A050232 a(n) is the number of n-tosses having a run of 4 or more heads for a fair coin (i.e., probability is a(n)/2^n).
0, 0, 0, 1, 3, 8, 20, 48, 111, 251, 558, 1224, 2656, 5713, 12199, 25888, 54648, 114832, 240335, 501239, 1042126, 2160676, 4468664, 9221281, 18989899, 39034824, 80103276, 164126496, 335808927, 686182387, 1400438814, 2854992080, 5814293120, 11829648225, 24046855887, 48840756608
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
- 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.
- 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,-1,-2).
Programs
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Magma
R
:= PowerSeriesRing(Integers(), 50); [0,0,0] cat Coefficients(R!( x^4/((1-2*x)*(1-x-x^2-x^3-x^4)) )); // G. C. Greubel, Jun 01 2025 -
Mathematica
Flatten[With[{tetrnos=LinearRecurrence[{1,1,1,1},{0,1,1,2},50]}, Table[ 2^n- Take[tetrnos,{n+3}],{n,40}]]] (* Harvey P. Dale, Dec 02 2011 *) LinearRecurrence[{3,-1,-1,-1,-2}, {0,0,0,1,3}, 31] (* Ray Chandler, Aug 03 2015 *)
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PARI
a(n)=([0,1,0,0,0; 0,0,1,0,0; 0,0,0,1,0; 0,0,0,0,1; -2,-1,-1,-1,3]^(n-1)*[0;0;0;1;3])[1,1] \\ Charles R Greathouse IV, Feb 09 2017
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Python
def a(n, adict={0:0, 1:0, 2:0, 3:1, 4:3}): if n in adict: return adict[n] adict[n]=3*a(n-1) - a(n-2) - a(n-3) - a(n-4) - 2*a(n-5) return adict[n] # David Nacin, Mar 07 2012
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SageMath
def A050232_list(prec): P.
= PowerSeriesRing(QQ, prec) return P( x^4/((1-2*x)*(1-x-x^2-x^3-x^4)) ).list() a=A050232_list(41); a[1:] # G. C. Greubel, Jun 01 2025
Formula
a(n) = 2^n - tetranacci(n+4), see A000078. - Vladeta Jovovic, Feb 23 2003
G.f.: x^4/((1-2*x)*(1-x-x^2-x^3-x^4)). - Geoffrey Critzer, Jan 29 2009
a(n) = 3*a(n-1) - a(n-2) - a(n-3) - a(n-4) - 2*a(n-5). - Wesley Ivan Hurt, Apr 23 2021
Comments