A052507 Take n-th palindromic prime p, let P = all primes having same digits; a(n) = q-p where q is smallest prime in P >p if q exists; otherwise a(n) = p-r where r is largest prime in P

0, 0, 0, 0, 0, 0, 18, 180, 180, 0, 630, 630, 720, 720, 18, 18, 0, 36, 360, 360, 0, 450, 450, 180, 180, 90, 90, 180, 180, 720, 72, 72, 0, 198, 702, 1998, 17010, 17010, 39600, 900, 900, 540, 5400, 44100, 900, 900, 18, 180, 180, 0, 180, 1800, 9900, 17100, 17100

Offset: 1

Author

Enoch Haga, Mar 17 2000

Keywords

• base
• easy
• nonn

Comments

The primes in P are required to have the same number of digits as p; thus internal 0's must remain internal 0's.

Computation of this sequence is more complicated than the Name implies. Taking each palindromic prime in turn (i.e., primes from A002385), find all permutations of its digits (without leading 0's) which are also prime (obviously there will be at least 1 such permutation). This gives the terms of A052480. Then considering each of those primes apply the rule in the Name to determine q or r or 0. - Sean A. Irvine, Nov 23 2021

Examples

			a(8)=180 because the distance from 131 to 311 is 180.
		

Links

• Sean A. Irvine, Java program (github)

Crossrefs

Cf. A002385, A052495, A052480.