[seqfan] R. Mathar to V. Librandi (by way of admin)

[Olivier Gerard]

> ---------- Forwarded message ----------
> From: Richard Mathar <mathar at strw.leidenuniv.nl>
> To: seqfan at seqfan.eu
> Date: Tue, 07 Jul 2009 00:01:29 +0200



vl> Date: Mon, 6 Jul 2009 14:24:29 +0200 (CEST)
vl> From: "vincenzo.librandi at tin.it" <vincenzo.librandi at tin.it>
vl> Subject: [seqfan]  sequence
vl> To: seqfan at list.seqfan.eu
vl> Cc: vincenzo.librandi at tin.it
vl>
vl> Hi all
vl>
vl> Ask you formula sequence:
vl>
vl> 3, 3, 3, 4, 4, 5, 5, 6, 8, 8, 10, 11, 11, 12, 14, 16, 16, 18, 19,
19, 21, 22, 24, 27, 28, 28, 29, 29, 30, 36, 37 ,...,
vl>
vl> or formula sequence:
vl>
vl> 0, 0, 0, 1, 1, 2, 2, 3, 5, 5, 7, 8, 8, 9, 11, 13, 13, 15, 16, 16,
18, 19, 21, 24, 25, 25, 26, 26, 27, 33,...,
vl> quando ai termini della prima sequenza si sottrae 3.
...
vl>
vl> Thans fot Hasler and Mathar.
vl>
vl> OK A008507, I now it.
vl>
vl> volevo però una sequenza senza utilizzare i numeri prmi.

I claim that the first sequence =3+A008507(n)=3+second sequence
is by definition bound to the primes, so it is not possible to define
3+A008507(n)
without mentioning implicitly or explicitly the primes in some form.

vl>
vl> Lo scopo:
vl>
vl> If K = A067076 (0,1,2,4,5,7,8,10, and so on) tali che 2*k+3=primo

This means that the k=A067076(n) are essentially a way of rewriting
the halves of the prime sequence.
vl>
vl> If S_K =A008507 (3,3,3,4,4,5,5,6,8, an so on)

So this is 3+A008507(n), another way of writing the halves of the
prime sequence.

vl> che chiamo "Salti di K", infatti per
vl>
vl> K       S_k
vl> 0        3
vl> 1        3  (nessun salto da 0 a 1)
vl> 2        3
vl> 4        4 (un salto perchè manca il 3)
vl> 5        4
vl> 7        5 (un salto, manca 6)
vl> 8        5
vl> 10      6
vl> 13      8 (due salti perch mancano 11e 12)
vl> 14      8
vl> 17     10 (due salti perchè 15 e 16)
vl> 19     11
vl> and so on
vl>
vl> se alle due colonne aggiungiamo la serie dei numeri naturali
(0,1,2,3,4,5,6,7,, ..) avremo
vl>
vl> 0+3+0=3
vl> 1+3+1=5
vl> 2+3+2=7
vl> 4+4+3=11
vl> 5+4+4=13
vl> 7+5+5=17
vl> 8+5+6=19
vl> 10+6+7=23
vl>
vl> and so on
vl>
vl> Come potremo legare il tutto ?

So these summations mean
  k+3+A008507(n)+n=prime(n+1),
where k=A067076(n).  Now this simply is a result of the definition
2k+3=prime(n),
You have essentially written 2k+3 as k+3+A008507(n)+n, which is equivalent to
  k+3=3+A008507(n)+n
->   k=A008507(n)+n
->   k="Number of odd composites less than n-th-odd prime" +n
And this is (up to fine work) almost trivial because on the right hand side
the number of the odd composites less than the n-th-odd prime plus n
(the prime index) is
ALL of the odd numbers up to the n-th prime. Now since k=(prime(n)-3)/2
is essentially half of all numbers (even and odd), we are practically done
with the proof. (I am obviously not interested in any details, because it's
9 minutes before midnight...) So basically the message in the summations above
is that in all numbers in the range up to prime(n)-3 there are even numbers, odd
composites and other primes, which may be counted in various ways
according to taste.

I suspect that can be summarized in some relation like A008507(n)+n-1=A067076(n)

Richard