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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A000366 Genocchi numbers of second kind (A005439) divided by 2^(n-1).

Original entry on oeis.org

1, 1, 2, 7, 38, 295, 3098, 42271, 726734, 15366679, 391888514, 11860602415, 420258768950, 17233254330343, 809698074358250, 43212125903877439, 2599512037272630686, 175079893678534943287, 13122303354155987156306, 1088559958829010054171343, 99456043127935948731527942
Offset: 1

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Keywords

Comments

The earliest known reference to these numbers is the Dellac Marseille memoir. - Don Knuth, Jul 11 2007
According to Ira Gessel, Dellac's interpretation is the following: start with a 2n X n array of cells and consider the set D of cells in rows i through i+n of column i, for i from 1 to n. Then a(n) is the number of subsets of D containing two cells in each column and one cell in each row.
Barsky proved that for even n>1, a(n) is congruent to 3 mod 4 and for odd n>1, congruent to 2 mod 4. Gessel shows that for even n>5, a(n) is congruent to 4n-1 mod 16 and for odd n>2 that a(n)/2 is congruent to 2-n mod 8.
The entry for A005439 has further information.
The number of sequences (I_1,...,I_{n-1}) consisting of subsets of the set {1,...,n} such that the number of elements in I_k is exactly k and I_k\subset I_{k+1}\cup {k+1}. The Euler characteristics of the degenerate flag varieties of type A. - Evgeny Feigin, Dec 15 2011
Kreweras proved that for n>2, a(n) is alternatively congruent to 2 and to 7 mod 36. - Michel Marcus, Nov 06 2012

Examples

			G.f. = x + x^2 + 2*x^3 + 7*x^4 + 38*x^5 + 295*x^6 + 3098*x^7 + ...
		

References

  • Anonymous, L'Intermédiaire des Mathématiciens, 7 (1900), p. 328.
  • Hippolyte Dellac, Problem 1735, L'Intermédiaire des Mathématiciens, Vol. 7 (1900), p. 9 ff.
  • E. Lemoine, L'Intermédiaire des Mathématiciens, 8 (1901), 168-169.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoulli'schen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.

Crossrefs

First column, first diagonal and row sums of triangle A014784.
Also row sums of triangle A239894.

Programs

  • Mathematica
    a[n_] = (-2^(-1))^(n-2)* Sum[ Binomial[n, k]*(1 - 2^(n+k+1))*BernoulliB[n+k+1], {k, 0, n}]; Table[a[n], {n,19}] (* Jean-François Alcover, Jul 18 2011, after PARI prog. *)
  • PARI
    a(n)=(-1/2)^(n-2)*sum(k=0,n,binomial(n,k)*(1-2^(n+k+1))*bernfrac(n+k+1))
    
  • PARI
    {a(n)=local(CF=1+x*O(x^n));if(n<1,return(0), for(k=1,n,CF=1/(1-((n-k)\2+1)*((n-k)\2+2)/2*x*CF));return(Vec(CF)[n]))} (Hanna)
    
  • PARI
    {a(n)=polcoeff( x*sum(m=0, n, m!*(m+1)!*(x/2)^m / prod(k=1, m,1 + k*(k+1)*x/2 +x*O(x^n)) ), n)} \\ Paul D. Hanna, Feb 03 2013
    
  • Python
    from math import comb
    from sympy import bernoulli
    def A000366(n): return (-1 if n&1 else 1)*sum(comb(n,k)*(1-(1<>n-2 if n>1 else 1 # Chai Wah Wu, Apr 14 2023
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A000366_list(n) :
        D = [0]*(n+2); D[1] = 1
        R = []; z = 1/2; b = False
        for i in(0..2*n-1) :
            h = i//2 + 1
            if b :
                for k in range(h-1, 0, -1) : D[k] += D[k+1]
                z *= 2
            else :
                for k in range(1, h+1, 1) :  D[k] += D[k-1]
            b = not b
            if not b : R.append(D[1]/z)
        return R
    A000366_list(19) # Peter Luschny, Jun 29 2012
    

Formula

From Don Knuth, Jul 11 2007: (Start)
The anonymous 1900 note in Interm. Math. gives a formula that is equivalent to a nice generating function:
For example, the first four terms on the right are
1
... 2x - 2x^2 + 2x^3 + ...
........ 9x^2 - 36x^3 + ...
............... 72x^3 + ...
summing to 1 + 2x + 7x^2 + 38x^3 + ... . Of course one can replace x with 2x and get a generating function for A005439. (End)
(-2)^(2-n) * Sum_{k=0..n} C(n, k)*(1-2^(n+k+1))*B(n+k+1), with B(n) the Bernoulli numbers.
O.g.f.: A(x) = x/(1-x/(1-x/(1-3*x/(1-3*x/(1-6*x/(1-6*x/(... -[n/2+1]*[n/2+2]/2*x/(1- ...)))))))) (continued fraction). - Paul D. Hanna, Oct 07 2005
Sum_{n>0} a(n)x^n = Sum_{n>0} (n!^2/2^{n-1}) (x^n/((1+x)(1+3x)...(1+binomial(n,2)x))).
a(n+1) = Sum_{k=0..n} A211183(n,k). - Philippe Deléham, Feb 03 2013
G.f.: Q(0)*2 - 2, where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 2/(1 - x*(k+1)^2/( x*(k+1)^2 - 2/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013
a(n) ~ 2^(n+5) * n^(2*n+3/2) / (exp(2*n) * Pi^(2*n+1/2)). - Vaclav Kotesovec, Oct 28 2014

Extensions

More terms from David W. Wilson, Jan 11 2001
Edited by Ralf Stephan, Apr 17 2004

A218826 Number of indecomposable (by concatenation) alternating n-anagrams.

Original entry on oeis.org

1, 1, 4, 25, 217, 2470, 35647, 637129, 13843948, 360022957, 11054457253, 396003680518, 16377463914091, 774714094061221, 41572230979229284, 2512149910125036865, 169831839578092130017, 12769241823369505582150, 1062122471116082751430087, 97264621940872013476357969
Offset: 1

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Author

Michel Marcus, Nov 07 2012

Keywords

Comments

Alternating anagrams enumeration is related to A000366 by a(n) = A000366(n+1).
For all n, a(n) are periodically congruent to 1, 1 and 4 modulo 6.

Crossrefs

Cf. A000366, A218827. First column of A239894.

Programs

  • PARI
    a000366(n)= {return((-1/2)^(n-2)*sum(k=0, n, binomial(n, k)*(1-2^(n+k+1))*bernfrac(n+k+1)));}
    bi(n,k) = {if (matb[n,k] == 0, if (n==k, v=1, if (k==1, v = b(n),v = sum(i=1, n-k+1, b(i)*bi(n-i,k-1)););); matb[n,k] = v;); return (matb[n,k]);}
    b(n) = {if (n==1, return(a000366(n+1)), return(a000366(n+1) - sum(i=2, n, bi(n,i))));}
    allb(m) = {matb = matrix(m,m); for (i=1, m, print1(b(i), ", "););}

Formula

G.f.: (1-Q(0))/x, where Q(k) = 1 - ((k+1)*(k+2)/2)*x/(1 - ((k+1)*(k+2)/2)*x/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: 1/Q(0), where Q(k) = 1 - ((k+1)*(k+2)/2)*x/(1 - ((k+2)*(k+3)/2)*x/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2013
a(n) = A000366(n + 1) - Sum_{k=2..n} A239894(n, k). - Andrew Howroyd, Feb 25 2020
Showing 1-2 of 2 results.