[seqfan] Re: confused about toothpick sequence A139250!
Maximilian Hasler
maximilian.hasler at gmail.com
Thu Apr 16 00:58:20 CEST 2009
> a(2^k) = (2^(2k+1)+1)/3
> a(2^k+4) = (2^(2k+1)+13)/3
sorry, the latter should of course read
a(2^k+1) = (2^(2k+1)+13)/3
since d(n)=a(n+1)-a(n) verifies
d(2^k) = 4 for k>0
Furthermore (without proofs)
d(2^k-1) = 2^k for k>1
d( 2^k + m ) = d( 2^(k-1) + m ) for 0 <= m < 2^(k-1)-1
d( 2^k + m ) = 2^k+4 for m = 2^(k-1)-1
d( 2^k + m ) = d( 2^(k-1) + m ) for 2^(k-1) <= m < 3*2^(k-2)-2
d( 2^k + m ) = d( 2^(k-1) + m ) + 4 for m = 3*2^(k-2)-2
d( 2^k + m ) = d( 2^k - 1 ) + 24 for m = 3*2^(k-2)-1
It remains only to find d( 2^k + m ) for 3*2^(k-2) <= m = 2^k-2
... and then to prove it.
M.
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