A135667 A028391

[Maximilian Hasler]

> Richard, couldn't you either put a direct link (even if search with
>  more hits than those you aim at) or some lines (say %N, %S) of the
>  sequences, in such mails ?
>  Thanks in advance (in the name of all seqfans...)!
>  Maximilian
PS: since it's my birthday today ;-)

%N A135667 a(n)=floor[n-n^(1/2)].
%N A028391 n - [sqrt(n)].
are indeed related by
A028391( n ) = A135667( n+1 )

Proof:
A135667( n+1 ) = [ n+1 - (n+1)^(1/2) ]
 = if issquare(n+1) then n+1 - sqrt(n+1) else n+1-[sqrt(n+1)]-1
 = if issquare(n+1) then n+1 - [sqrt(n)+1] else n+1-[sqrt(n)]-1
 = if issquare(n+1) then n - [sqrt(n)] else n-[sqrt(n)]
 = n - [sqrt(n)] = A028391( n )

Thus, A135667( n ) = A028391( n-1 )
and might be deleted, maybe adding
%F A028391 a(n) = floor[ (n+1) - (n+1)^(1/2)].
or
%F A028391 a(n) = b(n+1) for b(n) = floor[n-n^(1/2)]

PS: Now I see:
%F A028391 Also ceiling( n+1 - sqrt(n+1) ).
so one might just add to this line:
 = n+1 - floor( sqrt(n+1))

Maximilian