|
| |
|
|
A198442
|
|
Number of sequences of n coin flips that win on the last flip, if the sequence of flips ends with (1,1,0) or (1,0,0).
|
|
12
|
|
|
|
0, 0, 2, 3, 6, 8, 12, 15, 20, 24, 30, 35, 42, 48, 56, 63, 72, 80, 90, 99, 110, 120, 132, 143, 156, 168, 182, 195, 210, 224, 240, 255, 272, 288, 306, 323, 342, 360, 380, 399, 420, 440, 462, 483, 506, 528, 552, 575, 600, 624, 650, 675, 702, 728, 756, 783, 812
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
If the sequence ends with (1,1,0) Abel wins; if it ends with (1,0,0) Kain wins.
Abel(n) = A002620(n-1) = (2*n*(n - 2) + 1 - (-1)^n)/8.
Kain(n) = A004526(n-1) = floor((n - 1)/2).
Win probability for Abel = sum(Abel(n)/2^n) = 2/3.
Win probability for Kain = sum(Kain(n)/2^n) = 1/3.
Mean length of the game = sum(n*a(n)/2^n) = 16/3.
The sequence 2*a(n) is denoted as chi(n) by McKee (1994) and is the degree of the division polynomial f_n as a polynomial in x. He notes that "If x is given weight 1, a is given weight 2, and b is given weight 3, then all the terms in f_n(a, b, x) have weight chi(n)". - Michael Somos, Jan 09 2015
In Duistermaat (2010), at the end of section 11.2 The Elliptic Billiard, on page 492 the number of k-periodic fibers counted with multiplicities of the QRT root is given by equation (11.2.8) as "1/4 k^2 + 3{k/2}(1 - {k/2}) - 1 = n^2 - 1 when k = 2n, n^2 + n when k = 2n+1, for every integer k." - Michael Somos, Mar 14 2023
|
|
|
REFERENCES
|
J. J. Duistermaat, Discrete Integrable Systems, 2010, Springer Science+Business Media.
A. Engel, Wahrscheinlichkeitsrechnung und Statistik, Band 2, Klett, 1978, pages 25-26.
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = (2*n^2 - 5 - 3*(-1)^n)/8.
a(2*n) = n^2 - 1; a(2*n+1) = n*(n + 1).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) with n>=4.
G.f.: x^3*(2 - x)/((1 + x)*(1 - x)^3). - R. J. Mathar, Oct 27 2011
a(n) = a(-n) for all n in Z. a(0) = -1. - Michael Somos, Jan 09 2015
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-1 - a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Jan 09 2015
1 = a(n) - a(n+1) - a(n+2) + a(n+3), 2 = a(n) - 2*a(n+2) + a(n+4) for all n in Z. - Michael Somos, Jan 09 2015
|
|
|
EXAMPLE
|
For n = 6 the a(6) = 8 solutions are (0,0,0,1,1,0), (0,1,0,1,1,0),(0,0,1,1,1,0), (1,0,1,1,1,0), (0,1,1,1,1,0),(1,1,1,1,1,0) for Abel and
(0,0,0,1,0,0), (0,1,0,1,0,0) for Kain.
G.f. = 2*x^3 + 3*x^4 + 6*x^5 + 8*x^6 + 12*x^7 + 15*x^8 + 20*x^9 + ...
|
|
|
MAPLE
|
for n from 1 by 2 to 99 do
a(n):=(n^2-1)/4:
a(n+1):=(n+1)^2/4-1:
end do:
seq(a(n), n=1..100);
|
|
|
MATHEMATICA
|
a[ n_] := Quotient[ n^2 - 1, 4]; (* Michael Somos, Jan 09 2015 *)
|
|
|
PROG
|
(Perl) sub a {
my ($t, $n) = (0, shift);
for (0..((1<<$n)-1)) {
my $str = substr unpack("B32", pack("N", $_)), -$n;
$t++ if ($str =~ /1.0$/ and not $str =~ /1.0./);
}
return $t
(PARI) a(n)=([1, 1, 0, 0, 0, 0; 0, 0, 1, 1, 0, 0; 0, 1, 0, 0, 1, 0; 0, 0, 0, 1, 1, 0; 0, 0, 0, 0, 0, 2; 0, 0, 0, 0, 0, 2]^n)[1, 5] \\ Charles R Greathouse IV, Oct 26 2011
(Sage)
yield 0
x, y = 0, 2
while True:
yield x
x, y = x + y, x//y + 1
|
|
|
CROSSREFS
|
Cf. A000004, A002620, A004526, A004652, A005843, A008585, A008586, A023443, A028242, A047221, A047336, A052928, A242477, A265611.
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
