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.

A187245 Number of permutations of [n] having no cycle with 2 alternating runs (it is assumed that the smallest element of the cycle is in the first position).

Original entry on oeis.org

1, 1, 2, 5, 17, 78, 463, 3315, 27164, 247975, 2492539, 27422698, 328607417, 4266367567, 59686293284, 895068242601, 14320843215019, 243467476610732, 4382635181281015, 83272415871044649, 1665465961530365026, 34974843092354081119, 769445564105823722109
Offset: 0

Views

Author

Emeric Deutsch, Mar 07 2011

Keywords

Comments

a(n) = A187244(n,0).

Examples

			a(3)=5 because we have among the 6 permutations of {1,2,3} only 312=(132) has a cycle with 2 alternating runs.
		

Crossrefs

Cf. A187244.

Programs

  • Maple
    g := exp((4*exp(z)-exp(2*z)-3-2*z)*1/4)/(1-z): gser := series(g, z = 0, 25): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 22);
    # second Maple program:
    a:= proc(n) option remember;
          `if`(n=0, 1, add(a(n-j)*binomial(n-1, j-1)*
          `if`(j=1, 1, (j-1)!-(2^(j-2)-1)), j=1..n))
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, Apr 15 2017
  • Mathematica
    CoefficientList[Series[E^((4*E^x-E^(2*x)-3-2*x)/4)/(1-x), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Mar 15 2014 *)

Formula

E.g.f.: g(z)=exp[(4exp(z)-exp(2z)-3-2z)/4]/(1-z).
a(n) ~ exp(exp(1)-exp(2)/4-5/4) * n! = 0.68455780023755436... * n!. - Vaclav Kotesovec, Mar 15 2014

A187246 Number of cycles with 2 alternating runs in all permutations of [n] (it is assumed that the smallest element of the cycle is in the first position).

Original entry on oeis.org

0, 0, 0, 1, 7, 42, 267, 1900, 15263, 137494, 1375195, 15127656, 181532895, 2359929682, 33039019643, 495585302836, 7929364861759, 134799202682670, 2426385648353595, 46101327318849376, 922026546377249663, 19362557473922767210, 425976264426301927195
Offset: 0

Views

Author

Emeric Deutsch, Mar 07 2011

Keywords

Comments

a(n) = Sum_{k>=0} k*A187244(n,k).

Examples

			a(4)=7 because each of the following permutations of {1,2,3,4} has 1 cycle with 2 alternating runs: (132)(4), (142)(3), (143)(2), (1)(243), (1243), (1342), and (1432); the remaining 17 permutations have none.
		

Crossrefs

Cf. A187244.

Programs

  • Maple
    g := (1/4*(3+2*z+exp(2*z)-4*exp(z)))/(1-z): gser := series(g, z = 0, 25): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 22);
    # second Maple program:
    b:= proc(n) option remember; expand(
          `if`(n=0, 1, add(b(n-j)*binomial(n-1, j-1)*
          `if`(j=1, 1, (j-1)!+(2^(j-2)-1)*(x-1)), j=1..n)))
        end:
    a:= n-> (p-> add(coeff(p, x, i)*i, i=0..degree(p)))(b(n)):
    seq(a(n), n=0..30);  # Alois P. Heinz, Apr 15 2017
  • Mathematica
    CoefficientList[Series[(3+2*x+E^(2*x)-4*E^(x))/(4*(1-x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Mar 15 2014 *)

Formula

E.g.f.: g(z)=(1/4)[3+2z+exp(2z)-4exp(z)]/(1-z).
a(n) ~ (5/4-exp(1)+exp(2)/4) * n! = 0.378982196273617... * n!. - Vaclav Kotesovec, Mar 15 2014
D-finite with recurrence a(n) +(-n-3)*a(n-1) +(3*n-1)*a(n-2) +2*(-n+2)*a(n-3)=0. - R. J. Mathar, Jul 26 2022

A187250 Triangle read by rows: T(n,k) is the number of permutations of [n] having k cycles with at least 3 alternating runs (it is assumed that the smallest element of a cycle is in the first position), 0<=k<=floor(n/4).

Original entry on oeis.org

1, 1, 2, 6, 22, 2, 94, 26, 460, 260, 2532, 2508, 15420, 24760, 140, 102620, 254968, 5292, 739512, 2760432, 128856, 5729192, 31547344, 2640264, 47429896, 381339368, 50186136, 46200, 417429800, 4879612808, 926494712, 3483480, 3888426512, 66107044176, 17025751600, 157068912
Offset: 0

Views

Author

Emeric Deutsch, Mar 08 2011

Keywords

Comments

Number of entries in row n is 1+floor(n/4).
Sum of entries in row n is n!.
T(n,0)=A187251(n).
Sum(k*T(n,k), k>=0) = A187252(n).

Examples

			T(4,1)=2 because we have (1324) and (1423).
Triangle starts:
1;
1;
2;
6;
22,2;
94,26;
460,260;
		

Crossrefs

Programs

  • Maple
    G := exp((1/4*(1-t))*(2*z-1+exp(2*z)))/(1-z)^t: Gser := simplify(series(G, z = 0, 17)): for n from 0 to 14 do P[n] := sort(factorial(n)*coeff(Gser, z, n)) end do: for n from 0 to 14 do seq(coeff(P[n], t, k), k = 0 .. floor((1/4)*n)) end do; # yields sequence in triangular form

Formula

E.g.f.: G(t,z) = exp[(1/4)(1-t)(2z-1+exp(2z))]/(1-z)^t.
The 4-variate g.f. H(u,v,w,z) (exponential with respect z), where u marks number of cycles with 1 alternating run, v marks number of cycles with 2 alternating runs, w marks the number of all cycles, and z marks the size of the permutation, is given by H(u,v,w,z) = exp[(1/4)w((v-1)(exp(2z)+2z)+4(u-v)exp(z)+1-4u+3v)]/(1-z)^w.
We have G(t,z) = H(1/t,1/t,t,z).
Showing 1-3 of 3 results.