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For any integer n, let d(n) be the smallest k > 0 such that at least one of n-k or n+k is a prime number; we build an undirected graph G on top of the prime numbers as follows: two consecutive prime numbers p and q are connected iff at least one of d(p) or d(q) equals q-p; a(n) is the number of terms in the n-th connected component of G (ordered by least element).
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%I #50 Jun 15 2019 19:34:07

%S 4,2,3,2,7,3,3,3,3,2,2,8,2,7,2,5,4,4,2,4,5,3,2,2,3,4,3,3,2,2,5,8,7,4,

%T 2,5,3,2,2,2,2,3,4,4,3,5,4,2,2,2,3,2,3,6,3,2,2,4,6,2,3,2,4,3,4,2,5,4,

%U 3,7,4,2,2,2,3,4,4,4,2,5,4,2,2,5,3,3,2

%N For any integer n, let d(n) be the smallest k > 0 such that at least one of n-k or n+k is a prime number; we build an undirected graph G on top of the prime numbers as follows: two consecutive prime numbers p and q are connected iff at least one of d(p) or d(q) equals q-p; a(n) is the number of terms in the n-th connected component of G (ordered by least element).

%C Each connected component of G has at least two elements.

%C Is the sequence bounded?

%e The first terms, alongside the corresponding components, are:

%e n a(n) n-th component

%e -- ---- --------------

%e 1 4 {2, 3, 5, 7}

%e 2 2 {11, 13}

%e 3 3 {17, 19, 23}

%e 4 2 {29, 31}

%e 5 7 {37, 41, 43, 47, 53, 59, 61}

%e 6 3 {67, 71, 73}

%e 7 3 {79, 83, 89}

%e 8 3 {97, 101, 103}

%e 9 3 {107, 109, 113}

%e 10 2 {127, 131}

%o (PARI) d(p) = for (k=1, oo, if (p-k>0 && isprime(p-k), return (k), isprime(p+k), return (k)))

%o v=1; p=2; forprime (q=p+1, oo, if (d(p)==q-p || d(q)==q-p, v++, print1 (v", "); if (n++==87, break); v = 1); p=q)

%Y Cf. A000040, A051700.

%K nonn

%O 1,1

%A _Rémy Sigrist_, Jun 02 2019