A069727 - OEIS

%I #34 Mar 29 2021 15:06:22

%S 1,1,2,4,12,34,154,675,3534,18985,108070,632109,3807254,23411290,

%T 146734695,934382820,6034524474,39457153432,260855420489,

%U 1741645762265,11732357675908,79673115468562,545036528857605,3753642607424647,26010818244754788,181266500331748878

%N Number of nonisomorphic unrooted Eulerian planar maps with n edges (Eulerian means that all vertices are of even valency; there is an Eulerian cycle).

%C By duality, also the number of unrooted (sensed) bipartite maps with n edges. - _Andrew Howroyd_, Mar 29 2021

%H Vaclav Kotesovec, <a href="/A069727/b069727.txt">Table of n, a(n) for n = 0..1000</a>

%H M. Deryagina, <a href="ftp://ftp.pdmi.ras.ru/pub/publicat/znsl/v446/p031.pdf">On the enumeration of hypermaps which are self-equivalent with respect to reversing the colors of vertices</a>, Preprint 2016.

%H V. A. Liskovets and T. R. S. Walsh, <a href="http://dx.doi.org/10.1016/j.disc.2003.09.015">Enumeration of Eulerian and unicursal planar maps</a>, Discr. Math., 282 (2004), 209-221.

%F a(n) = (1/(2n))*(3*2^(n-1)*binomial(2n, n)/((n+1)(n+2)) + Sum_{k=1..n-1, k|n} phi(n/k)*d(n/k)*2^(k-2)*binomial(2k, k)) + q(n) where phi is the Euler function A000010, q(n) = 2^((n-4)/2)*binomial(n, n/2)/(n+2) if n is even, q(n) = 2^((n-1)/2)*binomial(n-1, (n-1)/2)/(n+1) if n is odd, d(n)=4, if n is even and d(n)=3 if n is odd. - _Valery A. Liskovets_, Mar 17 2005

%F a(n) ~ 3 * 2^(3*n-2) / (sqrt(Pi) * n^(7/2)). - _Vaclav Kotesovec_, Aug 28 2019

%t a[n_] := (1/(2n)) * (3*2^(n-1) * Binomial[2n, n]/((n+1)*(n+2)) + Sum[ EulerPhi[n/k] * d[n/k] * 2^(k-2) * Binomial[2k, k], {k, Most[ Divisors[n]]}]) + q[n]; a[0] = 1; q[n_?EvenQ] := 2^((n-4)/2)*Binomial[ n, n/2]/(n+2); q[n_?OddQ] := 2^((n-1)/2)*Binomial[(n-1), (n-1)/2]/(n+1); d[n_] := 4-Mod[n, 2]; Table[ a[n], {n, 0, 20}] (* _Jean-François Alcover_, Dec 19 2011, after _Valery A. Liskovets_ *)

%o (PARI) a(n) = {if(n==0, 1, sumdiv(n, d, if(d<n, 1, 2/((n+1)*(n+2))) * eulerphi(n/d) * (4-n/d%2) * 2^(d-2) * binomial(2*d,d))/(2*n) + if(n%2, 2^((n-1)/2)/(n+1), 2^((n-4)/2)/(n+2))*binomial(n\2*2, n\2))} \\ _Andrew Howroyd_, Mar 29 2021

%Y Cf. A000257 (rooted), A069720, A069724, A103939 (with distinguished face), A103940 (with distinguished vertex).

%K easy,nice,nonn

%O 0,3

%A _Valery A. Liskovets_, Apr 07 2002