OFFSET
1,1
COMMENTS
Recall that A119752 is the sequence defined recursively by a(1)=2 and a(k) is the first even number greater than a(k-1) such that 2a(k)+1 is prime and a(k)+a(j)+1 is prime for all 1<=j<k.
a(n)=A(n,1), the first element of each sequence A(n) defined recursively as follows. Let A(1)=A119752, that is, A(1,k)=A119752(k). Then A(n) is the sequence defined recursively as follows: (1) A(n,1) is the first even number not in any A(m), 1<=m<n, such that 2A(n,1)+1 is prime. (2) A(n,k) is the first even number greater than A(n,k-1), not in any A(m), 1<=m<n, such that 2A(n,k)+1 is prime. (3) A(n,k)+A(n,j)+1 is prime for all 1<=j<k.
Let us say that a positive integer t is eventually in A(n) if 2^p*t, p>=1, is in A(n). For example, if t=19, then A(36,1)=2^5*19. The only numbers not eventually in some A(n) are all powers 2^p such that 2^(p+1)+1 is not prime, (so p is not 0, 1, 3, 7, 15) and Sierpinski numbers of the second kind, namely odd integers t such that 2^p*t+1 is composite for all p>=1. The smallest known example is t=78,557.
LINKS
Eric Weisstein's World of Mathematics, Sierpinski Number of the Second Kind.
EXAMPLE
PROG
(PARI) isok(va, k, n) = if (isprime(2*k+1), for (i=1, n-1, if (! isprime(va[i]+k+1), return(0))); return(1));
avec(start, lim) = my(list=List(), n=2, ok=1); listput(list, start); while(ok, my(k=list[n-1]+2); while (!isok(list, k, n), k+=2; if (k>lim, ok=0; break)); if (ok, listput(list, k)); n++; ); Vec(list);
findfirst(v, lim) = forstep (n=2, lim, 2, if (isprime(2*n+1) && !vecsearch(v, n), return(n)));
lista(lim) = my(list = List()); listput(list, 2); my(v=avec(2, lim)); while (1, my(new = findfirst(v, lim)); if (! new, break); my(w = avec(new, lim)); v = vecsort(concat(v, w)); listput(list, new); ); Vec(list); \\ Michel Marcus, Mar 06 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
Walter Kehowski, Jan 10 2007
STATUS
approved