[seqfan] Re: computability of A080075 Proth numbers

Klaus Brockhaus klaus-brockhaus at t-online.de
Mon Aug 16 15:38:01 CEST 2010


Apparently A080075(n)-1 = A116882(n+1); found by Superseeker, checked 
for n  <= 1300. A080075 and A116882 do not crossreference each other.

%N A116882 A number n is included if (highest odd divisor of n)^2 <= n.

Georgi Guninski schrieb:
> A080075 Proth numbers: of the form k*2^m + 1 for k odd, m >= 1 and 2^m >
> k
>
> machine analysis seems to suggest these relations (verified to 10 000
> against the OEIS bfile)
>
> a(2*n-1) = (4*a(n - 1) - 3)
> a(2*n) = ( 2*a(n - 1) + 2*a(n) - 3 )
> a(2*n+1) = (4*a(n) - 3)
>
> a(n+1) = (-a(n - 1) + 2*a(n))
>
> OR (these are the roots of a quadratic relation, a possible approach may
> be to take the smaller proth of them)
>
> a(n+1) = (-2*a(n - 1) + 3*a(n))
>
> doubling formulas resemble "divide and conquer sequences" - do proth
> numbers have generating function ?
>
>   





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