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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A164652 Triangle read by rows: Hultman numbers: a(n,k) is the number of permutations of n elements whose cycle graph (as defined by Bafna and Pevzner) contains k cycles for n >= 0 and 1 <= k <= n+1.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 5, 0, 1, 8, 0, 15, 0, 1, 0, 84, 0, 35, 0, 1, 180, 0, 469, 0, 70, 0, 1, 0, 3044, 0, 1869, 0, 126, 0, 1, 8064, 0, 26060, 0, 5985, 0, 210, 0, 1, 0, 193248, 0, 152900, 0, 16401, 0, 330, 0, 1, 604800, 0, 2286636, 0, 696905, 0, 39963, 0, 495, 0, 1, 0, 19056960, 0, 18128396, 0, 2641925, 0, 88803, 0, 715, 0, 1
Offset: 0

Views

Author

Anthony Labarre, Aug 19 2009

Keywords

Comments

a(n,k) is also the number of ways to express a given (n+1)-cycle as the product of an (n+1)-cycle and a permutation with k cycles (see Doignon and Labarre). a(n,n+1-2k) is the number of permutations of n elements whose block-interchange distance is k (see Christie, Doignon and Labarre).
Named after the Swedish mathematician Axel Hultman. - Amiram Eldar, Jun 11 2021
a(2*n,1) is the number of spanning trees in certain graphs with 2*n+1 vertices and n*(n+1) edges (see Ishikawa, Miezaki, and Tanaka). - Tsuyoshi Miezaki, Feb 08 2023

Examples

			Triangle begins:
  n=0:  1;
  n=1:  0, 1;
  n=2:  1, 0, 1;
  n=3:  0, 5, 0, 1;
  n=4:  8, 0, 15, 0, 1;
  n=5:  0, 84, 0, 35, 0, 1;
  n=6:  180, 0, 469, 0, 70, 0, 1;
  n=7:  0, 3044, 0, 1869, 0, 126, 0, 1;
  n=8:  8064, 0, 26060, 0, 5985, 0, 210, 0, 1;
  n=9:  0, 193248, 0, 152900, 0, 16401, 0, 330, 0, 1;
  n=10: 604800, 0, 2286636, 0, 696905, 0, 39963, 0, 495, 0, 1;
  ...
From _Jon E. Schoenfield_, May 20 2023: (Start)
As a right-aligned triangle:
                                                      1; n=0
                                                   0, 1; n=1
                                                1, 0, 1; n=2
                                           0,   5, 0, 1; n=3
                                        8, 0,  15, 0, 1; n=4
                                 0,    84, 0,  35, 0, 1; n=5
                            180, 0,   469, 0,  70, 0, 1; n=6
                      0,   3044, 0,  1869, 0, 126, 0, 1; n=7
                8064, 0,  26060, 0,  5985, 0, 210, 0, 1; n=8
          0,  193248, 0, 152900, 0, 16401, 0, 330, 0, 1; n=9
  604800, 0, 2286636, 0, 696905, 0, 39963, 0, 495, 0, 1; n=10
  ...
(End)
		

References

  • Axel Hultman, Toric permutations, Master's thesis, Department of Mathematics, KTH, Stockholm, Sweden, 1999.

Crossrefs

Cf. A185263 (rows reversed without 0's).

Programs

  • Haskell
    a164652 n k = a164652_tabl !! n !! k
    a164652_row n = a164652_tabl !! n
    a164652_tabl = [0] : tail (zipWith (zipWith (*)) a128174_tabl $
       zipWith (map . flip div) (tail a000217_list) (map init $ tail a130534_tabl))
    -- Reinhard Zumkeller, Aug 01 2014
    
  • Maple
    A164652:= (n, k)-> `if`(n-k mod 2 = 1, -Stirling1(n+2, k)/binomial(n+2, 2), 0):
    for n from 0 to 7 do seq(A164652(n,k),k=1..n+1) od; # Peter Luschny, Mar 22 2015
  • Mathematica
    T[n_, k_] := If[OddQ[n-k], Abs[StirlingS1[n+2, k]]/Binomial[n+2, 2], 0];
    Table[T[n, k], {n, 0, 11}, {k, 1, n+1}] // Flatten (* Jean-François Alcover, Aug 10 2018 *)
  • PARI
    T(n,k)= my(s=(n-k)%2); (-1)^s*s*stirling(n+2,k,1)/binomial(n+2,2);
    concat(vector(12, n, vector(n, k, T(n-1,k)))) \\ Gheorghe Coserea, Jan 23 2018
  • Sage
    def A164652(n, k):
        return stirling_number1(n+2,k)/binomial(n+2,2) if is_odd(n-k) else 0
    for n in (0..7): print([A164652(n,k) for k in (1..n+1)]) # Peter Luschny, Mar 22 2015
    

Formula

T(n,k) = S1(n+2,k)/C(n+2,2) if n-k is odd, and 0 otherwise. Here S1(n,k) are the unsigned Stirling numbers of the first kind A132393 and C(n,k) is the binomial coefficient (see Bona and Flynn).
For n > 0: T(n,k) = A128174(n+1,k) * A130534(n+1,k-1) / A000217(n+1). - Reinhard Zumkeller, Aug 01 2014
n-th row polynomial R(n,x) = (x/2)*( P(n+1,x) + (-1)^n * P(n+1,-x) ) / binomial(n+2,2), where P(k,x) = (x + 1)*(x + 2)*...*(x + k). - Peter Bala, May 14 2023

Extensions

T(0,1) set to 1 by Peter Luschny, Mar 24 2015
Edited to match values of k to the range 1 to n+1. - Max Alekseyev, Nov 20 2020

A185259 Irregular triangle read by rows: coefficients in order of decreasing exponents of polynomials P_g(x) related to Hultman numbers.

Original entry on oeis.org

1, 1, 12, 8, 1, 72, 528, 704, 180, 1, 324, 8760, 53792, 98124, 56160, 8064, 1, 1344, 103040, 1759520, 9936360, 21676144, 19083456, 6356160, 604800, 1, 5436, 1054056, 41312704, 539233128, 2901894144, 7118351104, 8247838464, 4418632656, 988952832, 68428800, 1, 21816, 10106736, 823376896, 21574613676, 235937470944, 1230387808384, 3281254260864, 4608240745104, 3390175943424, 1247151098880, 204083712000, 10897286400
Offset: 1

Views

Author

N. J. A. Sloane, Jan 21 2012

Keywords

Comments

Row n contains 2*n-1 terms.
Evaluating the polynomials at 1 gives A035319.

Examples

			Triangle begins:
[1] 1
[2] 1   12      8
[3] 1   72    528     704     180
[4] 1  324   8760   53792   98124    56160     8064
[5] 1 1344 103040 1759520 9936360 21676144 19083456 6356160 604800
[6] ...
		

Crossrefs

Programs

  • Mathematica
    P[n_, x_] := (f = (1-x)^(4n+1); s = Sum[-StirlingS1[2n+2+k, k+1]/ Binomial[2n+2+k, 2] x^k, {k, 0, 2n-2}]; f s + O[x]^(2n-1) // Normal);
    row[n_] := CoefficientList[P[n, x], x] // Reverse;
    Array[row, 7] // Flatten (* Jean-François Alcover, Sep 05 2018, after Gheorghe Coserea *)
  • PARI
    P(n, v='x) = {
      my(x='x+O('x^(2*n-1)), f=(1-x)^(4*n+1),
         s=sum(k=0, 2*n-2, -stirling(2*n+2+k, k+1, 1)/binomial(2*n+2+k,2)*x^k));
      subst(Pol(f*s, 'x), 'x, v);
    };
    concat(vector(7, n, Vec(P(n))))
    \\ test: N=50; vector(N, n, P(n,1)) == vector(N, n, (4*n)!/((2*n+1)!*4^n))
    \\ Gheorghe Coserea, Jan 30 2018

Extensions

More terms from Gheorghe Coserea, Jan 30 2018
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