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A033504 a(n)/4^n is expected number of tosses of a coin required to obtain n heads or n tails. 5
1, 10, 66, 372, 1930, 9516, 45332, 210664, 960858, 4319100, 19188796, 84438360, 368603716, 1598231992, 6889682280, 29551095248, 126193235194, 536799072924, 2275560109868, 9616650989560, 40527780684972, 170368957887656 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

The number of rooted two-vertex n-edge maps in the plane (planar with a distinguished outside face). - Valery A. Liskovets, Mar 17 2005

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.

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) = 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 2^(2 n - k - l) Binomial(k+l, k), where the sum is from 0 to n for k and l

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

PROG

(MAGMA) [(n+1)*(2^(2*n+1)-Binomial(2*n+1, n+1)): n in [0..25]]; // Vincenzo Librandi, Jun 09 2011

CROSSREFS

Cf. A002457, A100511, A103943.

Cf. A000346, A130783.

Sequence in context: A004310 A026853 A177452 * A163615 A232062 A229003

Adjacent sequences:  A033501 A033502 A033503 * A033505 A033506 A033507

KEYWORD

easy,nonn,nice

AUTHOR

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

STATUS

approved

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Last modified November 24 08:43 EST 2014. Contains 249873 sequences.