[seqfan] Re: sum floor(n/(m+k)) = n
Peter Pein
petsie at dordos.net
Mon Feb 9 13:48:50 CET 2009
Peter Pein schrieb:
> Leroy Quet schrieb:
>> Consider sequence A145265.
>>
>> A145265: A positive integer n is included if there exists a positive integer m such that sum{k>=0} floor(n/(m+k)) = n.
>>
>>
>> Does A145265 contain all those positive integers that are congruent to
>> 0, 1, 4, 5, 8, 15 (mod 18), and no other positive integers?
>>
>> This might be easily proved. Still, I think this might be an interesting question.
>>
>> Thanks,
>> Leroy Quet
>
> Don't know about a proof but the sequence of the (least) m which belongs to
> A145265(n) starts:
> 1, 2, 2, 3, 5, 6, 6, 7, 7, 8, 10, 11, 11, 12, 12, 13, 15, 16, 16, 17, 17, 18,
> 20, 21, 21, 22, 22, 23, 25, 26, 26, 27, 27, 28, 30, 31, 31, 32, 32, 33, 35,
> 36, 36, 37, 37, 38, 40, 41, 41, 42, 42, 43, 45, 46, 46, 47, 47, 48, 50, 51,
> 51, 52, 52, 53, 55, 56, 56, 57, 57, 58, 60, 61, 61, 62, 62, 63, 65, 66, 66,
> 67, 67, 68, 70, 71, 71, 72, 72, 73, 75, 76, 76, 77, 77, 78, 80, 81, 81, 82,
> 82, 83
>
> m(n+1)-m(n) is apparently periodic: deltam(n)={1, 0, 1, 2, 1, 0,....)
>
> Peter
>
> P.S. I hope I did not overlook a similar post to this topic.
>
... under these assumptions the g.f. for A145265 is:
(1 + 2*x - x^2 + 4*x^3 + 3*x^4) / ((-1 + x)^2*(1 - x + x^2)*(1 + x + x^2))
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