Cleaning out the junk

David Wilson davidwwilson at comcast.net
Thu Nov 22 01:50:47 CET 2007


Well I don't either, since I apparently clipped away part of this old note.

I guess the whole thing can be forgotten.

----- Original Message ----- 
From: "Olivier Gerard" <olivier.gerard at gmail.com>
To: "David Wilson" <davidwwilson at comcast.net>
Sent: Wednesday, November 21, 2007 1:24 AM
Subject: Re: Cleaning out the junk


> Dear David,
>
> I may be dense, but I don't understand how you compute p_2 for
> instance.  What is the definition ? I guess it is a variation on
> integer partitions because of p_3 but I can't find which.
>
> regards,
>
> Olivier
>
>
> On Nov 18, 2007 7:31 PM, David Wilson <davidwwilson at comcast.net> wrote:
>> I'm cleaning up junk from my sequence researches. Here is an old note I
>> wrote to myself in case anyone is interested. I don't think I ever acted 
>> on
>> it by noting or creating sequences.
>>
>> When we compute a few terms of p_k for k = 1 through 10, we get
>>
>> p_1 = (1 0 0 0 0 0 0 0 0 0 0 0 0 0 ...)
>>
>> p_2 = (1 1 1 1 0 -1 -2 -3 -3 -1 2 6 10 11 ...)
>>
>> p_3 = (1 1 2 3 5 7 11 15 22 30 42 56 77 101 ...)
>>
>> p_4 = (1 1 2 4 7 13 23 42 75 135 242 434 779 1396 ...)
>>
>> p_5 = (1 1 2 4 8 15 29 55 106 202 387 739 1414 2702 ...)
>>
>> p_6 = (1 1 2 4 8 16 31 61 119 234 458 898 1759 3447 ...)
>>
>> p_7 = (1 1 2 4 8 16 32 63 125 247 490 970 1922 3806 ...)
>>
>> p_8 = (1 1 2 4 8 16 32 64 127 253 503 1002 1994 3970 ...)
>>
>> p_9 = (1 1 2 4 8 16 32 64 128 255 509 1015 2026 4042 ...)
>>
>> p_10 = (1 1 2 4 8 16 32 64 128 256 511 1021 2039 4074 ...)
>>
>> p_1 = A000007 and p_3 = A000041 and the termwise limit p_inf = A011782 
>> are
>> straightforward. None of the other sequences appear to be in the OEIS.
>>
>> ------------------------------------------------------------------------
>>
>> p_2 is interesting. It has the recurrence
>>
>> p_2(0) = 1
>>
>> p_2(n) = SUM(k with 1 <= k^2 <= n; (-1)^(k+1) * p_2(n-k^2)) (n >= 1)
>>
>> or more prettily
>>
>> p_2(0) = 1
>>
>> p_2(n) = p_2(n-1) - p_2(n-4) + p_2(n-9) - p_2(n-16) + p_2(n-25) - ...
>>
>> where terms with negative index are dropped in the latter sum.
>>
>> p_2 is the only one of the p_n having negative elements. Here I provide 
>> more
>> terms, formatted to emphasize the regularity of the sign changes:
>>
>> p_2 = (
>>
>> 1 1 1 1 0
>>
>> -1 -2 -3 -3 -1
>>
>> 2 6 10 11 8 0
>>
>> -14 -29 -39 -38 -18
>>
>> 22 74 123 144 110 6
>>
>> -161 -352 -491 -484 -251
>>
>> 235 896 1528 1825 1452 191
>>
>> -1892 -4317 -6164 -6243 -3488
>>
>> 2482 10788 18957 23140 19085 3858
>>
>> -22025 -52833 -77224 -80198 -47899
>>
>> 25330 129563 234774 292984 249938 66467
>>
>> -254632 -645419 -966200 -1028145 -651774
>>
>> 244756 1550984 2903014 3703662 3262048 1057768
>>
>> -2919617 -7868439 -12071433 -13154408 -8798790
>>
>> 2156064 18502260 35840006 46747073 42441704 16042331
>>
>> -33149312 -95721387 -150601090 -167980582 -117979248
>>
>> 15610769 219874960 441758527 589147690 550604495 235523164
>>
>> -371916746 -1161853462 -1876169578 -2141169812 -1572640509
>>
>> 52991977 2601849672 5435999338 7413974493 7123950308 3377603117
>>
>> ...)
>>
>> ------------------------------------------------------------------------
>>
>> Let P_k be the g.f. for p_k. I am fairly convinced that
>>
>> P_k = SUM(j = 0 to inf; (-(x^k))^j * Q_j)
>>
>> where the Q_j is the g.f. for q_j with
>>
>> q_0 = (1 1 2 4 8 16 32 64 128 256 512 ...) = A011782
>>
>> q_1 = (1 3 9 22 54 126 290 654 1458 3214 7026 ...)
>>
>> q_2 = (1 5 20 64 189 519 1367 3475 8611 20887 49807 ...)
>>
>> ...
>>
>> q_1 and q_2 are not in the OEIS. I have not looked at q_3 or beyond.
>>
>>
>>
>>
>
>
> -- 
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> 11/21/2007 10:01 AM
> 






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