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 1-8 of 8 results.

A047863 Number of labeled graphs with 2-colored nodes where black nodes are only connected to white nodes and vice versa.

Original entry on oeis.org

1, 2, 6, 26, 162, 1442, 18306, 330626, 8488962, 309465602, 16011372546, 1174870185986, 122233833963522, 18023122242478082, 3765668654914699266, 1114515608405262434306, 467221312005126294077442, 277362415313453291571118082, 233150477220213193598856331266
Offset: 0

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Author

Keywords

Comments

Row sums of A111636. - Peter Bala, Sep 30 2012
Column 2 of Table 2 in Read. - Peter Bala, Apr 11 2013
It appears that 5 does not divide a(n), that a(n) is even for n>0, that 3 divides a(2n) for n>0, that 7 divides a(6n+5), and that 13 divides a(12n+3). - Ralf Stephan, May 18 2013

Examples

			For n=2, {1,2 black, not connected}, {1,2 white, not connected}, {1 black, 2 white, not connected}, {1 black, 2 white, connected}, {1 white, 2 black, not connected}, {1 white, 2 black, connected}.
G.f. = 1 + 2*x + 6*x^2 + 26*x^3 + 162*x^4 + 1442*x^5 + 18306*x^6 + ...
		

References

  • H. S. Wilf, Generatingfunctionology, Academic Press, NY, 1990, p. 79, Eq. 3.11.2.

Crossrefs

Column k=2 of A322280.
Cf. A135079 (variant).

Programs

  • Magma
    A047863:= func< n | (&+[Binomial(n,k)*2^(k*(n-k)): k in [0..n]]) >;
    [A047863(n): n in [0..40]]; // G. C. Greubel, Nov 03 2024
    
  • Mathematica
    Table[Sum[Binomial[n,k]2^(k(n-k)),{k,0,n}],{n,0,20}] (* Harvey P. Dale, May 09 2012 *)
    nmax = 20; CoefficientList[Series[Sum[E^(2^k*x)*x^k/k!, {k, 0, nmax}], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Jun 05 2019 *)
  • PARI
    {a(n)=n!*polcoeff(sum(k=0,n,exp(2^k*x +x*O(x^n))*x^k/k!),n)} \\ Paul D. Hanna, Nov 27 2007
    
  • PARI
    {a(n)=polcoeff(sum(k=0, n, x^k/(1-2^k*x +x*O(x^n))^(k+1)), n)} \\ Paul D. Hanna, Mar 08 2008
    
  • PARI
    N=66; x='x+O('x^N); egf = sum(n=0, N, exp(2^n*x)*x^n/n!);
    Vec(serlaplace(egf))  \\ Joerg Arndt, May 04 2013
    
  • Python
    from sympy import binomial
    def a(n): return sum([binomial(n, k)*2**(k*(n - k)) for k in range(n + 1)]) # Indranil Ghosh, Jun 03 2017
    
  • SageMath
    def A047863(n): return sum(binomial(n,k)*2^(k*(n-k)) for k in range(n+1))
    [A047863(n) for n in range(41)] # G. C. Greubel, Nov 03 2024

Formula

a(n) = Sum_{k=0..n} binomial(n, k)*2^(k*(n-k)).
a(n) = 4 * A000683(n) + 2. - Vladeta Jovovic, Feb 02 2000
E.g.f.: Sum_{n>=0} exp(2^n*x)*x^n/n!. - Paul D. Hanna, Nov 27 2007
O.g.f.: Sum_{n>=0} x^n/(1 - 2^n*x)^(n+1). - Paul D. Hanna, Mar 08 2008
From Peter Bala, Apr 11 2013: (Start)
Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)) = 1 + x + x^2/(2!*2) + x^3/(3!*2^3) + .... Then a generating function is E(x)^2 = 1 + 2*x + 6*x^2/(2!*2) + 26*x^3/(3!*2^3) + .... In general, E(x)^k, k = 1, 2, ..., is a generating function for labeled k-colored graphs (see Stanley). For other examples see A191371 (k = 3) and A223887 (k = 4).
If A(x) = 1 + 2*x + 6*x^2/2! + 26*x^3/3! + ... denotes the e.g.f. for this sequence then sqrt(A(x)) = 1 + x + 2*x^2/2! + 7*x^3/3! + ... is the e.g.f. for A047864, which counts labeled 2-colorable graphs. (End)
a(n) ~ c * 2^(n^2/4+n+1/2)/sqrt(Pi*n), where c = Sum_{k = -infinity..infinity} 2^(-k^2) = EllipticTheta[3, 0, 1/2] = 2.128936827211877... if n is even and c = Sum_{k = -infinity..infinity} 2^(-(k+1/2)^2) = EllipticTheta[2, 0, 1/2] = 2.12893125051302... if n is odd. - Vaclav Kotesovec, Jun 24 2013

Extensions

Better description from Christian G. Bower, Dec 15 1999

A191371 Number of simple labeled graphs with (at most) 3-colored nodes such that no edge connects two nodes of the same color.

Original entry on oeis.org

1, 3, 15, 123, 1635, 35043, 1206915, 66622083, 5884188675, 830476531203, 187106645932035, 67241729173555203, 38521811621470420995, 35161184767296890265603, 51112793797110111859802115, 118291368253025368001553530883, 435713124846749574718274002747395, 2553666761436949125065383836043837443
Offset: 0

Views

Author

Geoffrey Critzer, Jun 01 2011

Keywords

Comments

Cf. A213442, which counts colorings of labeled graphs on n vertices using exactly 3 colors. For 3-colorable labeled graphs on n vertices see A076315. - Peter Bala, Apr 12 2013

Examples

			a(2) = 15: There are two labeled 3-colorable graphs on 2 nodes, namely
A)  1    2   B)  1    2
    o    o       o----o
Using 3 colors there are 9 ways to color the graph of type A and 3*2 = 6 ways to color the graph of type B so that adjacent vertices do not share the same color. Thus there are in total 15 labeled 3-colored graphs on 2 vertices.
		

References

  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, 1973, 16-18.

Crossrefs

Column k=3 of A322280.

Programs

  • Mathematica
    f[{k_,r_,m_}]:= Binomial[m+r+k,k] Binomial[m+r,r] 2^ (k r + k m + r m); Table[Total[Map[f,Compositions[n,3]]],{n,0,20}]
  • PARI
    seq(n)={my(p=(sum(j=0, n, x^j/(j!*2^binomial(j, 2))) + O(x*x^n))^3); Vecrev(sum(j=0, n, j!*2^binomial(j,2)*polcoef(p,j)*x^j))} \\ Andrew Howroyd, Dec 03 2018
  • Python
    from sympy import binomial
    def a047863(n): return sum([binomial(n, k)*2**(k*(n - k)) for k in range(n + 1)])
    def a(n): return sum([binomial(n, k)*2**(k*(n - k))*a047863(k) for k in range(n + 1)]) # Indranil Ghosh, Jun 03 2017
    

Formula

a(n) = Sum(C(n,k)*C(n-k,r)*2^(k*r+k*m+r*m)) where the sum is taken over all nonnegative solutions to k + r + m = n.
From Peter Bala, Apr 12 2013: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*2^(k*(n-k))*A047863(k).
a(n) = 3*A000685(n) for n >= 1.
Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)). Then a generating function for this sequence is E(x)^3 = sum {n >= 0} a(n)*x^n/(n!*2^C(n,2)) = 1 + 3*x + 15*x^2/(2!*2) + 123*x^3/(3!*2^3) + 1635*x^4/(4!*2^6) + ....
In general, for k = 1, 2, ..., E(x)^k is a generating function for labeled k-colored graphs (see Read). For other examples see A047863 (k = 2) and A223887 (k = 4). (End)

A322280 Array read by antidiagonals: T(n,k) is the number of graphs on n labeled nodes, each node being colored with one of k colors, where no edge connects two nodes of the same color.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 6, 1, 0, 1, 4, 15, 26, 1, 0, 1, 5, 28, 123, 162, 1, 0, 1, 6, 45, 340, 1635, 1442, 1, 0, 1, 7, 66, 725, 7108, 35043, 18306, 1, 0, 1, 8, 91, 1326, 20805, 254404, 1206915, 330626, 1, 0, 1, 9, 120, 2191, 48486, 1058885, 15531268, 66622083, 8488962, 1, 0
Offset: 0

Views

Author

Andrew Howroyd, Dec 01 2018

Keywords

Comments

Not all colors need to be used.

Examples

			Array begins:
===============================================================
n\k| 0 1      2        3          4           5           6
---+-----------------------------------------------------------
0  | 1 1      1        1          1           1           1 ...
1  | 0 1      2        3          4           5           6 ...
2  | 0 1      6       15         28          45          66 ...
3  | 0 1     26      123        340         725        1326 ...
4  | 0 1    162     1635       7108       20805       48486 ...
5  | 0 1   1442    35043     254404     1058885     3216486 ...
6  | 0 1  18306  1206915   15531268    95261445   386056326 ...
7  | 0 1 330626 66622083 1613235460 15110296325 83645197446 ...
...
		

Crossrefs

Columns k=0..4 are A000007, A000012, A047863, A191371, A223887.
Main diagonal gives A372920.

Programs

  • Mathematica
    nmax = 10;
    T[n_, k_] := n!*2^Binomial[n, 2]*SeriesCoefficient[Sum[ x^i/(i!* 2^Binomial[i, 2]), {i, 0, nmax}]^k, {x, 0, n}];
    Table[T[n - k, k], {n, 0, nmax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 23 2019 *)
  • PARI
    M(n)={
      my(p=sum(j=0, n, x^j/(j!*2^binomial(j, 2))) + O(x*x^n));
      my(q=sum(j=0, n, x^j*j!*2^binomial(j, 2)) + O(x*x^n));
      matconcat([1, Mat(vector(n, k, Col(serconvol(q, p^k))))]);
    }
    my(T=M(7)); for(n=1, #T, print(T[n,]))

Formula

T(n,k) = n!*2^binomial(n,2) * [x^n](Sum_{i>=0} x^i/(i!*2^binomial(i,2)))^k.
T(n,k) = Sum_{j=0..k} binomial(k,j)*j!*A058843(n,j).

A213441 Number of 2-colored graphs on n labeled nodes.

Original entry on oeis.org

0, 4, 24, 160, 1440, 18304, 330624, 8488960, 309465600, 16011372544, 1174870185984, 122233833963520, 18023122242478080, 3765668654914699264, 1114515608405262434304, 467221312005126294077440, 277362415313453291571118080, 233150477220213193598856331264, 277465561009648882424932436803584, 467466753447825987214906927108587520
Offset: 1

Views

Author

N. J. A. Sloane, Jun 11 2012

Keywords

Comments

From Peter Bala, Apr 10 2013: (Start)
A coloring of a simple graph is a choice of color for each graph vertex such that no two vertices sharing the same edge have the same color. This sequence counts only those colorings of labeled graphs on n vertices that use exactly two colors.
Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)) = 1 + x + x^2/(2!*2) + x^3/(3!*2^3) + .... Then Read has shown that (E(x) - 1)^k is a generating function for counting labeled graphs colored using precisely k colors. This is the case k = 2. For other cases see A213442 (k = 3) and A224068 (k = 4).
A coloring of a graph G that uses k or fewer colors is called a k-coloring of G. The graph G is k-colored if a k-coloring of G exists.
Then E(x)^k is a generating function for the enumeration of labeled k-colored graphs on n vertices (see Stanley). For cases see A047863 (k = 2), A191371 (k = 3) and A223887 (k = 4). (End)

Examples

			a(2) = 4: Denote the vertex coloring by o and *. The 4 labeled graphs on 2 vertices that can be colored using exactly two colors are
....1....2............1....2
....o....*............*----o
...........................
....1....2............1....2
....*....o............o----*
- _Peter Bala_, Apr 10 2013
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

Crossrefs

Programs

  • Maple
    F2:=n->add(binomial(n,r)*2^(r*(n-r)), r=1..n-1);
    [seq(F2(n),n=1..20)];
  • Mathematica
    nn=20;a[x_]:=Sum[x^n/(n!*(2^(n^2/2))),{n,0,nn}];Drop[Table[n!*(2^(n^2/2)),{n,0,nn}]CoefficientList[Series[(a[x]-1)^2,{x,0,nn}],x],1] (* Geoffrey Critzer, Aug 05 2014 *)
  • PARI
    a(n) = sum(k=1,n-1, binomial(n,k)*2^(k*(n-k)) ); /* Joerg Arndt, Apr 10 2013 */

Formula

From Peter Bala, Apr 10 2013: (Start)
a(n) = Sum_{k = 1..n-1} binomial(n,k)*2^(k*(n-k)). a(n) = A047863(n) - 2.
Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)) = 1 + x + x^2/(2!*2) + x^3/(3!*2^3) + x^4/(4!*2^6) + .... Then a generating function is (E(x) - 1)^2 = 4*x^2/(2!*2) + 24*x^3/(3!*2^3) + 160*x^4/(4!*2^6) + ....
Sequence is 1/2*(column 2) from A058843. (End)
E.g.f.: Sum_{n>=1}(exp(2^n*x)-1)*x^n/n!. - Geoffrey Critzer, Aug 11 2014

A224068 Number of labeled graphs on n vertices that can be colored using exactly 4 colors.

Original entry on oeis.org

0, 0, 0, 1536, 122880, 10813440, 1348730880, 261070258176, 81787921367040, 42364317235937280, 36686317873382031360, 53408511909378681470976, 131046345314766385022238720, 542471805171085602081503969280, 3789399960645715708906355231293440
Offset: 1

Views

Author

Peter Bala, Apr 10 2013

Keywords

Comments

A223887 counts labeled 4-colored graphs on n vertices, that is, colorings of labeled graphs on n vertices using 4 or fewer colors.
This sequence differs in that it counts only those colorings of labeled graphs on n vertices that use exactly 4 colors. Cf. A213441 and A213442.

Crossrefs

Programs

  • Mathematica
    nn=20;e[x_]:=Sum[x^n/(n!*2^Binomial[n,2]),{n,0,nn}];Table[n!*2^Binomial[n,2],{n,0,nn}]CoefficientList[Series[(e[x]-1)^4,{x,0,nn}],x] (* Geoffrey Critzer, Aug 11 2014 *)
  • PARI
    N=16;  x='x+O('x^N);
    E=sum(n=0, N, x^n/(n!*2^binomial(n,2)) );
    tgf=(E-1)^4;
    v=concat([0,0,0], Vec(tgf));
    v=vector(#v, n, v[n] * n! * 2^(n*(n-1)/2) )
    /* Joerg Arndt, Apr 10 2013 */

Formula

a(n) = Sum_{k=2..n-2} C(n,k)*2^(k*(n-k))*A213441(k)*A213441(n-k).
Let E(x) = Sum_{n>=0} x^n/(n!*2^C(n,2)) = 1 + x + x^2/(2!*2) + x^3/(3!*2^3) + x^4/(4!*2^6) + .... Then a generating function is (E(x) - 1)^4 = 1536*x^4/(4!*2^6) + 122880*x^5/(5!*2^10) + 10813440*x^6/(6!*2^15) + ... + a(n)*x^n/(n!*2^(n*(n-1)/2)) + ... (see Read).

A000685 Number of 3-colored labeled graphs on n nodes, divided by 3.

Original entry on oeis.org

1, 5, 41, 545, 11681, 402305, 22207361, 1961396225, 276825510401, 62368881977345, 22413909724518401, 12840603873823473665, 11720394922432296755201, 17037597932370037286600705
Offset: 1

Views

Author

Keywords

Comments

Sequence represents 1/3 of the number of 3-colored labeled graphs on n nodes. Indeed, on p. 413 of the Read paper, column 3 is 3, 15, 123, 1635, ...; or see A047863. - Emeric Deutsch, May 06 2004

References

  • R. C. Read, personal communication.
  • 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

Programs

  • Maple
    c[0]:=1: for n from 1 to 30 do c[n]:=sum(binomial(n,i)*2^(i*(n-i)),i=0..n) od: a:=n->(1/3)*sum(binomial(n,j)*2^(j*(n-j))*c[j],j=0..n): seq(a(n),n=1..19);
  • Mathematica
    a[n_] := 1/3*Sum[ 2^((i-j)*j + i*(n-i))*Binomial[n, i]*Binomial[i, j], {i, 0, n}, {j, 0, i}]; Table[ a[n], {n, 1, 14}] (* Jean-François Alcover, Dec 07 2011, after Emeric Deutsch *)

Formula

a(n) = (1/3)Sum_{j=0..n} binomial(n, j)*2^(j(n-j))*c(j) where c(n) = Sum_{i=0..n} binomial(n, i)*2^(i(n-i)) = A047863(n). - Emeric Deutsch, May 06 2004
From Peter Bala, Apr 12 2013: (Start)
a(n) = 1/3*A191371(n). Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)). Then a generating function for this sequence is 1/3*E(x)^3 - 1/3 = Sum_{n >= 1} a(n)*x^n/(n!*2^C(n,2)) = x + 5*x^2/(2!*2) + 41*x^3/(3!*2^3) + .... In general, E(x)^k, k = 1, 2, ..., is a generating function for labeled k-colored graphs (see Read). For examples see A047863 (k = 2), A191371 (k = 3) and A223887 (k = 4). (End)

Extensions

More terms from Pab Ter (pabrlos(AT)yahoo.com) and Emeric Deutsch, May 05 2004

A000686 Number of 4-colored labeled graphs on n nodes, divided by 4.

Original entry on oeis.org

1, 7, 85, 1777, 63601, 3882817, 403308865, 71139019777, 21276992674561, 10778161937857537, 9238819435213784065, 13390649605615389843457, 32796747486424209782108161, 135669064080920007649863745537, 947468281528010179181982467702785, 11166618111585805201637975219611631617
Offset: 1

Views

Author

Keywords

Comments

Sequence represents 1/4 of the number of 4-colored labeled graphs on n nodes. Indeed, on p. 413 of the Read paper, column 4 is 4, 28, 340, 7108, ... - Emeric Deutsch, May 06 2004

References

  • R. C. Read, personal communication.
  • 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

Programs

  • Mathematica
    b[n_] := Sum[ 2^((i-j)*j + i*(n-i))*Binomial[n, i]*Binomial[i, j], {i, 0, n}, {j, 0, i}]; a[n_] := 1/4*Sum[ Binomial[n, k]*2^(k*(n-k))*b[k], {k, 0, n}]; Table[a[n], {n, 1, 14}] (* Jean-François Alcover, Dec 07 2011, after Emeric Deutsch *)
  • PARI
    N=66;  x='x+O('x^N);
    E=sum(n=0, N, x^n/(n!*2^binomial(n,2)) );
    tgf=E^4-1;  v=Vec(tgf);
    v=vector(#v, n, v[n] * n! * 2^(n*(n-1)/2) ) / 4
    /* Joerg Arndt, Apr 10 2013 */

Formula

a(n) = (1/4)*Sum_{k=0..n} binomial(n, k)*2^(k(n-k))*b(k), where b(0)=1 and b(k) = 3*A000685(k) for k > 0. - Emeric Deutsch, May 06 2004
From Peter Bala, Apr 12 2013: (Start)
a(n) = (1/4)*A223887(n).
a(n) = (1/4)*Sum_{k = 0..n} binomial(n,k)*2^(k*(n-k))*b(k)*b(n-k), where b(n) := Sum_{k = 0..n} binomial(n,k)*2^(k*(n-k)).
Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)). Then a generating function for this sequence is (1/4)*(E(x)^4 - 1) = Sum_{n >= 0} a(n)*x^n/(n!*2^C(n,2)) = x + 7*x^2/(2!*2) + 85*x^3/(3!*2^3) + .... (End)

Extensions

More terms from Pab Ter (pabrlos(AT)yahoo.com) and Emeric Deutsch, May 05 2004

A002029 Number of connected graphs on n labeled nodes, each node being colored with one of 4 colors, such that no edge joins nodes of the same color.

Original entry on oeis.org

1, 4, 12, 132, 3156, 136980, 10015092, 1199364852, 234207001236, 75018740661780, 39745330657406772, 35073541377640231092, 51798833078501480220756, 128412490016744675540378580, 535348496386845235339961362932, 3757366291145650829115977555259252
Offset: 0

Views

Author

Keywords

References

  • 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).
  • R. C. Read, personal communication.

Crossrefs

Column k=4 of A322279.
Cf. A002032.

Programs

  • Mathematica
    m = 16;
    serconv = (CoefficientList[Sum[x^j*2^Binomial[j, 2], {j, 0, m}] + O[x]^m, x]*CoefficientList[(Sum[x^j/(j!*2^Binomial[j, 2]), {j, 0, m}] + O[x]^m)^4, x]) . x^Range[0, m-1];
    CoefficientList[1 + Log[serconv] + O[x]^m, x]*Range[0, m-1]! (* Jean-François Alcover, Sep 04 2019, after Andrew Howroyd *)
  • PARI
    seq(n)={Vec(serlaplace(1 + log(serconvol(sum(j=0, n, x^j*2^binomial(j, 2)) + O(x*x^n), (sum(j=0, n, x^j/(j!*2^binomial(j, 2))) + O(x*x^n))^4))))} \\ Andrew Howroyd, Dec 03 2018

Formula

E.g.f.: log(b(x)+1)+1 where b(x) = 4 * e.g.f. of A000686. - Sean A. Irvine, May 27 2013
a(n) = m_n(4) using the functions defined in A002032. - Sean A. Irvine, May 29 2013
Logarithmic transform of A223887. - Andrew Howroyd, Dec 03 2018

Extensions

More terms from Sean A. Irvine, May 27 2013
Name clarified and offset corrected by Andrew Howroyd, Dec 03 2018
Showing 1-8 of 8 results.