A094077

[Henry in Rotherhithe]

We have
(a) each positive integer is in exactly one of A001951 and A001952. The
dichotomy comes from:
if sqrt(2)*ceiling[n/sqrt(2)]<n+1 then A001951(ceiling[n/sqrt(2)])=n,
otherwise (2+sqrt(2))*ceiling[n/(2+sqrt(2))]<n+1 and
A001952(ceiling[n/(2+sqrt(2))])=n
(b) A001952(n)=A001951(n)+2n from the formulae

So inductively, if A094077(2n-1)=A001951(n)
then A094077(2n) =A094077(2n-1)+2n =A001951(n)+2n =A001952(n)
while A094077(2n+1) is the smallest number not in A001951(k) or A001952(k)
for k=1,2,...,n i.e. A001951(n+1) since this is less than A001952(n+1)

Incidentally, this and the formulae demonstrate the conjecture
A094077(2n)/A094077(2n-1) = A001952(n)/A001951(n) is about
(2+sqrt(2))/sqrt(2)=1+sqrt(2)

Henry Bottomley


> -----Original Message-----
> From: Klaus Brockhaus
> Sent: 3 February 2006 22:27
> To: seqfan
> Subject: A094077
>
>
>
> Apparently
>
> A001951 a(n) = floor[n*sqrt 2], 	(1, 2, 4, 5, 7, 8, 9, 11,
> 12, 14, 15, 16, 18, ...)
>
> and
>
> A001952: a(n) = floor[n*(2 + sqrt 2)], (3, 6, 10, 13, 17, 20, 23,
> 27, 30, 34, 37, 40, 44, ...)
>
> are the bisection of
>
> A094077: a(1)=1 and, for n>1, a(n)=a(n-1)+n if n is even, and
> a(n)=smallest positive integer which has not yet appeared in the
> sequence if n is odd, (1, 3, 2, 6, 4, 10, 5, 13, 7, 17, 8, 20, 9,
> 23, 11, 27, 12, 30, 14, 34, 15, 37, 16, 40, 18, 44, ...).
>
> >From the definitions this is by no means obvious, but perhaps
> someone has an idea how it can be proved.
>
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