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November 2013

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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: webmster@us.ibm.com