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 1-10 of 13 results. Next

A130168 a(n) = (b(n) + b(n+1))/3, where b(n) = A000366(n).

Original entry on oeis.org

1, 3, 15, 111, 1131, 15123, 256335, 5364471, 135751731, 4084163643, 144039790455, 5884504366431, 275643776229531, 14673941326078563, 880908054392169375, 59226468571935857991, 4432461082611507366531, 367227420727722013775883, 33514867695588319595233095
Offset: 2

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Author

Don Knuth, Aug 02 2007

Keywords

Comments

As remarked by Gessel, A000366 has a combinatorial interpretation via a certain 2n X n array; this sequence is for a similar array of size (2n-1) X (n-1).
In effect, Dellac gives a combinatorial reason why the elements of A000366 are alternately -1 and +1 modulo 3. Dellac also shows that all the terms of this sequence are odd.

Crossrefs

Programs

  • Mathematica
    b[n_] := (-2^(-1))^(n-2)*Sum[Binomial[n, k]*(1-2^(n+k+1))* BernoulliB[n+k+1], {k, 0, n}];
    a[n_] := (b[n] + b[n+1])/3;
    a /@ Range[2, 20] (* Jean-François Alcover, Apr 08 2021 *)
  • Python
    from math import comb
    from sympy import bernoulli
    def A130168(n): return (abs((2-(2<>n-1)//3 # Chai Wah Wu, Apr 14 2023

Formula

G.f.: 2*(1+x)/(3*x^3)*Q(0) - 2/(3*x) - 1/x^2 - 2/(3*x^3), 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) = |(2-2^(n+2))*Bernoulli(n+1) - (n+1)*(1-2^(2n+2))*Bernoulli(2n+2) - (1-2^(2n+3))*Bernoulli(2n+3) + Sum_{k=0..n-1} (2*binomial(n,k+1)-binomial(n+1,k))*(1-2^(n+k+2))*Bernoulli(n+k+2)|/(3*2^(n-1)). - Chai Wah Wu, Apr 14 2023

A005439 Genocchi medians (or Genocchi numbers of second kind).

Original entry on oeis.org

1, 1, 2, 8, 56, 608, 9440, 198272, 5410688, 186043904, 7867739648, 401293838336, 24290513745920, 1721379917619200, 141174819474169856, 13266093250285568000, 1415974941618255921152, 170361620874699124637696, 22948071824232932086513664, 3439933090471867097102680064
Offset: 0

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Author

Keywords

Comments

a(n) is the number of Boolean functions of n variables whose ROBDD (reduced ordered binary decision diagram) contains exactly n branch nodes, one for each variable. - Don Knuth, Jul 11 2007
The earliest known reference for these numbers is Seidel (1877, pages 185 and 186). - Don Knuth, Jul 13 2007
Hankel transform of 1,1,2,8,... is A168488. - Paul Barry, Nov 27 2009
According to Hetyei [2017], alternation acyclic tournaments "are counted by the median Genocchi numbers"; an alternation acyclic tournament "does not contain a cycle in which descents and ascents alternate." - Danny Rorabaugh, Apr 25 2017
The n-th Genocchi number of the second kind is also the number of collapsed permutations in (2n) letters. A permutation pi of size 2n is said to be collapsed if 1+floor(k/2) <= pi^{-1}(k) <= n + floor(k/2). There are 2 collapsed permutations of size 4, namely 1234 and 1324. - Arvind Ayyer, Oct 23 2020
For any positive integer n, a(n) is (-1)^n times the permanent of the 2n X 2n matrix M with M(j, k) = floor((2*j-k-1)/(2*n)). This former conjecture of Luschny, inspired by a conjecture of Zhi-Wei Sun in A036968, was proven by Fu, Lin and Sun (see link). - Peter Luschny, Sep 07 2021 [updated Sep 24 2021]

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    List([1..20],n->2*(-1)^n*Sum([0..n],k->Binomial(n,k)*(1-2^(n+k+1))*Bernoulli(n+k+1))); # Muniru A Asiru, Nov 29 2018
    
  • Magma
    [2*(-1)^n*(&+[Binomial(n, k)*(1-2^(n+k+1))*Bernoulli(n+k+1): k in [0..n]]): n in [1..20]]; // G. C. Greubel, Nov 28 2018
    
  • Maple
    seq(2*(-1)^n*add(binomial(n,k)*(1 - 2^(n+k+1))*bernoulli(n+k+1), k=0..n), n=0..20); # G. C. Greubel, Oct 18 2019
  • 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,16}] (* Jean-François Alcover, Jul 18 2011, after PARI prog. *)
  • PARI
    a(n)=2*(-1)^n*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+2)));if(n<0,return(0), for(k=1,n+1,CF=1/(1-((n-k+1)\2+1)^2*x*CF));return(Vec(CF)[n+2])) \\ Paul D. Hanna
    
  • Python
    from math import comb
    from sympy import bernoulli
    def A005439(n): return (-2 if n&1 else 2)*sum(comb(n,k)*(1-(1<Chai Wah Wu, Apr 14 2023
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A005439_list(n) :
        D = []; [D.append(0) for i in (0..n+2)]; D[1] = 1
        R = [] ; b = True
        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]
            else :
                for k in range(1,h+1,1) :  D[k] += D[k-1]
            if b : R.append(D[1])
            b = not b
        return R
    A005439_list(18) # Peter Luschny, Apr 01 2012
    
  • Sage
    [2*(-1)^n*sum(binomial(n,k)*(1-2^(n+k+1))*bernoulli(n+k+1) for k in (0..n)) for n in (1..20)] # G. C. Greubel, Oct 18 2019
    

Formula

a(n) = T(n, 1) where T(1, x) = 1; T(n, x) = (x+1)*((x+1)*T(n-1, x+1)-x*T(n-1, x)); see A058942.
a(n) = A000366(n)*2^(n-1).
a(n) = 2 * (-1)^n * Sum_{k=0..n} binomial(n, k)*(1-2^(n+k+1))*B(n+k+1), with B(n) the Bernoulli numbers. - Ralf Stephan, Apr 17 2004
O.g.f.: 1 + x*A(x) = 1/(1-x/(1-x/(1-4*x/(1-4*x/(1-9*x/(1-9*x/(... -[(n+1)/2]^2*x/(1-...)))))))) (continued fraction). - Paul D. Hanna, Oct 07 2005
G.f.: (of 1,1,2,8,...) 1/(1-x-x^2/(1-5*x-16*x^2/(1-13*x-81*x^2/(1-25*x-256*x^2/(1-41*x-625*x^2/(1-... (continued fraction). - Paul Barry, Nov 27 2009
O.g.f.: Sum_{n>=0} n!*(n+1)! * x^(n+1) / Product_{k=1..n} (1 + k*(k+1)*x). - Paul D. Hanna, May 10 2012
From Sergei N. Gladkovskii, Dec 14 2011, Dec 27 2012, May 29 2013, Oct 09 2013, Oct 24 2013, Oct 27 2013: (Start)
Continued fractions:
G.f.: A(x) = 1/S(0), S(k) = 1 - x*(k+1)*(k+2)/(1 - x*(k+1)*(k+2)/S(k+1)).
G.f.: A(x) = -1/S(0), S(k) = 2*x*(k+1)^2 - 1 - x^2*(k+1)^2*(k+2)^2/S(k+1).
G.f.: A(x) = 1/G(0) where G(k) = 1 - x*(k+1)^2/(1 - x*(k+1)^2/G(k+1)).
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 1/(1 - 1/(4*x*(k+1)) + 1/G(k+1))).
G.f.: Q(0)/x - 1/x, where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/(1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1)))).
G.f.: T(0)/(1-2*x), where T(k) = 1 - x^2*((k + 2)*(k+1))^2/(x^2*((k + 2)*(k+1))^2 - (1 - 2*x*k^2 - 4*x*k - 2*x)*(1 - 2*x*k^2 - 8*x*k - 8*x)/T(k+1)).
G.f.: R(0), where R(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/(1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/R(k+1) ))). (End)
a(n) ~ 2^(2*n+4) * n^(2*n+3/2) / (exp(2*n) * Pi^(2*n+1/2)). - Vaclav Kotesovec, Oct 28 2014
Rewriting the above: a(n) ~ 4*(2*n+1)! / Pi^(2*n+1). Compare to Genocchi numbers A110501(n) = g_n ~ 4*(2*n)! / Pi^(2*n). So these are indeed like "Genocchi medians" g_{n + 1/2}. - Alan Sokal, May 13 2022
Asymptotic expansion: a(n) ~ 4*(2*n+1)! * Pi^(-(2*n+1)) * (1 + (Pi^2/16)/n + (Pi^2 (Pi^2 - 16)/512)/n^2 + (Pi^2 (Pi^4 + 384)/24576)/n^3 + (Pi^2 (Pi^6 + 96*Pi^4 + 768*Pi^2 - 12288)/1572864)/n^4 + (Pi^2 (Pi^8 + 320*Pi^6 + 12800*Pi^4 + 491520)/125829120)/n^5 + ...) --- Proof uses binomial sum for Genocchi medians in terms of Genocchi or Bernoulli numbers, combined with leading term of convergent sum (with exponentially small corrections) for the latter. Can also check against the 10000 term a-file. - Alan Sokal, May 23 2022.
a(n) = n!^2 * [x^n*y^n] exp(x)*f(x-y), where f(x) is the derivative of the Genocchi number generating function 2*x/(exp(x)+1). - Ira M. Gessel, Jul 23 2024

Extensions

More terms and additional comments from David W. Wilson, Jan 11 2001
a(0)=1 prepended by Peter Luschny, Apr 14 2023

A211183 Triangle T(n,k), 0<=k<=n, read by rows, given by (0, 1, 1, 3, 3, 6, 6, 10, 10, 15, ...) DELTA (1, 0, 2, 0, 3, 0, 4, 0, 5, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 4, 1, 0, 7, 19, 11, 1, 0, 38, 123, 107, 26, 1, 0, 295, 1076, 1195, 474, 57, 1, 0, 3098, 12350, 16198, 8668, 1836, 120, 1, 0, 42271, 180729, 268015, 176091, 52831, 6549, 247, 1, 0, 726734, 3290353, 5369639, 4105015, 1564817, 287473, 22145
Offset: 0

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Author

Philippe Deléham, Feb 02 2013

Keywords

Examples

			Triangle begins :
1;
0, 1;
0, 1, 1;
0, 2, 4, 1;
0, 7, 19, 11, 1;
0, 38, 123, 107, 26, 1;
0, 295, 1076, 1195, 474, 57, 1;
0, 3098, 12350, 16198, 8668, 1836, 120, 1;
0, 42271, 180729, 268015, 176091, 52831, 6549, 247, 1;
0, 726734, 3290353, 5369639, 4105015, 1564817, 287473, 22145, 502, 1; ...
		

Crossrefs

Programs

  • PARI
    T(n,k)=polcoeff(polcoeff(sum(m=0, n, m!*x^m*prod(k=1, m, (y + (k-1)/2)/(1+(k*y+k*(k-1)/2)*x+x*O(x^n)))), n,x),k,y)
    for(n=0,12,for(k=0,n,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Feb 03 2013

Formula

Sum_{k, 0<=k<=n}T(n,k)*x^(n-k) = A000012(n), A000366(n+1), A110501(n+1), A211194(n), A221972(n) for x = 0, 1, 2, 3, 4 respectively.
T(n,n-1) = A000295(n).
T(n,1) = A000366(n).
G.f.: A(x,y) = Sum_{n>=0} n! * x^n * Product_{k=1..n} (y + (k-1)/2) / (1 + (k*y + k*(k-1)/2)*x). - Paul D. Hanna, Feb 03 2013

A014784 Triangle of numbers associated with Genocchi numbers.

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 7, 12, 12, 7, 38, 69, 81, 69, 38, 295, 552, 702, 702, 552, 295, 3098, 5901, 7857, 8559, 7857, 5901, 3098, 42271, 81444, 111618, 128034, 128034, 111618, 81444, 42271, 726734, 1411197, 1971945, 2339631, 2467665, 2339631, 1971945
Offset: 0

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Author

Keywords

Crossrefs

Cf. A035003.
Columns include A000366 and A102078. Row sums are in A000366.

Programs

  • Mathematica
    a[ 1, 1 ]=1; a[ n_, k_ ] := 0 /; (k>n || k<=0); a[ n_, 1 ]=Sum[ a[ n-1, i ], {i, 1, n-1} ] a[ n_, k_ ] := a[ n, k ]=2a[ n, k-1 ]-a[ n, k-2 ]-a[ n-1, k-1 ]-a[ n-1, k-2 ]; Flatten[ Table[ a[ n, i ], {n, 1, 10}, {i, 1, n} ] ]

Extensions

More terms from Erich Friedman.

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

Views

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

A239894 Triangle read by rows: T(n,k) (n>=1, 1 <= k <= n) = number of alternating anagrams on n letters (of length 2n) which are decomposable into at most k slices.

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 25, 9, 3, 1, 217, 58, 15, 4, 1, 2470, 500, 100, 22, 5, 1, 35647, 5574, 861, 152, 30, 6, 1, 637129, 78595, 9435, 1313, 215, 39, 7, 1, 13843948, 1376162, 130159, 14192, 1870, 290, 49, 8, 1, 360022957, 29417919, 2232792, 191850, 20001, 2547, 378, 60, 9, 1
Offset: 1

Views

Author

N. J. A. Sloane, Apr 04 2014

Keywords

Examples

			Triangle begins:
1
1 1
4 2 1
25 9 3 1
217 58 15 4 1
...
		

Crossrefs

Row sums are A000366.
First column is A218826.

Programs

  • PARI
    \\ here G(n) is A000366(n).
    G(n)={(-1/2)^(n-2)*sum(k=0, n, binomial(n, k)*(1-2^(n+k+1))*bernfrac(n+k+1))}
    A(n)={my(M=matrix(n,n)); for(n=1, n, for(k=2, n, M[n,k] = sum(i=1, n-k+1, M[i,1]*M[n-i, k-1])); M[n,1]=G(n+1)-sum(i=2, n, M[n,i])); M}
    {my(T=A(10)); for(n=1, #T, print(T[n, 1..n]))} \\ Andrew Howroyd, Feb 24 2020

Formula

T(n,k) = b(1)*T(n-1,k-1)+b(2)*T(n-2,k-1)+...+b(n-k+1)*T(k-1,k-1), where b(i) = A218826(i) for k > 1.
T(n,k) = Sum_{i=1..n-k+1} A218826(i)*T(n-i, k-1) for k > 1. - Andrew Howroyd, Feb 24 2020

Extensions

Terms a(16) and beyond from Andrew Howroyd, Feb 19 2020

A098278 D(n,0)/2^n, where D(n,x) is triangle A098277.

Original entry on oeis.org

1, 1, 3, 21, 267, 5349, 154923, 6120741, 316271787, 20701782309, 1673934058923, 163850823271461, 19093313058395307, 2611858473935397669, 414452507370456337323, 75508557963926980473381
Offset: 0

Views

Author

Ralf Stephan, Sep 07 2004

Keywords

Comments

This is related to formula (1.7) in Lazar and Wachs reference.
Apparently all terms (except the initial 1s) have 3-valuation 1. - F. Chapoton, Jul 31 2021

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 267*x^4 + 5349*x^5 + ...
where A(x) = 1 + x/(1+x) + 2!^2*x^2/((1+x)*(1+3*x)) + 3!^2*x^3/((1+x)*(1+3*x)*(1+6*x)) + 4!^2*x^4/((1+x)*(1+3*x)*(1+6*x)*(1+10*x)) + ... - _Paul D. Hanna_, Sep 05 2012
		

Crossrefs

Cf. A000366.

Programs

  • Mathematica
    d[0, ] = 1; d[n, x_] := d[n, x] = (x+1)(x+2)d[n-1, x+2]-x(x+1)d[n-1, x];
    a[n_] := d[n, 0]/2^n;
    Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Oct 26 2018 *)
  • PARI
    {a(n)=polcoeff(sum(m=0, n, m!^2*x^m/prod(k=1, m, 1+k*(k+1)/2*x +x*O(x^n))), n)} \\ Paul D. Hanna, Sep 05 2012

Formula

G.f.: Sum_{n>=0} a(n)*x^n = 1/(1-1*1*x/(1-1*2*x/(1-2*3*x/(1-2*4*x/...)))).
G.f.: Sum_{n>=0} n!^2 * x^n / Product_{k=1..n} (1 + k*(k+1)/2*x). - Paul D. Hanna, Sep 05 2012
G.f.: 1/G(0) where G(k) = 1 - x*(k+1)*(2*k+1)/(1 - x*(k+1)*(2*k+2)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Jan 14 2013.
a(n+1) = Sum_{k=0..n} A098277(n,k)*(1/2)^k. - Philippe Deléham, Feb 08 2013

A130169 a(1) = 1; for n>1, a(n) = (c(n) + c(n+1))/2, where c(n) = A130168(n).

Original entry on oeis.org

1, 2, 9, 63, 621, 8127, 135729, 2810403, 70558101, 2109957687, 74061977049, 3014272078443, 140764140297981, 7474792551154047, 447790997859123969, 30053688313164013683, 2245843775591721612261, 185829940905166760571207, 16941047558158020804504489
Offset: 1

Views

Author

Don Knuth, Aug 02 2007

Keywords

Crossrefs

Programs

  • Mathematica
    b[n_] := (-2^(-1))^(n-2)*Sum[Binomial[n, k]*(1-2^(n+k+1))* BernoulliB[n+k+1], {k, 0, n}];
    c[n_] := (b[n] + b[n+1])/3;
    a[n_] := If[n == 1, 1, (c[n] + c[n+1])/2];
    a /@ Range[1, 19] (* Jean-François Alcover, Apr 08 2021 *)

Formula

G.f.: (1+x)^2/(3*x^3)*Q(0) + (x^3 - 5*x^2 - 5*x - 2)/(6*x^3), 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

A218827 Number of indecomposable (by shuffling) alternating n-anagrams.

Original entry on oeis.org

1, 1, 3, 16, 129, 1438, 20955, 384226, 8623101, 231978454, 7359117591, 271673905642, 11543742745689, 559348370431630, 30659822500574739, 1887796293833267746, 129757032076160998677, 9900820197631733600518, 834421415151529202479023, 77318409826165250051727514
Offset: 1

Views

Author

Michel Marcus, Nov 07 2012

Keywords

Comments

Alternating anagrams enumeration is related to A000366 by a(n) = A000366(n+1).
For n>2, a(n) are alternatively congruent to 3 and -2 modulo 18.

Crossrefs

Cf. A000366, A218826. First column of A239895.

Programs

  • Mathematica
    m = 20(*terms*); matc = Array[0&, {m, m}];
    a366[n_] := (-2^(-1))^(n - 2)*Sum[Binomial[n, k]*(1 - 2^(n + k + 1))* BernoulliB[n + k + 1], {k, 0, n}];
    ci[n_, k_] := ci[n, k] = Module[{v}, If[matc[[n, k]] == 0, If[n == k, v = 1, If[k == 1, v = c[n], v = Sum[Binomial[n - 1, i - 1]*c[i]*ci[n - i, k - 1], {i, 1, n - k + 1}]]]; matc[[n, k]] = v]; Return[matc[[n, k]]]];
    a[n_] := a366[n + 1] - If[n == 1, 0, Sum[ci[n, i], {i, 2, n}]];
    Array[a, m] (* Jean-François Alcover, Aug 03 2019, from PARI *)
  • PARI
    a000366(n)= {return((-1/2)^(n-2)*sum(k=0, n, binomial(n, k)*(1-2^(n+k+1))*bernfrac(n+k+1)));}
    ci(n,k) = {if (matc[n,k] == 0, if (n==k, v = 1, if (k==1, v = c(n), v = sum(i=1, n-k+1, binomial(n-1,i-1)*c(i)*ci(n-i,k-1)););); matc[n,k] = v;); return(matc[n,k]);}
    c(n) = {if (n==1, return(a000366(n+1)), return(a000366(n+1) - sum(i=2, n, ci(n,i))));}
    allc(m) = {matc = matrix(m,m); for (i=1, m, print1(c(i), ", "););}

A230740 O.g.f.: Sum_{n>=0} x^n * Product_{k=1..n} (k*(k+1)/2 + x) / (1 + k*(k+1)/2*x).

Original entry on oeis.org

1, 1, 3, 10, 51, 370, 3691, 48525, 812089, 16832928, 422860609, 12649706416, 444120983433, 18078156682309, 844323149201499, 44838127594166770, 2686250544297734323, 180295858504407010026, 13473490672899749784979, 1114874245392058455432873
Offset: 0

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Author

Paul D. Hanna, Oct 28 2013

Keywords

Comments

Compare to an o.g.f. of Genocchi numbers of the second kind (A000366):
Sum_{n>=0} x^n * Product_{k=1..n} k*(k+1)/2 / (1 + k*(k+1)/2*x).

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 10*x^3 + 51*x^4 + 370*x^5 + 3691*x^6 +...
where
A(x) = 1 + x*(1+x)/(1+x) + x^2*(1+x)*(3+x)/((1+x)*(1+3*x)) + x^3*(1+x)*(3+x)*(6+x)/((1+x)*(1+3*x)*(1+6*x)) + x^4*(1+x)*(3+x)*(6+x)*(10+x)/((1+x)*(1+3*x)*(1+6*x)*(1+10*x)) +...
		

Crossrefs

Cf. A230682.

Programs

  • PARI
    {a(n)=polcoeff(sum(m=0, n, x^m*prod(k=1, m, k*(k+1)/2+x+x*O(x^n))/prod(k=1, m, 1+k*(k+1)/2*x+x*O(x^n))), n)}
    for(n=0, 20, print1(a(n), ", "))

Formula

a(n) ~ 2^(n+6) * n^(2*n+7/2) / (exp(2*n) * Pi^(2*n+5/2)). - Vaclav Kotesovec, Oct 28 2014
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