, 900, 943, 987, 1032, 1078
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Essentially the triangular numbers (A000217) minus 3.
Also essentially the same as A055998.
Cf. A000217, A034856, A046691.

Programs

Formula

a(n) = n*(n+5)/2 = A000217(n+2) - 3 for n>=1. - Emeric Deutsch, Mar 01 2004
a(n) = n + a(n-1) + 2, n>1. - Vincenzo Librandi, Dec 06 2009
E.g.f.: 1 + x*(x + 6)*exp(x)/2. - G. C. Greubel, May 14 2017

A060092 Triangle T(n,k) of k-block ordered bicoverings of an unlabeled n-set, n >= 2, k = 3..n+floor(n/2).

Original entry on oeis.org

3, 7, 16, 12, 63, 125, 90, 18, 162, 722, 1716, 1680, 25, 341, 2565, 11350, 27342, 29960, 7560, 33, 636, 7180, 49860, 208302, 503000, 631512, 302400, 42, 1092, 17335, 173745, 1099602, 4389875, 10762299, 14975730, 9632700, 1247400
Offset: 2

Views

Author

Vladeta Jovovic, Feb 26 2001

Keywords

Comments

All columns are polynomials of order binomial(k, 2). - Andrew Howroyd, Jan 30 2020

Examples

			[3],
[7, 16],
[12, 63, 125, 90],
[18, 162, 722, 1716, 1680],
[25, 341, 2565, 11350, 27342, 29960, 7560],
[33, 636, 7180, 49860, 208302, 503000, 631512, 302400],
[42, 1092, 17335, 173745, 1099602, 4389875, 10762299, 14975730, 9632700, 1247400], ...
There are 23=7+16 ordered bicoverings of an unlabeled 3-set: 7 3-block bicoverings and 16 4-block bicoverings, cf. A060090.

Links

Crossrefs

Row sums are A060090.
Columns k=3..7 are A055998(n-1), A060091, A060093, A060094, A060095.
Cf. A059443, A060052, A060487, A060492, A094573, A331571.

Programs

  • PARI
    \\ gives g.f. of k-th column.
ColGf(k) = k!*polcoef(exp(-x - x^2*y/(2*(1-y)) + O(x*x^k))*sum(j=0, k, 1/(1-y)^binomial(j, 2)*x^j/j!), k) \\ Andrew Howroyd, Jan 30 2020
  • PARI
    T(n)={my(m=(3*n\2), y='y + O('y^(n+1))); my(g=serlaplace(exp(-x - x^2*y/(2*(1-y)) + O(x*x^m))*sum(k=0, m, 1/(1-y)^binomial(k, 2)*x^k/k!))); Mat([Col(p/y^2, -n) | p<-Vec(g)[2..m+1]])}
{ my(A=T(8)); for(n=2, matsize(A)[1], print(A[n, 3..3*n\2])) } \\ Andrew Howroyd, Jan 30 2020

Formula

E.g.f. for k-block ordered bicoverings of an unlabeled n-set is exp(-x-x^2/2*y/(1-y))*Sum_{k=0..inf} 1/(1-y)^binomial(k, 2)*x^k/k!.

A209268 Inverse permutation A054582.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 10, 7, 15, 9, 21, 8, 28, 14, 36, 11, 45, 20, 55, 13, 66, 27, 78, 12, 91, 35, 105, 19, 120, 44, 136, 16, 153, 54, 171, 26, 190, 65, 210, 18, 231, 77, 253, 34, 276, 90, 300, 17, 325, 104, 351, 43, 378, 119, 406, 25, 435, 135, 465, 53, 496, 152
Offset: 1

Views

Author

Boris Putievskiy, Jan 15 2013

Keywords

Comments

Permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.

Examples

			The start of the sequence for n = 1..32 as table, distributed by exponent of highest power of 2 dividing n:
   |   Exponent of highest power of 2 dividing n
n  |--------------------------------------------------
   |    0      1      2       3      4         5    ...
------------------------------------------------------
1  |....1
2  |...........2
3  |....3
4  |..................4
5  |....6
6  |...........5
7  |...10
8  |..........................7
9  |...15
10 |...........9
11 |...21
12 |..................8
13 |...28
14 |..........14
15 |...36
16 |................................11
17 |...45
18 |..........20
19 |...55
20 |.................13
21 |...66
22 |..........27
23 |...78
24 |................................12
25 |...91
26 |..........35
27 |..105
28 |.................19
29 |..120
30 |..........44
31 |..136
32 |.........................................16
. . .
Let r_c be number row inside the column number c.
r_c = (n+2^c)/2^(c+1).
The column number 0 contains numbers r_0*(r_0+1)/2,     A000217,
The column number 1 contains numbers r_1*(r_1+3)/2,     A000096,
The column number 2 contains numbers r_2*(r_2+5)/2 + 1, A034856,
The column number 3 contains numbers r_3*(r_3+7)/2 + 3, A055998,
The column number 4 contains numbers r_4*(r_4+9)/2 + 6, A046691.

Links

Crossrefs

Cf. A054582, A003602, A007814, A000217, A000096, A034856, A055998, A046691, A014132.

Programs

  • Mathematica
    a[n_] := (v = IntegerExponent[n, 2]; (1/2)*(((1/2)*(n/2^v + 1) + v)^2 + (1/2)*(n/2^v + 1) - v)); Table[a[n], {n, 1, 55}] (* Jean-François Alcover, Jan 15 2013, from 1st formula *)
  • Python
    f = open("result.csv", "w")
def A007814(n):
### author        Richard J. Mathar 2010-09-06 (Start)
### http://oeis.org/wiki/User:R._J._Mathar/oeisPy/oeisPy/oeis_bulk.py
        a = 0
        nshft = n
        while (nshft %2 == 0):
                a += 1
                nshft >>= 1
        return a
###(End)
for  n in range(1,10001):
     x = A007814(n)
     y = (n+2**x)/2**(x+1)
     m = ((x+y)**2-x+y)/2
     f.write('%d;%d;%d;%d;\n' % (n, x, y, m))
f.close()

Formula

a(n) = (((A003602)+A007814(n))^2 - A007814(n) + A003602(n))/2.
a(n) = ((x+y)^2-x+y)/2, where x = max {k: 2^k | n}, y = (n+2^x)/2^(x+1).

A090452 Scaled array A078740 ((3,2)-Stirling2).

Original entry on oeis.org

1, 1, 3, 2, 1, 7, 16, 15, 5, 1, 12, 51, 105, 114, 63, 14, 1, 18, 118, 396, 771, 910, 644, 252, 42, 1, 25, 230, 1110, 3235, 6083, 7580, 6240, 3270, 990, 132, 1, 33, 402, 2600, 10365, 27483, 50464, 65331, 59625, 37620, 15642, 3861, 429, 1, 42, 651, 5390, 27825, 97188
Offset: 1

Views

Author

Wolfdieter Lang, Dec 23 2003

Keywords

Comments

This scaled Stirling2 array will be called s2_{3,2}(n,m).
The sequence of row lengths is [1,3,5,7,...]=A005408(n-1).
The generating function for the sequence from column no. m is G(m,x)=(x^ceiling(m/2))*P(m,x)/(1-x)^(2*m-3) with the row polynomials of array A091029(m,k).
The generating functions of the column sequences obey the hypergeometric differential-difference eq.:x*(1-x)*G''(m,x) + 2*(1-m*x)*G'(m,x) - m*(m-1)*G(m,x) = 2*m*x*G'(m-1,x) + 2*m*(m-1)*G(m-1,x) + m*(m-1)*G(m-2,x), m>=3; with G(2,x)=x/(1-x) and G(1,x)=0. The primes denote differentiation w.r.t. x.

Examples

			Triangle begins:
  [1];
  [1,3,2];
  [1,7,16,15,5];
  [1,12,51,105,114,63,14];
  ...

Links

Crossrefs

a(n, 2*n)=A000108(n) (Catalan), n>=1, a(n, 2*n-1)=3*A002054(n-1), n>=2, a(n, 2*n-2)=A091031(n), n>=2.
The column sequences (without leading zeros) are: A000012 (powers of 1), A055998, A090453-4, A091026-7, etc.
Cf. A090442 (row sums). The alternating row sums are 0 except for row n=1 which gives 1.

Programs

Formula

a(n, m) = (m!/((n+1)!*n!))*A078740(n, m), n>=1, 2<= m <=2*n.
Recursion: a(n, m) = ((n+m-1)*(n+m-2)*a(n-1, m)+2*(n+m-2)*m*a(n-1, m-1)+m*(m-1)*a(n-1, m-2))/((n+1)*n), n>=2, 2<=m<=2*n, a(1, 2)=1, a(n, 0) := 0, a(n, 1) := 0 (from the recursion of array A078740).

A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

Original entry on oeis.org

0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601
Offset: 0

Views

Author

Paul Curtz, Nov 18 2009

Keywords

Comments

From Paul Curtz, Mar 26 2011: (Start)
Successive A026741(n) * A026741(n+p):
p=0: 0, 1, 1, 9, 4, 25, 9, a(n),
p=1: 0, 1, 3, 6, 10, 15, 21, A000217,
p=2: 0, 3, 2, 15, 6, 35, 12, A142705,
p=3: 0, 2, 5, 9, 14, 20, 27, A000096,
p=4: 0, 5, 3, 21, 8, 45, 15, A171621,
p=5: 0, 3, 7, 12, 18, 25, 33, A055998,
p=6: 0, 7, 4, 27, 10, 55, 18,
p=7: 0, 4, 9, 15, 22, 30, 39, A055999,
p=8: 0, 9, 5, 33, 12, 65, 21, (see A061041),
p=9: 0, 5, 11, 18, 26, 35, 45, A056000. (End)
The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016
Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

Links

Crossrefs

Cf. A040001, A168068, A181318.

Programs

  • Magma
    I:=[0,1,1,9,4,25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016
  • Maple
    a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016
  • Mathematica
    LinearRecurrence[{0,3,0,-3,0,1},{0,1,1,9,4,25},60] (* Harvey P. Dale, May 14 2011 *)
f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *)
CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x,0,60}], x] (* Vincenzo Librandi, Jul 10 2016 *)
  • PARI
    concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016
  • PARI
    a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018
  • Sage
    (x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019

Formula

From R. J. Mathar, Jan 22 2011: (Start)
G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(n) = n^2*(5 - 3*(-1)^n)/8. (End)
a(n) = A026741(n)^2.
a(2*n) = A000290(n); a(2*n+1) = A016754(n).
a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011
4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011
a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n.
O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3.
Cf. A181318. (End)
From Werner Schulte, Aug 30 2020: (Start)
Multiplicative with a(2^e) = 2^(2*e-2) for e > 0, and a(p^e) = p^(2*e) for prime p > 2.
Dirichlet g.f.: zeta(s-2) * (1 - 3/2^s).
Dirichlet convolution with A259346 equals A000290.
Sum_{n>0} 1/a(n) = Pi^2 * 7 / 24. (End)
Sum_{k=1..n} a(k) ~ (5/24) * n^3. - Amiram Eldar, Nov 28 2022

A067724 a(n) = 5*n^2 + 10*n.

Original entry on oeis.org

15, 40, 75, 120, 175, 240, 315, 400, 495, 600, 715, 840, 975, 1120, 1275, 1440, 1615, 1800, 1995, 2200, 2415, 2640, 2875, 3120, 3375, 3640, 3915, 4200, 4495, 4800, 5115, 5440, 5775, 6120, 6475, 6840, 7215, 7600, 7995, 8400, 8815, 9240, 9675
Offset: 1

Views

Author

Robert G. Wilson v, Feb 05 2002

Keywords

Comments

Positive numbers m such that 5*(5 + m) is a perfect square.

Links

Crossrefs

Cf. numbers k such that k*(k + m) is a perfect square: A028560 (k=9), A067728 (k=8), A067727 (k=7), A067726 (k=6), A028347 (k=4), A067725 (k=3), A054000 (k=2), A067998 (k=1).
Cf. A055998.

Programs

  • Magma
    [5*n*(n+2): n in [1..50]]; // Vincenzo Librandi, Jul 08 2012
  • Mathematica
    Select[Range[10000], IntegerQ[ Sqrt[5 (5 + # )]] &]
CoefficientList[Series[5 (3 - x)/(1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 08 2012 *)
Table[5n^2+10n,{n,60}] (* or *) LinearRecurrence[{3,-3,1},{15,40,75},60] (* Harvey P. Dale, May 22 2018 *)
  • PARI
    a(n)=5*n*(n+2) \\ Charles R Greathouse IV, Dec 07 2011

Formula

From Vincenzo Librandi, Jul 08 2012: (Start)
G.f.: 5*x*(3 - x)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
a(n) = A055998(3*n) + A055998(n). - Bruno Berselli, Sep 23 2016
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 3/20.
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/20. (End)
E.g.f.: 5*exp(x)*x*(3 + x). - Stefano Spezia, Oct 01 2023

A350116 Number of ways to partition the set of vertices of a convex {n+8}-gon into 3 non-intersecting polygons.

Original entry on oeis.org

0, 12, 45, 110, 220, 390, 637, 980, 1440, 2040, 2805, 3762, 4940, 6370, 8085, 10120, 12512, 15300, 18525, 22230, 26460, 31262, 36685, 42780, 49600, 57200, 65637, 74970, 85260, 96570, 108965, 122512, 137280, 153340, 170765, 189630, 210012, 231990, 255645, 281060, 308320
Offset: 0

Views

Author

Janaka Rodrigo, Dec 21 2021

Keywords

Comments

Equivalently, the number of noncrossing set partitions of an {n+8}-set into 3 blocks with 3 or more elements in each block.

Examples

			The a(1) = 12 solutions are:
   {123}{456}{789}, {234}{567}{891}, {345}{678}{912},
   {156}{234}{567}, {267}{345}{891}, {378}{456}{912},
   {489}{567}{123}, {591}{678}{234}, {612}{789}{345},
   {723}{891}{456}, {834}{912}{567}, {945}{123}{678}.
In the above, the numbers can be considered to be the partition of a 9-set into 3 blocks or the partition of the vertices of a convex 9-gon into 3 triangles (with the vertices labeled 1..9 in order).
a(2) = 45 corresponding to the number of ways to partition the vertices of a 10-gon into two triangles and one quadrilateral.

Links

Crossrefs

Column k=3 of A350248.
The case of any number of parts for an n-gon is A114997.
The case of exactly 2 parts for a {n+5}-gon is A055998.

Programs

  • Mathematica
    a[n_] := n*(n + 1)*(n + 7)*(n + 8)/12; Array[a, 40, 0] (* Amiram Eldar, Dec 21 2021 *)

Formula

a(n) = n*(n+1)*(n+7)*(n+8)/12.
G.f.: -x*(12-15*x+5*x^2)/(x-1)^5 . - R. J. Mathar, Aug 03 2022

A060488 Number of 4-block ordered tricoverings of an unlabeled n-set.

Original entry on oeis.org

4, 13, 28, 50, 80, 119, 168, 228, 300, 385, 484, 598, 728, 875, 1040, 1224, 1428, 1653, 1900, 2170, 2464, 2783, 3128, 3500, 3900, 4329, 4788, 5278, 5800, 6355, 6944, 7568, 8228, 8925, 9660, 10434, 11248, 12103, 13000, 13940, 14924, 15953, 17028, 18150, 19320
Offset: 3

Views

Author

Vladeta Jovovic, Mar 20 2001

Keywords

Comments

A covering of a set is a tricovering if every element of the set is covered by exactly three blocks of the covering.
If Y is a 4-subset of an n-set X then, for n>=6, a(n-3) is the number of 3-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007
Also the number of balls in a triangular pyramid of which all balls located on the edges have been removed such that the remaining pyramid's edges each consist of two adjacent balls. The layers of pyramids of this form start (from the top) 3, 7, 12, 18, 25, 33,... (A055998) with one smaller additional layer 1, 3, 6, 10, 15, 21,... (A000217) at the bottom. Thus, a(n) = A000217(n) + Sum_{k=1..n} A055998(k). Example: a(4) = (3+7+12+18)+10 = 50. - K. G. Stier, Dec 12 2012

Links

Crossrefs

Essentially the same as A026054. - Vladeta Jovovic, Jun 15 2006
Column k=4 of A060492.
Cf. A060091, A060491.
Fourth column (m=3) of (1, 4)-Pascal triangle A095666.

Programs

Formula

a(n) = binomial(n+3, 3) - 6*binomial(n+1, 1) + 8*binomial(n, 0) - 3*binomial(n-1, -1).
G.f.: -y^3*(-4+3*y)/(-1+y)^4.
E.g.f. for ordered k-block tricoverings of an unlabeled n-set is exp(-x+x^2/2+x^3/3*y/(1-y)) * sum(k>=0, 1/(1-y)^binomial(k, 3)*exp(-x^2/2*1/(1-y)^n)*x^k/k! ).
a(n) = (n+9)*binomial(n-1, 2)/3.
a(n) = (n-2)*(n-1)*(n+9)/6. - Zak Seidov, Jun 15 2006
a(3)=4, a(4)=13, a(5)=28, a(6)=50, a(n) = 4*a(n-1)-6*a(n-2)+ 4*a(n-3)- a(n-4). - Harvey P. Dale, Jul 21 2012
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