SEQUENCES AND SERIES INVOLVING THE SEQUENCE OF COMPOSITE NUMBERS

Denoting by pn and cn the nth prime number and the nth composite number, respectively, we prove that both the sequence (xn)n≥1, defined by xn=∑k=1n (ck


Introduction. We use the following notation:
(i) π(x) is the number of prime numbers less than or equal to x, (ii) p n is the nth prime number, (iii) c n is the nth composite number; c 1 = 4,c 2 = 6,..., (iv) log 2 n = log(log n).
In 1967, Bojarincev [1] estimated c n and found out that If n is large enough, then . (1. 3) It was shown in [3] that, for n large enough we have Of course, the irregularities in the distribution of the prime numbers imply irregularities in the distribution of the composed numbers.Although in the sequence of the composite numbers, large "gaps" cannot be found.For the sequence (p n ) n≥1 we have lim sup n→∞ (p n+1 − p n ) = ∞, while in the case of the sequence (c n ) n≥1 we have 1 ≤ c k+1 − c k ≤ 2, with the specification that c k+1 − c k = 2 if and only if the number c k+1 is prime.In this case, denoting c k + 1 = p m , it is proved in [3] that Since c n ∼ n, we can expect that "in mean" the sequence (c n ) n≥1 behaves as the sequence of the natural numbers.It is readily seen that, since the series The situations to be analyzed in the present paper are, however, more complicated and lean on a series of facts, namely (1.7) , for every k > 0; (1.8) (see [2]); (1.9) (see [5]); (1.10) [4]). (1.11)

Asymptotic behavior of certain series
Proposition 2.1.The sequence Proof.We have where extends to all values of k such that c k +1 is a prime number, that is, where extends to the indices k such that c k + 1 is prime.Now (1.6) and (2.3) imply that and the proof ends.
Proof.The following cases can arise: (a) if both c k+1 and c k+1 +1 are composite numbers, then Next, denote by π 2 (x) the number of the prime numbers p ≤ x such that p + 2 is prime.It is known (see [2]) that

.7)
By taking into consideration the above four cases, it follows that Proof.By analyzing the cases occurring in the proof of the preceding proposition, we see that c k+2 − 2c k+1 + c k = 0 in the cases (a) and (d), while (c k+2 − 2c k+1 + c k ) 2 = 1 in the cases (b) and (d).Consequently, t n = 2π(n)+ O(1). is convergent.

Studying the convergence of certain series
First we prove the following fact.
Proof.By (1.10), it follows that which implies the desired conclusion.
Proof of Proposition 3.1.By (1.1), we get and then we have, in view of the above lemma, Now, replacing p n by means of (1.9), we get If we let x = p cn in (1.8), then we deduce For x = c pn , also (1.8) implies It is readily seen that π(c m ) + m + 1 = c m and for m = p n this implies that π(c pn ) + p n + 1 = c pn .Then, (3.9) becomes  (3.20) then it follows that the sequence ( n i=1 ε i ) n≥1 is bounded.Consequently, the convergence of our series follows by Dirichlet's criterion.

Proposition 3 . 4 .
The series ∞ n=1 c n+1 − c n − 1 p n (3.21) is convergent.Proof.Denoting S n = n k=1 (c k+1 − c k − 1)/p k , it follows that S n = p k ≤n 1/p k , where extends to the indices k such that c k + 1 = p m with prime p m .In view of (1.5), it follows that k = k(m) ∼ p m ∼ m log m, hence p k ∼ k log k ∼ m log 2 m. then (1.7) implies that the series ∞ n=1 (c n+1 − c n − 1)/p n is convergent.
In view of(1.11), it then follows that It follows by Proposition 2.1 that both the series Let s be a real number.If y n .2) Since p m ∼ m log m, we get mx m /p m (p m + mx m ) ∼ 1m log 2 m and then (1.7) implies that the series ∞ m=1 mx m /p m (p m + mx m ) is convergent; denote its sum by a. n→∞ → b − a. (2.3)By (1.10) lim n→∞ (p n /n − log n − log 2 n) = −1, hence (1.6) and (2.3) imply that lim n→∞ x n = γ + b − a + 1.(2.4)Theproof is completed.Remark 2.2.