

A165646


The revolver sequence.


0



0, 0, 0, 0, 2, 3, 4, 5, 18, 43, 60, 84, 294, 472, 724, 1077, 3504, 5807, 10396, 15944, 43664, 84308, 137004, 217728, 582966, 1134304, 1883360, 3225812, 8964134, 15461472, 27796942, 45814765, 123307634, 233218013, 396304692, 663041846
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OFFSET

1,5


COMMENTS

a(n) is in how many ways you can load a revolver with n chambers, so that if a person survives after the first shot, he/she has better chances of survival for the second shot if the shooter continues rather than spins. a(n) <= A000031(n) as A000031(n) gives the number of possible revolver loadings.


LINKS

Table of n, a(n) for n=1..36.
Tanya Khovanova, Heard on the Street


EXAMPLE

Based on the famous interview question where the revolver has six chambers and the shooter loads two adjacent bullets. As the answer to this question is to continue, a(6) must be at least 1.


MATHEMATICA

<< Combinatorica` colors[x_] := Table[PadLeft[Table[1, {n, i}], x], {i, 0, x}] continueBetter[list_] := (len = Length[list]; c0 = 0; c00 = 0; rotateN = 0; While[rotateN < len, newList = RotateLeft[list, rotateN]; If[newList[[1]] == 0, c0++; If[newList[[2]] == 0, c00++ ]]; rotateN++ ]; If[c0 > 0, c00/c0 > c0/len, False]) continueNum[num_] := (neckNum = Length[colors[num]]; ans = 0; count = 1; While[count <= neckNum, ans = ans + Length[Select[ListNecklaces[num, colors[num][[count]], Cyclic], continueBetter[ # ] &]]; count++ ]; ans) Table[continueNum[n], {n, 2, 15}]


PROG

(PARI) { a(n) = sum(z=0, n, sum(r=1, min(ceil(zz^2/n)1, nz), sumdiv(gcd([n, z, r]), d, eulerphi(d) * binomial(z/d  1, r/d  1) * binomial((nz)/d  1, r/d  1) )/r )) } [From Max Alekseyev, Sep 25 2009]


CROSSREFS

A000031
Sequence in context: A162657 A145520 A061412 * A261639 A024635 A217679
Adjacent sequences: A165643 A165644 A165645 * A165647 A165648 A165649


KEYWORD

nonn,uned,changed


AUTHOR

Tanya Khovanova, Sep 23 2009


EXTENSIONS

Extended by Max Alekseyev, Sep 25 2009


STATUS

approved



