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A033504 a(n)/4^n is the expected number of tosses of a coin required to obtain n+1 heads or n+1 tails.

%I A033504 #53 Oct 01 2023 10:52:44
%S A033504 1,10,66,372,1930,9516,45332,210664,960858,4319100,19188796,84438360,
%T A033504 368603716,1598231992,6889682280,29551095248,126193235194,
%U A033504 536799072924,2275560109868,9616650989560,40527780684972,170368957887656,714556104675736,2990728476330672
%N A033504 a(n)/4^n is the expected number of tosses of a coin required to obtain n+1 heads or n+1 tails.
%C A033504 The number of rooted two-vertex n-edge maps in the plane (planar with a distinguished outside face). - _Valery A. Liskovets_, Mar 17 2005
%D A033504 M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.
%D A033504 V. A. Liskovets and T. R. Walsh, Enumeration of unrooted maps on the plane, Rapport technique, UQAM, No. 2005-01, Montreal, Canada, 2005.
%H A033504 Vincenzo Librandi, <a href="/A033504/b033504.txt">Table of n, a(n) for n = 0..172</a>
%H A033504 V. A. Liskovets and T. R. Walsh, <a href="http://dx.doi.org/10.1016/j.aam.2005.03.006">Counting unrooted maps on the plane</a>, Advances in Applied Math., 36, No.4 (2006), 364-387.
%F A033504 With a different offset: Sum_{j=0..n} Sum_{k=0..n} binomial(n, j)*binomial(n, k)*min(j, k) = n*2^(n-1) + (n/2)*binomial(2*n, n). [see Klamkin]
%F A033504 a(n-1) = 4^(n-1)*b(n, n), where b(n, m) = b(n-1, m)/2 + b(n, m-1)/2 + 1; b(n, 0)=b(0, n)=0.
%F A033504 a(n) = Sum_{k=0..n, l=0..n} 2^(2n - k - l) binomial(k+l, k).
%F A033504 a(n) = (2n+1)*Sum_{0<=i,j<=n} binomial(2n, i+j)/(i+j+1). - _Benoit Cloitre_, Mar 05 2005
%F A033504 a(n) = (n+1)*(2^(2*n+1) - binomial(2*n+1,n+1)). - _Vladeta Jovovic_, Aug 23 2007
%F A033504 n*a(n) + 6*(-2*n+1)*a(n-1) + 48*(n-1)*a(n-2) + 32*(-2*n+3)*a(n-3) = 0. - _R. J. Mathar_, Dec 22 2013
%F A033504 a(n) ~ 2^(2*n+1)*n. - _Ilya Gutkovskiy_, Jul 21 2016
%e A033504 From _Jeremy Tan_, Mar 13 2018: (Start)
%e A033504 For n=1 the sequences of flips ending at two heads or two tails are:
%e A033504 HH, TT (probability 1/4 each)
%e A033504 HTH, HTT, THH, THT (1/8 each)
%e A033504 The expected number of flips is 2*2*1/4 + 3*4*1/8 = 10/4 = a(1)/4^1. (End)
%t A033504 a[n_]:=(n+1)*(2^(2*n+1)-Binomial[2*n+1,n+1])
%t A033504 a /@ Range[0,50] (* _Julien Kluge_, Jul 21 2016 *)
%o A033504 (Magma) [(n+1)*(2^(2*n+1)-Binomial(2*n+1,n+1)): n in [0..25]]; // _Vincenzo Librandi_, Jun 09 2011
%Y A033504 Cf. A002457, A100511, A103943.
%Y A033504 Cf. A000346, A130783.
%K A033504 easy,nonn,nice
%O A033504 0,2
%A A033504 Michael Ulm (ulm(AT)mathematik.uni-ulm.de)
%E A033504 Name corrected by _Jeremy Tan_, Mar 13 2018