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%I #17 Feb 14 2022 06:35:22
%S 1,2,1,5,4,8,7,7,14,11,16,20,19,19,22,29,26,31,35,32,37,37,38,46,50,
%T 49,53,52,44,61,61,68,61,74,71,74,79,79,82,89,82,95,94,98,89,95,109,
%U 113,112,112,119,112,121,124,127,134,131,136,140,133,134,151
%N a(n) = (3/2)*(1+prime(n)) - prime(n+1).
%C Conjecture: a(n) > 0 for all n (cf. A062234).
%C Note that a(1) = 3/2 hence offset is 2.
%C There are many cases of two successive terms of the same value, the first case is a(8)=a(9)=7: p(8)=19, p(9)=23, p(10)=29, (3/2)*(1+19)-23 = (3/2)*(1+23)-29 = 7.
%C The first case of 3 equal successive terms is a(691..693)=2588 for corresponding 4 consecutive primes primes p(691..694)= 5189, 5197, 5209, 5227.
%C The first case of 4 equal successive terms is a(12702874..12702878)=15579672 for corresponding 5 consecutive primes primes p(12702874..12702878)= 231159373,231159389,231159413,231159449,231159503.
%C Also of interest are cases with a(n)>a(n-1), e.g., a(27..29): 53, 52, 44 (the general tendency is, of course, increasing a(n) with n).
%H Zak Seidov, <a href="/A207480/b207480.txt">Table of n, a(n) for n = 2..1001</a>
%p a:= n-> 3*(1+ithprime(n))/2-ithprime(n+1):
%p seq(a(n), n=2..63); # _Alois P. Heinz_, Feb 14 2022
%t (3(#[[1]]+1)/2)-#[[2]]&/@Partition[Prime[Range[2,70]],2,1] (* _Harvey P. Dale_, Jul 27 2016 *)
%o (PARI) a(n) = my(p=prime(n)); (3/2)*(1+p) - nextprime(p+1); \\ _Michel Marcus_, Feb 14 2022
%Y Cf. A062234.
%K nonn
%O 2,2
%A _Zak Seidov_, Feb 18 2012
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