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A331944
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a(n)/ceiling(6^(n-7)) is the expected number of rolls of a fair 6-sided die in a game where the player starts at 0, advances the position by the outcome of the die's roll until exactly position n is reached. Positions beyond n are avoided by staying at the last visited position, but counting the rolls.
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1
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6, 6, 6, 6, 6, 6, 7, 43, 265, 1639, 10177, 63463, 397585, 2456503, 15189313, 93961351, 581260273, 3594003799, 22197096865, 136829952295, 843199062097, 5193720847351, 31972185139201, 196686016677319, 1209120275495089, 7428214177132183, 45613560985649761
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OFFSET
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1,1
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COMMENTS
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a(100)/6^93 = 33.333333333333370756088277230775... is the expected playing time of the "Snakes and Ladders" game on the empty board with all snakes and ladders removed. Althoen et al. (see link p. 74) cite this as "almost exactly 33 moves". One can assume that the omission of the addend of 1/3 was an obvious oversight.
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LINKS
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FORMULA
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G.f.: x*(6 - 36*x - 36*x^2 - 36*x^3 - 36*x^4 - 36*x^5 - 35*x^6 + 279930*x^7 + 279900*x^8 + 279720*x^9 + 278640*x^10 + 272160*x^11 + 233280*x^12) / ((1 - 6*x)^2*(1 + 5*x + 24*x^2 + 108*x^3 + 432*x^4 + 1296*x^5)).
a(n) = 7*a(n-1) - 46656*a(n-7) for n>13.
(End)
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PROG
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(PARI) xpected(n, m)={my(M=matrix(n+1, n+1, i, j, 0)); for(i=1, n+1, my(kadd=0); for(j=i+1, i+m, if(j>n+1, kadd++, M[i, j]=1)); M[i, i]+=kadd); vecsum((1/(matid(n)-M[1..n, 1..n]/m))[1, ])};
for(k=1, 27, my(x=xpected(k, 6)); print1(numerator(x), ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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