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a(n) = prime((a(n-1) + a(n-2)) / 2) for n>1, with a(0) = 3, a(1) = 5.
1

%I #14 Nov 15 2020 01:57:55

%S 3,5,7,13,29,73,233,883,4049,22013,140269,1036183,8771489,84270997,

%T 910550899,10977409981,146618640533,2155882170713,34701120973507,

%U 608312884303213,11560371513545909,237170393590520353,5232858369551752099,123736201926723358439

%N a(n) = prime((a(n-1) + a(n-2)) / 2) for n>1, with a(0) = 3, a(1) = 5.

%t nxt[{a_, b_}] := {b, Prime[(a + b)/2]}; Transpose[NestList[nxt,{3,5},17]][[1]] (* The program generates only the first 18 terms of the sequence; the 19th term is too large for Mathematica to generate it *) (* _Harvey P. Dale_, Apr 24 2016 *)

%o (C)

%o #include <stdio.h>

%o #define TOP (1ULL<<32)

%o typedef unsigned long long U64;

%o int main() {

%o U64 i, j, p = 2, prpr = 3, prev = 5;

%o unsigned int *primes = (unsigned int*)malloc(TOP/4);

%o char *c = (char*)malloc(TOP/2);

%o memset(c, 0, TOP/2);

%o for (primes[1] = 2, i = 3; i < TOP; i+=2)

%o if (c[i>>1]==0)

%o for (primes[p++]=i, j=i*i>>1; j<TOP/2; j+=i) c[j]=1;

%o for (printf("3, 5, "); (prev+prpr)/2<p; prpr=prev, prev=i)

%o i = primes[(prev+prpr)/2], printf("%llu, ", i);

%o }

%Y Cf. A000040, A107327.

%K nonn,hard,more

%O 0,1

%A _Alex Ratushnyak_, May 09 2013

%E a(15)-a(18) from _Max Alekseyev_, Jan 30 2014

%E a(19)-a(23) from _Jinyuan Wang_, Nov 14 2020