Explanation
Let a_1=c. Then a_2=a_1+1=c+1,a_4=a_2+2=c+3. Since the sequence is increasing,it follows that a_3=c+2. We Prove that a_n=c+n1 for all n1. Indeed, if n=2k for some integer k this follows by induction on k. Suppose that a_{2^k}=c+2^k1. Then a_{2^{k+1}}=a_{2.2^k}=a_{2^k}+2^k=c+2^{k+1}1. If 2^k<n<2^{k+1}, then c+2^{k}1=a_{2^k}<a_{2^k+1}<...<a_n<...<a_{2^k+1}=c+2^{k+1}1 add this is possible only if a_n=c+n1.
Next we prove that c=1. Suppose that c2 and let p<q be two consecutive prime numbers greater than c. We have a_{qc+1}=c+qc=q, hence qc+1 is a prime and clearly qc+1p. It follows that for any consecutive prime numbers p<q we have qc+1. The numbers (c+1)!+2,(c+1)!+3...,(c+1)!+c+1 are all composite, hence if p and q are the consecutive primes such that p<(c+1)!+2<(c+1)!+c+1<q then qp>c1, a contradiction. It follows that c=0 and a_n=n for all n1.
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