login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

a(n)/4^n is the expected number of tosses of a coin required to obtain n+1 heads or n+1 tails.
6

%I #53 Oct 01 2023 10:52:44

%S 1,10,66,372,1930,9516,45332,210664,960858,4319100,19188796,84438360,

%T 368603716,1598231992,6889682280,29551095248,126193235194,

%U 536799072924,2275560109868,9616650989560,40527780684972,170368957887656,714556104675736,2990728476330672

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

%C 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 M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.

%D V. A. Liskovets and T. R. Walsh, Enumeration of unrooted maps on the plane, Rapport technique, UQAM, No. 2005-01, Montreal, Canada, 2005.

%H Vincenzo Librandi, <a href="/A033504/b033504.txt">Table of n, a(n) for n = 0..172</a>

%H 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 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 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 a(n) = Sum_{k=0..n, l=0..n} 2^(2n - k - l) binomial(k+l, k).

%F a(n) = (2n+1)*Sum_{0<=i,j<=n} binomial(2n, i+j)/(i+j+1). - _Benoit Cloitre_, Mar 05 2005

%F a(n) = (n+1)*(2^(2*n+1) - binomial(2*n+1,n+1)). - _Vladeta Jovovic_, Aug 23 2007

%F 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 a(n) ~ 2^(2*n+1)*n. - _Ilya Gutkovskiy_, Jul 21 2016

%e From _Jeremy Tan_, Mar 13 2018: (Start)

%e For n=1 the sequences of flips ending at two heads or two tails are:

%e HH, TT (probability 1/4 each)

%e HTH, HTT, THH, THT (1/8 each)

%e The expected number of flips is 2*2*1/4 + 3*4*1/8 = 10/4 = a(1)/4^1. (End)

%t a[n_]:=(n+1)*(2^(2*n+1)-Binomial[2*n+1,n+1])

%t a /@ Range[0,50] (* _Julien Kluge_, Jul 21 2016 *)

%o (Magma) [(n+1)*(2^(2*n+1)-Binomial(2*n+1,n+1)): n in [0..25]]; // _Vincenzo Librandi_, Jun 09 2011

%Y Cf. A002457, A100511, A103943.

%Y Cf. A000346, A130783.

%K easy,nonn,nice

%O 0,2

%A Michael Ulm (ulm(AT)mathematik.uni-ulm.de)

%E Name corrected by _Jeremy Tan_, Mar 13 2018