A341932 a(n) = largest k < n such that the decimal concatenation n||n-1||n-2||...||n-k is prime, or -1 if no such prime exists.

-1, -1, 0, 0, 1, 0, -1, 4, -1, -1, 3, 0, -1, 0, -1, -1, -1, 0, -1, 0, -1, -1, 3, 0, 1, 12, -1, 4, -1, 0, -1, 0, -1, -1, 1, -1, -1, 0, -1, -1, -1, 0, 1, 0, -1, -1, 43, 0, 31, -1, -1, 4, 15, 12, -1, 28, 9, -1, 1, 0, 13, 0, -1, -1, -1, -1, -1, 0, 57, -1, 1, 0, -1

Offset: 0

Author

Chai Wah Wu, Feb 23 2021

Keywords

• sign
• base

Comments

a(82) = 81, are there any other n such that a(n) = n-1?

Primes p such that a(p) > 0: 7, 53, 73, 79, 89, 103, ...

n such that a(n) > A341702(n): 7, 10, 22, 46, 48, 53, 55, 73, ...

Similar argument as in A341716 shows that if n > 3 and a(n) >= 0, then n-a(n) is odd, a(n) !== 2 (mod 3) and 2n-a(n) !== 0 (mod 3).

Examples

			a(22) = 3 since 22212019 is prime.
		

Crossrefs

Cf. A052088, A052089, A341701, A341702, A341715, A341716, A341717, A341931.

Programs

Python

from sympy import isprime
def A341932(n):
    k, m, r = n, n-1, 0 if isprime(n) else -1
    while m > 0:
        k = int(str(k)+str(m))
        if isprime(k):
            r = n-m
        m -= 1
    return r

Formula

a(n) = n-A341931(n) >= A341702(n).