[seqfan] Re: confused about toothpick sequence A139250!

[Maximilian Hasler]

>  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.