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Number of labeled interval orders on n elements: (2+2)-free posets.
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%I #69 Jan 05 2022 13:53:11

%S 1,1,3,19,207,3451,81663,2602699,107477247,5581680571,356046745023,

%T 27365431508779,2494237642655487,266005087863259291,

%U 32815976815540917183,4636895313201764853259,743988605732990946684927

%N Number of labeled interval orders on n elements: (2+2)-free posets.

%C From _Peter Bala_, Dec 26 2021: (Start)

%C We make the following conjectures:

%C 1) Taking the sequence modulo an integer k gives an eventually periodic sequence with period dividing phi(k). For example, the sequence taken modulo 8 begins [1, 1, 3, 3, 7, 3, 7, 3, 7, ...] and appears to have a pre-period of length 3 and a period of length 2 = (1/2)*phi(8).

%C 2) Let i >= 0 and define a_i(n) = a(n+i). Then for each i the Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i the expansion of exp( Sum_{n >= 1} a_i(n)*x^n/n ) has integer coefficients. (End)

%H Alois P. Heinz, <a href="/A079144/b079144.txt">Table of n, a(n) for n = 0..260</a>

%H Peter Bala, <a href="/A002439/a002439.pdf">Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)</a>

%H Graham Brightwell and Mitchel T. Keller, <a href="http://arXiv.org/abs/1111.6766">Asymptotic Enumeration of Labelled Interval Orders</a>, arXiv:1111.6766 [math.CO], 2011.

%H Anders Claesson, Mark Dukes and Martina Kubitzke, <a href="http://arXiv.org/abs/1006.1312">Partition and composition matrices</a>, arXiv:1006.1312 [math.CO], 2010-2011.

%H Hsien-Kuei Hwang and Emma Yu Jin, <a href="https://arxiv.org/abs/1911.06690">Asymptotics and statistics on Fishburn matrices and their generalizations</a>, arXiv:1911.06690 [math.CO], 2019.

%H Don Zagier, <a href="http://dx.doi.org/10.1016/S0040-9383(00)00005-7">Vassiliev invariants and a strange identity related to the Dedekind eta-function</a>, Topology, vol.40, pp.945-960 (2001); see p. 952.

%H Yan X Zhang, <a href="http://arxiv.org/abs/1508.00318">Four Variations on Graded Posets</a>, arXiv preprint arXiv:1508.00318 [math.CO], 2015.

%F a(n) = (1/(24^n))*Sum_{k=0..n} binomial(n, k)*A002439(k). Zagier 2001, p. 954.

%F G.f.: Sum(Product(1-exp(-k*x),k = 1 .. n),n = 0 .. infinity). a(n) = Sum_{k=0..n} k!*Stirling2(n,k)*A138265(k). - _Vladeta Jovovic_, Mar 11 2008

%F From _Peter Bala_, Mar 19 2009: (Start)

%F Conjectural form for the o.g.f. as a continued fraction:

%F 1/(1-x/(1-2*x/(1-5*x/(1-7*x/(1-12*x/(1-15*x/(1- ...))))))) = 1 + x + 3*x^2 + 19*x^3 + 207*x^4 + ..., where the sequence [1,2,5,7,12,15,..] is the sequence of generalized pentagonal numbers A001318. Compare with the continued fraction form of the o.g.f. of A002105. (End)

%F E.g.f.: 1+(exp(x)-1)/(G(0)+1-exp(x)), where G(k)= 2*exp(x*(k+1))-1-exp(x*(k+1))*(exp(x*(k+2))-1)/G(k+1); (continued fraction, Euler's kind, 1-step). - _Sergei N. Gladkovskii_, Jan 06 2012

%F Asymptotics (Brightwell and Keller, 2011): a(n) ~ 12*sqrt(3)/Pi^(5/2) * (n!)^2 * sqrt(n) * (6/Pi^2)^n. - _Vaclav Kotesovec_, May 03 2014

%F From _Peter Bala_, May 11 2017: (Start)

%F For a proof of above conjectural continued fraction representation of the o.g.f. see the Bala link.

%F G.f.: 1/(1 + x - 2*x/(1 - 1*x/(1 + x - 7*x/(1 - 5*x/(1 + x - 15*x/(1 - 12*x/(1 + x - 26*x/(1 - 22*x/(1 + x - ...))))))))), where the sequence of unsigned partial numerators [2, 1, 7, 5, 15, 12, ...] is obtained from A001318 by swapping adjacent terms.

%F E.g.f.: F(q) = Sum_{n >= 0} q^(n+1)*Product_{i = 1..n} (1 - q^i)^2 at q = exp(t). Note that F(q) at q = 1/(1 - t) is a g.f. for unlabeled interval orders A022493, and at q = 1 - t gives a g.f. for A138265. (End)

%F From _Peter Bala_, Dec 26 2021: (Start)

%F a(6*n + 5) == 0 (mod 7); a(10*n + 7) == 0 (mod 11);

%F a(12*n + 11) == 0 (mod 13); a(16*n + 5) == a(16*n + 8) == 0 (mod 17);

%F a(18*n + 3) == 0 (mod 19); a(22*n + 4) == 0 (mod 23). (End)

%e 1 + x + 3*x^2 + 19*x^3 + 207*x^4 + 3451*x^5 + 81663*x^6 + 2602699*x^7 + ...

%p A002439 := proc(n) option remember; if n = 0 then 1; else (-4)^n-add((-9)^k*binomial(2*n+1,2*k)*procname(n-k),k=1..n+1) ; end if;end proc:

%p seq(1/(24^n)*add(binomial(n,k)*A002439(k), k = 0..n), n = 0..20); # _Peter Bala_, Dec 26 2021

%t nmax=20; rk=Rest[CoefficientList[Series[Sum[Product[1-1/(1+x)^j,{j,1,n}],{n,0,nmax}],{x,0,nmax}],x]]; Flatten[{1,Table[Sum[rk[[k]] * k! * StirlingS2[n,k],{k,1,n}],{n,1,nmax}]}] (* _Vaclav Kotesovec_, May 03 2014 *)

%o (PARI) {a(n) = if( n<0, 0, n! * polcoeff( subst( sum( i=0, n, prod( j=1, i, 1 - (1 - x + O(x^(n - i + 2)))^j )), x, 1 - exp( -x + x * O(x^n))), n))} /* _Michael Somos_, Apr 01 2012 */

%Y Cf. A022493 (unlabeled interval orders).

%Y Cf. A002439 (Glaisher's T numbers), A002114 (Glaisher's H numbers).

%Y Cf. A001318, A138265.

%K nonn,easy

%O 0,3

%A Detlef Pauly (dettodet(AT)yahoo.de), Dec 27 2002