1,1

The sequence could begin with 1 by convention. The sequence in which d can be 1 is a subsequence. The elements are assumed composite so as to exclude the Sophie Germain primes (A005384) and (A045536). All terms except 8 and 9 are odd numbers in square-free semiprimes (A006881) or 3-almost-primes (A014612). The only square is 9, the first few cubes are 8, 125, 357911=71^2, 6967871=191^3 and the first few 3-almost primes are 935=5*11*17, 1859=11*13^2, 11123=7^2*227, 305015=5*53*1151. The first 3-almost-prime divisible by 9 is 149049=3^2*16561. All elements not divisible by 3 are 5 or 11 mod 12. I have been unable to find an element with more than 3 prime factors. If one exists, it must be very large. One reason is that the number of divisors grows rapidly with the number of factors. For example, if n is square-free with k factors, then tau(n)=2^k. The condition that the 2^k-1 numbers n+d+1 be prime is then quite strong. Another reason is that one or more of the numbers n+d+1 may always be composite. For example, if n=p^5, p prime, then both p^5+p^4+1 and p^5+p+1 are composite.

a(n)=nth number such that n is composite and n+d+1 is prime for all divisors d greater than 1.

a(9)=935=5*11*17 since the divisors d greater than one are {5,11,17,55,85,187,935} and all elements in the set of n+d+1, {941,947,953,991,1021,1123,1871}, are primes.

with(numtheory); P:=[]: for w to 1 do for k from 2 do #start at 1, get first element 1 if not isprime(k) and isprime(2*k+1) then S:=divisors(k) minus {1, k}; Q:=map(z-> z+k+1, S); if andmap(isprime, Q) then P:=[op(P), k]; print(nops(P), k, ifactor(k)) fi; fd:=fopen("C:/temp/n+d+1=prime-1st-1000.txt", APPEND); fprintf(fd, "%d ", x); fclose(fd); if nops(P)=1000 then break fi; fi; od od;

Cf. A005384, A045536, A006881, A014612.

Sequence in context: A046874 A038344 A050771 this_sequence A041136 A041915 A036764

Adjacent sequences: A120773 A120774 A120775 this_sequence A120777 A120778 A120779

nonn

Walter Kehowski (wkehowski(AT)cox.net), Jul 05 2006