= <1,1,3,7,17,41,99,239,577,1393,...>,
= <0,1,2,5,12,29,70,169,408, 985,...>.
Formulars:
p(n) = ((1+sqrt(2))^n + (1-sqrt(2))^n)/2,
q(n) = ((1+sqrt(2))^n - (1-sqrt(2))^n)/(2sqrt(2)).
For n odd we get solutions with the minus sign. As p and q are both odd
we always get a solution of the original problem.
Solutions: (R,B) = (3,1), (15, 6), (85, 35), (493, 204), ...
Appendix:
Find Binomial Coefficients with 2*C(n,k) = C(m,k).
These come in pairs as we have the identity:
C(r+s,s) / C(r+s+t,s) = C(r+t,t) / C(r+s+t,t).
So 1/2 = C(n,1) / C(2n,1) = C(2n-1,n) / C(2n,n)
which we call the trivial solutions.
References:
Martin Gardner;
Wheels, Life, and Other Mathematical Amusements
Freeman (1983) New York
(german: Martin Gardner's Mathematische Denkspiele, Hugendubel)
Chap 5: Nontransitive Dice and Other Probability Paradoxes
n-card monte: R red and B black cards are given pick a pair
prob(same color) = (C(R,2)+C(B,2))/C(R+B,2)
Karl Josef Jacquemain;
Denkspiele f\"ur Tennisspieler,
it 1727, Insel Verlag, 1995
Problem 11: Linksh\"anderduell
Find R, L with 2*C(L,2) = C(R+L,2).
Ivan Morris;
99 neunmalkluge Denkspiele,
dtv 1243, Deutscher Taschenbuch Verlag, 1977
Problem 75: Gaststudenten
Find R, L with 2*C(L,5) = C(R+L,5).
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