A052507 - OEIS

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 <p if r exists; otherwise a(n) = 0.

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

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OFFSET

1,7

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

LINKS

Table of n, a(n) for n=1..55.

Sean A. Irvine, Java program (github)

EXAMPLE

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

CROSSREFS

Cf. A002385, A052495, A052480.

Sequence in context: A227023 A239581 A213350 * A071910 A121038 A341370

Adjacent sequences: A052504 A052505 A052506 * A052508 A052509 A052510

KEYWORD

base,easy,nonn

AUTHOR

Enoch Haga, Mar 17 2000

STATUS

approved