About cookies on this site Our websites require some cookies to function properly (required). In addition, other cookies may be used with your consent to analyze site usage, improve the user experience and for advertising. For more information, please review your options. By visiting our website, you agree to our processing of information as described in IBM’sprivacy statement. To provide a smooth navigation, your cookie preferences will be shared across the IBM web domains listed here.
November 2013
<<October November December>>
This month's challenge was both too easy and too hard:
It was too easy, since Wikipedia has the formula (http://en.wikipedia.org/wiki/Hypercube) and one solution is very simple: f(5,0)=f(4,1)=f(16,15)=32
There are several families of solutions:
F(x,x-1) = 2x, so we can complete every non-trivial pair into a triplet. One that many of you discovered is f(m+2^m-1,0) = f(2^m,1)
Yevgeny Pyasetskiy sent us another sequence: f(3*k+2,k) = f(3*k+2,k+1)
Oleg Vlasii gave us a more complicated sequence:
f(p_n - 1; q_n) = f(p_n; q_n - 1)
where p_1=5, q_1=1
p_2 = 174, q_2 = 30
p_{n+1} = 34p_n - p_{n-1} + 4
q_{n+1} = 34q_n - q_{n-1} - 4
And there are also more sporadic solutions, such as:
f(7, 2) = f(9, 6) = f(336, 335)
f(10, 5)= f(64, 62)= f(4032, 4031)
f(8, 1)= f(10, 0)= f(512, 511)
f(13, 8)= f(144, 142)= f(20592, 20591)
f(7, 1)= f(8, 5)= f(224, 223)
f(15, 11)= f(105, 103) =f(10920, 10919)
f(13, 4) = f(66, 63) = f(183040, 183039)�
f(22, 16) = f(154, 151) = f(2387616, 2387615)
f(83, 48) = f(85, 51) = f(5284246911660441908808151266754560, 5284246911660441908808151266754559)
But the challenge was also too difficult, since the bonus question was an open one, See Singmaster's conjecture (http://en.wikipedia.org/wiki/Singmaster%27s_conjecture) as a reference to a similar problem.
If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com