CF A110036 vs. A090678 Non-Squashing Partitions mod 2
Paul D. Hanna
pauldhanna at juno.com
Sat Jul 9 04:50:41 CEST 2005
Seqfans,
I have verified that A110036 and 2*A090678 coincide,
at least for the first 4095 terms (n>=1), by
|A110036(n)| = 2*A090678(n) for n>1.
Further, the g.f. for A090678 led me to the following g.f. for A110036:
A(x) = (1-x+3*x^2+x^3)/(1+x^2) - 2*Sum_{k>=1}
x^(3*2^(k-1))/Product_{j=0..k} (1+x^(2^j)).
The PARI code I used to test this is as follows:
{ a(n)=subst(contfrac(1+sum(k=0,#binary(n+1),1/x^(2^k)))[n+1],x,0) }
{
A=(1-x+3*x^2+x^3)/(1+x^2)-2*sum(k=1,12,x^(3*2^(k-1))/prod(j=0,k,1+x^(2^j)
)) }
sum(n=0,4096,a(n)*x^n) - A +O(x^4097)
which returns all zeros to verify equivalence.
Paul
> Below I paste Neil's "non-squashing" partitions mod 2 (A090678).
>
> If true, I wonder why.
> Thanks,
> Paul
>
> URL: http://www.research.att.com/projects/OEIS?Anum=A090678
> Sequence:
> 1,1,1,0,0,1,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0,1,0,1,1,0,1,0,1,0,
>
> 0,1,0,1,1,0,0,1,0,1,0,1,1,0,1,0,0,1,1,0,1,0,0,1,0,1,1,0,1,0,
>
> 1,0,0,1,0,1,1,0,0,1,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0,1,0,1,1,0,
> 1,0,1,0,0,1,1,0,1,0,0,1,0,1,0
> Name: A088567 mod 2.
> %I A110036
> %S A110036
> 1,-1,2,0,0,-2,0,2,0,-2,2,0,-2,0,0,2,0,-2,2,0,0,-2,0,2,-2,0,
> 2,0,-2,0,0,2,0,-2,2,0,0,-2,0,2,0,-2,2,0,-2,0,0,2,-2,0,2,0,0,
> -2,0,2,-2,0,2,0,-2,0,0,2,0,-2,2,0,0,-2,0,2,0,-2,2,0,-2,0,0,
> 2,0,-2,2,0,0,-2,0,2,-2,0,2,0,-2,0,0,2,-2,0,2,0,0,-2,0,2,0
> %N A110036 Constant term of the partial quotients of the continued
> fraction
> expansion of 1 + Sum_{n>=0} 1/x^(2^n), where each partial quotient
> has the form {x + a(n)} after the initial constant term of 1.
> %C A110036 Suggested by Ralf Stephan.
> %e A110036 1 + 1/x + 1/x^2 + 1/x^4 + 1/x^8 + 1/x^16 + ... =
> [1; x - 1, x + 2, x, x, x - 2, x, x + 2, x, x - 2, ...].
> %o A110036 (PARI) contfrac(1+sum(n=0,10,1/x^(2^n)))
> %O A110036 0
> %K A110036 ,cofr,sign,
> %A A110036 Paul D Hanna (pauldhanna at juno.com), Jul 08 2005
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