[seqfan] Re: Divisibilty sequences

Richard Guy rkg at cpsc.ucalgary.ca
Sun Mar 29 18:56:04 CEST 2009

Thanks, Tony!  Nothing new under the sun.  However,
I pass the reference on to other seqfans, point out
the omission of a couple of squares in

A003739 = 5 * (A001906)^2 * (b=15, c=105)^2

[as Comrie said, an uncomfortable trap for the
unwary plagiarist]

and give a further fine factorization:

A139400 = (A001906) * (A001353) * (A004254) * (b=13, c=33).

This last purports to be an order 32 recurrence.
It was a bootstrap operation.   Nothing is
proved.  It was obtained without even use of a
hand calculator, by assuming what I was trying
to show, factoring up to 25-digit numbers, using
the algebraic factors from earlier terms and
the expected form of the primitive factors
[cf Euler's `knowing' that, to factor the fifth
Fermat number, he only needed to try 257, 641,
769, ...]      R.

On Sun, 29 Mar 2009, T. D. Noe wrote:

>> Towards a multiplicative theory of divisibility sequences.
>>
>> A001542 = 2 * (A001109)
>> A003690 = 3 * (A004254)^2
>> A003696 = (A001353) * (b=14,c=68)  latter not in OEIS?
>> A003733 = 5 * (A143699)^2
>> A003739 = 5 * (A001906) * (A006238)
>> A003745 = 3 * 5^2 * (A004254) * (A004187)^3
>> A003751 = 5^3 * (A004187)^4
>> A003753 = 2^2 * (A001109) * (A001353)^2
>>         = 2 * (A001542) * (A001353)^2
>> A003755 = (A001109) * (A001906)^2
>> A003761 = (A001906) * (A004254) * (A001109)
>> A003767 = 2^3 * (A001353) * (A001109)^2
>> A003773 = 2 * (A001542)^3 = 2^4 * (A001109)^3
>> A092136 = (A004187) * (A001906)^3
>
> I image that these relations are examples of the result in the paper
>
> J.-P. Bezivin, A. Petho, and A. J. van der Poorten, A full characterisation
> of divisibility sequences, Amer. J. Math. 112 (1990), no. 6, 985-1001.
>
> that states that "if a linear recurrence sequence is a divisibility
> sequence, then it it a divisor of a product of such sequences."
>
> Tony