possiby double

[Ralf Stephan]

> Q for Paul Barry:  you said:

I'm not Paul but him not answering I'll try to.

> Me:  do you have proofs for these facts?  Or are they empirical
> observations only, based on the initial terms?

> %F A042965 G.f.: x(1+x)^2/(1-x-x^3+x^4); a(n)=sum{k=0..floor(n/2), binomial(n-k-1, k)A001045(n-2k)}, n>0. - Paul Barry (pbarry(AT)wit..ie), Jan 16 2005

This is 
%N A042965 Numbers not congruent 2 mod 4.

The identity with the g.f. is easy to see as both are the partial sums
of a period-3 sequence. Hint: the g.f. denominator factors to
(1-x)(1-x^3).

> %F A074227 G.f.: (1+x)^2/(1-x-x^3+x^4); a(n)=sum{k=0..floor((n+1)/2), C(n-k, k)(-1)^k*A001045(n-2k+2)}; - Paul Barry (pbarry(AT)wit.ie), Apr 26 2005

This is
%N Numbers n such that Kronecker(4,n)==mu(gcd(4,n)).

Since gcd(4,n) is 4-periodic: (1,2,1,4,...), mu of this is (1,-1,1,0,...)
and Kronecker(4,n) is (1,0,1,0...) you can see at once it's the same as
%N A042965 Numbers not congruent 2 mod 4.

so both sequences are identical.


ralf