1,1

From Jon E. Schoenfield, Jul 03 2023: (Start)

Equivalently, primes whose reversal is one less than the cube of a positive integer whose last digit is not a 1.

Since no prime starts or ends with a 0, reversing the prime will not change the number of digits, and since no prime consists only of 9's, adding 1 to its reversal will not change the number of digits, either, so the cube will have the same number of digits as the prime. Since the prime cannot begin with a 0, its reversal cannot end in 0, so the cube cannot end in 1 (and a cube ends in 1 if and only if its cube root ends in 1). Since cubes are less dense than primes, a reasonably efficient but simple way to search for all terms having at most D digits is to test each positive integer r < 10^(D/3) such that r mod 10 != 1: if the reversal of r^3 - 1 is a prime, then that prime is a term of the sequence. (End)

Michael S. Branicky, Table of n, a(n) for n = 1..10000

421 is prime and reversal(421) + 1 = 124 + 1 = 125 = 5^3.

364751 is in the sequence because it is prime and reversal(364751) + 1 = 157463 + 1 = 157464 = 54^3.

Select[Prime@Range@1000000, IntegerQ@CubeRoot@(FromDigits@Reverse@IntegerDigits@#+1) &]

r = Select[Range@300, Mod[#, 3] != 1 && Mod[#, 10] != 1 &];

s = Sort@Select[FromDigits /@ Reverse /@ IntegerDigits@(r^3 - 1), PrimeQ]

Select[Prime[Range[514*10^4]], IntegerQ[CubeRoot[1+IntegerReverse[#]]]&] (* Harvey P. Dale, Apr 03 2025 *)

(Python)

from sympy import isprime

s=[int(str(k**3-1)[::-1]) for k in range(1, 301)if k%10!=1 and k%3!=1]

t=[p for p in s if isprime(p)]

t.sort

print(t)

Cf. A007488, A167217.

Sequence in context: A099742 A287033 A367867 * A331338 A269555 A027876

Adjacent sequences: A362674 A362675 A362676 * A362678 A362679 A362680

base,nonn,easy

Zhining Yang, Jul 03 2023

a(18)-a(30) from Jon E. Schoenfield, Jul 03 2023

approved