A033504 a(n)/4^n is the expected number of tosses of a coin required to obtain n+1 heads or n+1 tails.
1, 10, 66, 372, 1930, 9516, 45332, 210664, 960858, 4319100, 19188796, 84438360, 368603716, 1598231992, 6889682280, 29551095248, 126193235194, 536799072924, 2275560109868, 9616650989560, 40527780684972, 170368957887656, 714556104675736, 2990728476330672
Offset: 0
Examples
From _Jeremy Tan_, Mar 13 2018: (Start) For n=1 the sequences of flips ending at two heads or two tails are: HH, TT (probability 1/4 each) HTH, HTT, THH, THT (1/8 each) The expected number of flips is 2*2*1/4 + 3*4*1/8 = 10/4 = a(1)/4^1. (End)
References
- M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.
- V. A. Liskovets and T. R. Walsh, Enumeration of unrooted maps on the plane, Rapport technique, UQAM, No. 2005-01, Montreal, Canada, 2005.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..172
- V. A. Liskovets and T. R. Walsh, Counting unrooted maps on the plane, Advances in Applied Math., 36, No.4 (2006), 364-387.
Programs
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Magma
[(n+1)*(2^(2*n+1)-Binomial(2*n+1,n+1)): n in [0..25]]; // Vincenzo Librandi, Jun 09 2011
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Mathematica
a[n_]:=(n+1)*(2^(2*n+1)-Binomial[2*n+1,n+1]) a /@ Range[0,50] (* Julien Kluge, Jul 21 2016 *)
Formula
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]
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.
a(n) = Sum_{k=0..n, l=0..n} 2^(2n - k - l) binomial(k+l, k).
a(n) = (2n+1)*Sum_{0<=i,j<=n} binomial(2n, i+j)/(i+j+1). - Benoit Cloitre, Mar 05 2005
a(n) = (n+1)*(2^(2*n+1) - binomial(2*n+1,n+1)). - Vladeta Jovovic, Aug 23 2007
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
a(n) ~ 2^(2*n+1)*n. - Ilya Gutkovskiy, Jul 21 2016
Extensions
Name corrected by Jeremy Tan, Mar 13 2018
Comments