[seqfan] Re: Seq. needed: linear comb. of primes

[Jim Nastos]

Using a starting sequence of floor(1.99^k)
1 3 7 15 31 62 123 245 489 973 1938 3856 7675 15273 30394 60485 120365
239527 476659
The sequence a(i) of smallest numbers that can be made of +/- linear
combinations of the first i numbers in the above list becomes:
1 2 3 4 5 5 4 3 2 1 3 1 0 1 1 2 3 0 1

Using a starting sequence of 2^k + k^2 (starting with k=1)
3 8 17 32 57 100 177 320 593 1124 2169 4240 8361 16580 32993 65792
131361 262468 524649
we get the smallest-possible-combinations to be:
3 5 6 4 3 1 0 4 1 1 2 0 5 3 0 6 1 3 0
which kind of looks nontrivial.

In trying to tread the 2^k line, I tried with initial sequence 2^k -
k*(-1)^k which is:
3 2 11 12 37 58 135 248 521 1014 2059 4084 8205 16370 32783 65520
131089 262126 524307
and that yields:
3 1 6 0 9 1 12 0 15 3 18 6 21 3 24 0 27 1 30

J



On Sun, Aug 5, 2012 at 4:49 PM,  <franktaw at netscape.net> wrote:
> I won't say absolutely not, but you have to choose your ground very
> precisely. Sequences growing slower than 2^n will tend to exhibit this sort
> of behavior; anything growing faster than 2^n will generally get you the
> most recent term minus all the previous ones.
>
>
> Franklin T. Adams-Watters
>
> -----Original Message-----
> From: Jim Nastos <nastos at gmail.com>
>
> Now change the problem by swapping out the set of primes with another
> set (something less dense) and there's probably some good sequences
> that can be found there.
> J
>
> On Sun, Aug 5, 2012 at 1:28 PM,  <franktaw at netscape.net> wrote:
>>
>> Just a sketch:
>>
>> If S_n is the set of possible values for the first n primes, then
>
> S_{n+1} =
>>
>> S_n U (S_n + prime(n+1)) U (S_n - prime(n+1)). Beyond about n=4, this
>
> will
>>
>> be everything even or everything odd in an interval around zero, and
>
> then a
>>
>> fringe on either side; the size of the interval will be 2 *
>
> A007504(n) - k
>>
>> for some small k. Recursively, since prime(n) << A007504(n), this will
>> continue to hold. Hence the sequence continues to alternate 0's and
>
> 1's.
>>
>>
>> A quite modest estimate on the distribution of primes suffices to
>
> complete
>>
>> the proof.
>>
>> Franklin T. Adams-Watters
>>
>>
>> -----Original Message-----
>> From: Neil Sloane <njasloane at gmail.com>
>>
>> Dear Seqfans,
>> take the first n primes and combine them with coefficients +1 and -1;
>> a(n) is the smallest number (in absolute value) that can be obtained.
>> I get by hand (for n>=1)
>> 2 1 0 1 0 1 0 1 0 1 0 1
>> How does it continue? Does 1,0 repeat for ever?
>> E.g. a(3) = 0 from -2-3+5
>>
>> a(11)=0 from 2-3-5-7+11-13+17+19-23-29+31= 0
>>
>> A214912 (which I just rescued from the "waiting to be submitted"
>
> stack)
>>
>> is an upper bound, but has a(11)=2 so is a different sequence.
>>
>> This must be a classical problem! Richard?
>>
>> Neil
>>
>>
>> --
>> Dear Friends, I have now retired from AT&T. New coordinates:
>>
>> Neil J. A. Sloane, President, OEIS Foundation
>> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA
>> Phone: 732 828 6098; home page: http://NeilSloane.com
>> Email: njasloane at gmail.com
>>
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