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 1-3 of 3 results.

A002697 a(n) = n*4^(n-1).

Original entry on oeis.org

0, 1, 8, 48, 256, 1280, 6144, 28672, 131072, 589824, 2621440, 11534336, 50331648, 218103808, 939524096, 4026531840, 17179869184, 73014444032, 309237645312, 1305670057984, 5497558138880, 23089744183296
Offset: 0

Views

Author

Keywords

Comments

Coefficient of x^(2n-2) in Chebyshev polynomial T(2n) is -a(n).
Let M_n be the n X n matrix m_(i,j) = 1 + 2*abs(i-j); then det(M_n) = (-1)^(n-1)*a(n-1). - Benoit Cloitre, May 28 2002
Number of subsequences 00 in all words of length n+1 on the alphabet {0,1,2,3}. Example: a(2)=8 because we have 000,001,002,003,100,200,300 (the other 57=A125145(3) words of length 3 have no subsequences 00). a(n) = Sum_{k=0..n} k*A128235(n+1, k). - Emeric Deutsch, Feb 27 2007
Let P(A) be the power set of an n-element set A. Then a(n) = the sum of the size of the symmetric difference of x and y for every subset {x,y} of P(A). - Ross La Haye, Dec 30 2007 (See the comment from Bernard Schott below.)
Let P(A) be the power set of an n-element set A and B be the Cartesian product of P(A) with itself. Then remove (y,x) from B when (x,y) is in B and x != y and call this R35. Then a(n) = the sum of the size of the symmetric difference of x and y for every (x,y) of R35. [proposed edit of comment just above; by Ross La Haye]
The numbers in this sequence are the Wiener indices of the graphs of n-cubes (Boolean hypercubes). For example, the 3-cube is the graph of the standard cube whose Wiener index is 48. - K.V.Iyer, Feb 26 2009
From Gary W. Adamson, Sep 06 2009: (Start)
Starting (1, 8, 48, ...) = 4th binomial transform of [1, 4, 0, 0, 0, ...].
Equals the sum of terms in 2^n X 2^n semi-magic square arrays in which each row and column is composed of a binomial frequency of terms in the set (1, 3, 5, 7, ...).
The first few such arrays = [1] [1,3; 3,1]; /Q.
[1, 3, 5, 3;
3, 1, 3, 5;
5, 3, 1, 3;
3, 5, 3, 1]
(sum of terms = 48, with a binomial frequency of (1, 2, 1) as to (1, 3, 5) in each row and column)
[1, 3, 5, 3, 5, 7, 5, 3;
3, 1, 3, 5, 7, 5, 3, 5;
5, 3, 1, 3, 5, 3, 5, 7;
3, 5, 3, 1, 3, 5, 7, 5;
5, 7, 5, 3, 1, 3, 5, 3;
7, 5, 3, 5, 3, 1, 3, 5;
5, 3, 5, 7, 5, 3, 1, 3;
3, 5, 7, 5, 3, 5, 3, 1]
(sum of terms = 256, with each row and column composed of one 1, three 3's, three 5's, and one 7)
... (End)
Let P(A) be the power set of an n-element set A and B be the Cartesian product of P(A) with itself. Then a(n) = the sum of the size of the intersection of x and y for every (x,y) of B. - Ross La Haye, Jan 05 2013
Following the last comment of Ross, A002699 is the similar sequence when "intersection" is replaced by "symmetric difference" and A212698 is the similar sequence when "intersection" is replaced by "union". - Bernard Schott, Jan 04 2013
Also, following the first comment of Ross, A082134 is the similar sequence when "symmetric difference" is replaced by "intersection" and A133224 is the similar sequence when "symmetric difference" is replaced by "union". - Bernard Schott, Jan 15 2013
Let [n] denote the set {1,2,3,...,n} and denote an n-permutation of the elements of [n] by p = p(1)p(2)p(3)...p(n), where p(i) is the i-th entry in the linear order given by p. Then (p(i),p(j)) is an inversion of p if i < j but p(i) > p(j). Denote the number of inversions of p by inv(p) and call a 2n-permutation p = p(1)p(2)...p(2n) 2-ordered if p(1) < p(3) < ... < p(2n-1) and p(2) < p(4) < ... < p(2n). Then Sum(inv(p)) = n*4^(n-1), where the sum is taken over all 2-ordered 2n-permutations of p. See Bona reference below. - Ross La Haye, Jan 21 2014
Sum over all peaks of Dyck paths of semilength n of the product of the x and y coordinates. - Alois P. Heinz, May 29 2015
Sum of the number of all edges over all j-dimensional subcubes of the boolean hypercube graph of dimension n, Q_n, for all j, so a(n) = Sum_{j=1..n} binomial(n,j)*2^(n-j) * j*2^(j-1). - Constantinos Kourouzides, Mar 24 2024

Examples

			From _Bernard Schott_, Jan 04 2013: (Start)
See the comment about intersection of X and Y.
If A={b,c}, then in P(A) we have:
{b}Inter{b}={b},
{b}Inter{b,c}={b},
{c}Inter{c}={c},
{c}Inter{b,c}={c},
{b,c}Inter{b}={b},
{b,c}Inter{c}={c},
{b,c}Inter{b,c}={b,c}
and : #{b}+ #{b}+ #{c}+ #{c}+ #{b}+ #{c}+ #{b,c} = 8 = 2*4^(2-1) = a(2).
The other intersections are empty.
(End)
		

References

  • Miklos Bona, Combinatorics of Permutations, Chapman and Hall/CRC, 2004, pp. 1, 43, 64.
  • C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 516.
  • 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

Formula

a(n) = n*4^(n-1).
G.f.: x/(1-4x)^2. a(n+1) is the convolution of powers of 4 (A000302). - Wolfdieter Lang, May 16 2003
Third binomial transform of n. E.g.f.: x*exp(4x). - Paul Barry, Jul 22 2003
a(n) = Sum_{k=0..n} k*binomial(2*n, 2*k). - Benoit Cloitre, Jul 30 2003
For n>=0, a(n+1) = Sum_{i+j+k+l=n} binomial(2i, i)*binomial(2j, j)*binomial(2k, k)*binomial(2l, l). - Philippe Deléham, Jan 22 2004
a(n) = Sum_{k=0..n} 4^(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2. - Paul Barry, Oct 15 2004
Sum_{n>0} 1/a(n) = 8*log(2) - 4*log(3). - Jaume Oliver Lafont, Sep 11 2009
a(0) = 0, a(n) = 4*a(n-1) + 4^(n-1). - Vincenzo Librandi, Dec 31 2010
a(n+1) is the convolution of A000984 with A002457. - Rui Duarte, Oct 08 2011
a(0) = 0, a(1) = 1, a(n) = 8*a(n-1) - 16*a(n-2). - Harvey P. Dale, Jan 18 2012
a(n) = A002699(n)/2 = A212698(n)/3. - Bernard Schott, Jan 04 2013
G.f.: W(0)*x/2 , where W(k) = 1 + 1/( 1 - 4*x*(k+2)/( 4*x*(k+2) + (k+1)/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(5/4). - Amiram Eldar, Oct 28 2020
a(n) = (1/2)*Sum_{k=0..n} k*binomial(2*n, k). Compare this with the formula of Benoit Cloitre above. - Wolfdieter Lang, Nov 12 2021
a(n) = (-1)^(n-1)*det(M(n)) for n > 0, where M(n) is the n X n symmetric Toeplitz matrix whose first row consists of 1, 3, ..., 2*n-1. - Stefano Spezia, Aug 04 2022

A212698 Main transitions in systems of n particles with spin 3/2.

Original entry on oeis.org

3, 24, 144, 768, 3840, 18432, 86016, 393216, 1769472, 7864320, 34603008, 150994944, 654311424, 2818572288, 12079595520, 51539607552, 219043332096, 927712935936, 3917010173952, 16492674416640, 69269232549888, 290271069732864, 1213860837064704, 5066549580791808
Offset: 1

Views

Author

Stanislav Sykora, May 25 2012

Keywords

Comments

Please refer to the general explanation in A212697. This particular sequence is obtained for base b=4, corresponding to spin S = (b-1)/2 = 3/2.
Let P(A) be the power set of an n-element set A and let B be the Cartesian product of P(A) with itself. Then a(n) = the sum of the size of the union of x and y for every (x,y) in B. [See Relation (28): U(n) in document of Ross La Haye in reference.] - Bernard Schott, Jan 04 2013
A002697 is the analogous sequence if "union" is replaced by "intersection" and A002699 is the analogous sequence if "union" is replaced by "symmetric difference". Here, X union Y and Y union X are considered as two distinct Cartesian products, if we want to consider that X Union Y = Y Union X are the same Cartesian product, see A133224. - Bernard Schott Jan 11 2013

Crossrefs

Cf. A001787, A212697, A212699, A212700, A212701, A212702, A212703, A212704 (for b = 2, 3, 5, 6, 7, 8, 9, 10).

Programs

  • Magma
    [3*n*4^(n-1): n in [1..30]]; // Vincenzo Librandi, Nov 29 2015
  • Mathematica
    Table[Sum[Binomial[n,i] i 3^i,{i,0,n}],{n,1,21}] (* Geoffrey Critzer, Aug 08 2013 *)
  • PARI
    mtrans(n, b) = n*(b-1)*b^(n-1);
    for (n=1, 100, write("b212698.txt", n, " ", mtrans(n, 4)))
    

Formula

a(n) = n*(b-1)*b^(n-1). For this sequence, set b=4.
a(n) = 3*n*4^(n-1).
a(n) = 3*A002697(n).
From Geoffrey Critzer, Aug 08 2013: (Start)
a(n) = Sum_{i>=0} binomial(n,i)*i*3^i.
E.g.f.: 3*x*exp(4*x). (End)
G.f.: 3*x/(4*x-1)^2. - Colin Barker, Nov 03 2014
From Elmo R. Oliveira, May 24 2025: (Start)
a(n) = 8*a(n-1) - 16*a(n-2) for n > 2.
a(n) = A008585(n)*A000302(n-1). (End)

A082134 Expansion of e.g.f. x*exp(3*x)*cosh(x).

Original entry on oeis.org

0, 1, 6, 30, 144, 680, 3168, 14560, 66048, 296064, 1313280, 5772800, 25178112, 109078528, 469819392, 2013388800, 8590196736, 36507779072, 154620002304, 652837519360, 2748784312320, 11544883101696, 48378534690816
Offset: 0

Views

Author

Paul Barry, Apr 06 2003

Keywords

Comments

Binomial transform of A082133. 3rd binomial transform of (0,1,0,3,0,5,0,7,...)
Let P(A) be the power set of an n-element set A and B be the Cartesian product of P(A) with itself. Then remove (y,x) from B when (x,y) is in B and x <> y and call this R35. Then a(n) = the sum of the size of the intersection of x and y for every (x,y) of R35. - Ross La Haye, Dec 30 2007; edited Jan 05 2013
A133224 is the analogous sequence if "Intersection" is replaced by "Union" and A002697 is the analogous sequence if "Intersection" is replaced by "Symmetric difference". Here, X Intersection Y = Y Intersection X is considered as the same set [Relation (37): T_Q(n) in document of Ross La Haye in reference]. If we want to consider that X Intersection Y and Y Intersection X are two distinct formula for describing the same set, see A002697. - Bernard Schott, Jan 19 2013

Crossrefs

Cf. A057711 (x*exp(x)*cosh(x)), A082133 (x*exp(2*x)*cosh(x)).
Cf. A082135 (x*exp(4*x)*cosh(x)), A082136 (x*exp(5*x)*cosh(x)).

Programs

  • Magma
    [n*2^(n-2)*(1+2^(n-1)): n in [0..25]]; // G. C. Greubel, Feb 05 2018
    
  • Maple
    a:= n -> n*binomial(2^(n-1) +1, 2); seq(a(n), n=0..25); # G. C. Greubel, Apr 16 2020
  • Mathematica
    Table[n(2^(n-1) +4^(n-1))/2, {n, 0, 22}] (* Michael De Vlieger, Nov 29 2015 *)
    With[{nmax = 25}, CoefficientList[Series[x*Exp[3*x]*Cosh[x], {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Feb 05 2018 *)
  • PARI
    a(n)=n*(2^n--+4^n)/2 \\ Charles R Greathouse IV, Jan 14 2013
    
  • Sage
    [n*binomial(2^(n-1)+1, 2) for n in (0..25)] # G. C. Greubel, Apr 16 2020

Formula

a(n) = n*(2^(n-1) + 4^(n-1))/2.
E.g.f.: x*exp(3*x)*cosh(x).
Conjecture: (n+28)*a(n) + (n-282)*a(n-1) + 2*(-17*n+423)*a(n-2) + 8*(7*n-94)*a(n-3) = 0. - R. J. Mathar, Nov 29 2012
G.f.: x*(10*x^2-6*x+1) / ((2*x-1)^2*(4*x-1)^2). - Colin Barker, Dec 10 2012
Showing 1-3 of 3 results.