A050231 a(n) is the number of n-tosses having a run of 3 or more heads for a fair coin (i.e., probability is a(n)/2^n).
0, 0, 1, 3, 8, 20, 47, 107, 238, 520, 1121, 2391, 5056, 10616, 22159, 46023, 95182, 196132, 402873, 825259, 1686408, 3438828, 6999071, 14221459, 28853662, 58462800, 118315137, 239186031, 483072832, 974791728, 1965486047, 3960221519, 7974241118, 16047432332, 32276862265
Offset: 1
References
- W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 2nd ed. New York: Wiley, p. 300, 1968.
Links
- T. D. Noe, Table of n, a(n) for n = 1..300
- David Broadhurst, Multiple Landen values and the tribonacci numbers, arXiv:1504.05303 [hep-th], 2015.
- Simon Cowell, A Formula for the Reliability of a d-dimensional Consecutive-k-out-of-n:F System, arXiv preprint arXiv:1506.03580 [math.CO], 2015.
- Erich Friedman, Illustration of initial terms
- T. Langley, J. Liese, and J. Remmel, Generating Functions for Wilf Equivalence Under Generalized Factor Order, J. Int. Seq. 14 (2011) # 11.4.2.
- Eric Weisstein's World of Mathematics, Run
- Index entries for linear recurrences with constant coefficients, signature (3,-1,-1,-2).
Programs
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Magma
R
:= PowerSeriesRing(Integers(), 60); [0,0] cat Coefficients(R!( x^3/((1-2*x)*(1-x-x^2-x^3)) )); // G. C. Greubel, Jun 01 2025 -
Mathematica
LinearRecurrence[{3, -1, -1, -2}, {0, 0, 1, 3}, 50] (* David Nacin, Mar 07 2012 *)
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PARI
concat([0,0], Vec(1/(1-2*x)/(1-x-x^2-x^3)+O(x^99))) \\ Charles R Greathouse IV, Feb 03 2015
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Python
def a(n, adict={0:0, 1:0, 2:1, 3:3}): if n in adict: return adict[n] adict[n]=3*a(n-1)-a(n-2)-a(n-3)-2*a(n-4) return adict[n] # David Nacin, Mar 07 2012
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SageMath
def A050231_list(prec): P.
= PowerSeriesRing(QQ, prec) return P( x^3/((1-2*x)*(1-x-x^2-x^3)) ).list() a=A050231_list(41); a[1:] # G. C. Greubel, Jun 01 2025
Formula
a(n) = 2^n - tribonacci(n+3), see A000073. - Vladeta Jovovic, Feb 23 2003
G.f.: x^3/((1-2*x)*(1-x-x^2-x^3)). - Geoffrey Critzer, Jan 29 2009
a(n) = 2 * a(n-1) + 2^(n-4) - a(n-4) since we can add T or H to a sequence of n-1 flips which has HHH, and H to one which ends in THH and does not have HHH among the first (n-4) flips. - Toby Gottfried, Nov 20 2010
a(n) = 3*a(n-1) - a(n-2) - a(n-3) - 2*a(n-4), a(0)=0, a(1)=0, a(2)=1, a(3)=3. - David Nacin, Mar 07 2012
Comments