cool problem! the bijection is: for the first spot, A=0 and B=1 after the first spot, if the letter changes, the number doesn't. this gives the pairing: AAAAAA OIOIOI AAAABB OIOIIO AAABBA OIOOII AABAAB OIIIOO AABBAA OIIOOI AABBBB OIIOIO ABAABA OOOIII ABBAAA OOIIOI ABBABB OOIIIO ABBBBA OOIOII BAAAAB IIOIOO BAABAA IIOOOI BAABBB IIOOIO BABBAB IIIOOO BBAAAA IOOIOI BBAABB IOOIIO BBABBA IOOOII BBBAAB IOIIOO BBBBAA IOIOOI BBBBBB IOIOIO erich On Mar 15, 2012, at 7:47 PM, Fred lunnon wrote:
From the formula for A(n, k) given in http://oeis.org/A183135 it can easily be deduced that the number of "doublet strings" of length 2 n on k = 2 symbols equals the number of binary strings with n zeros and n ones.
But if you were not told this, what constructive bijection between these sets might establish it directly?
For n = 3 the sets are (in lexicographical order!) :
AAAAAA OOOIII AAAABB OOIOII AAABBA OOIIOI AABAAB OOIIIO AABBAA OIOOII AABBBB OIOIOI ABAABA OIOIIO ABBAAA OIIOOI ABBABB OIIOIO ABBBBA OIIIOO BAAAAB IOOOII BAABAA IOOIOI BAABBB IOOIIO BABBAB IOIOOI BBAAAA IOIOIO BBAABB IOIIOO BBABBA IIOOOI BBBAAB IIOOIO BBBBAA IIOIOO BBBBBB IIIOOO
Fred Lunnon
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