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 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. Essentially the same as A035106. - R. J. Mathar, Oct 27 2011 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 REFERENCES A. Engel, Wahrscheinlichkeitsrechnung und Statistik, Band 2, Klett, 1978, pages 25-26. LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..10000 J. McKee, Computing division polynomials, Math. Comp. 63 (1994), 767-771.  MR1248973 (95a:11110) Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1). 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 a(n) = A002620(n+2) - A052928(n+2) for n >= 1. (Note A265611(n) = A002620(n+1) + A052928(n+1) for n >= 1.) - Peter Luschny, Dec 22 2015 a(n+1) = A110654(n)^2 + A110654(n)*(2 - (n mod 2)), n >= 0. - Fred Daniel Kline, Jun 08 2016 a(n) = A004526(n)*A004526(n+3). - Fred Daniel Kline, Aug 04 2016 a(n) = floor((n^2 - 1)/4). - Bruno Berselli, Mar 15 2021 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 } # Charles R Greathouse IV, Oct 26 2011 (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 (PARI) {a(n) = (n^2 - 1) \ 4}; /* Michael Somos, Jan 09 2015 */ (MAGMA) [(2*n^2-5-3*(-1)^n)/8: n in [1..60]]; // Vincenzo Librandi, Oct 28 2011 (Sage) def A198442():     yield 0     x, y = 0, 2     while True:        yield x        x, y = x + y, x//y + 1 a = A198442(); print([next(a) for i in range(57)]) # Peter Luschny, Dec 22 2015 CROSSREFS Cf. A000004, A002620, A004526, A004652, A005843, A008585, A008586, A023443, A028242, A047221, A047336, A052928, A242477, A265611. Cf. A035106, A079524, A238377. Sequence in context: A103567 A277913 A131723 * A035106 A122378 A181687 Adjacent sequences:  A198439 A198440 A198441 * A198443 A198444 A198445 KEYWORD nonn,easy AUTHOR Paul Weisenhorn, Oct 25 2011 EXTENSIONS a(12) inserted by Charles R Greathouse IV, Oct 26 2011 STATUS approved

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Last modified June 14 15:46 EDT 2021. Contains 345026 sequences. (Running on oeis4.)