cp's OEIS Frontend

A362677 Primes whose reversal + 1 is a cube.

%I A362677 #44 Apr 03 2025 11:11:36
%S A362677 7,421,827,4733,32831,57571,228301,364751,892079,1932677,2256713,
%T A362677 3684211,4213591,6751853,7218259,7887707,8497033,15720487,19925251,
%U A362677 21055813,28756943,29547961,47369149,51881849,55033973,57954643,59677001,63062963,74415157,88535987
%N A362677 Primes whose reversal + 1 is a cube.
%C A362677 From _Jon E. Schoenfield_, Jul 03 2023: (Start)
%C A362677 Equivalently, primes whose reversal is one less than the cube of a positive integer whose last digit is not a 1.
%C A362677 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)
%H A362677 Michael S. Branicky, <a href="/A362677/b362677.txt">Table of n, a(n) for n = 1..10000</a>
%e A362677 421 is prime and reversal(421) + 1 = 124 + 1 = 125 = 5^3.
%e A362677 364751 is in the sequence because it is prime and reversal(364751) + 1 = 157463 + 1 = 157464 = 54^3.
%t A362677 Select[Prime@Range@1000000,IntegerQ@CubeRoot@(FromDigits@Reverse@IntegerDigits@#+1) &]
%t A362677 r = Select[Range@300, Mod[#, 3] != 1 && Mod[#, 10] != 1 &];
%t A362677 s = Sort@Select[FromDigits /@ Reverse /@ IntegerDigits@(r^3 - 1),PrimeQ]
%t A362677 Select[Prime[Range[514*10^4]],IntegerQ[CubeRoot[1+IntegerReverse[#]]]&] (* _Harvey P. Dale_, Apr 03 2025 *)
%o A362677 (Python)
%o A362677 from sympy import isprime
%o A362677 s=[int(str(k**3-1)[::-1]) for k in range(1,301)if k%10!=1 and k%3!=1]
%o A362677 t=[p for p in s if isprime(p)]
%o A362677 t.sort
%o A362677 print(t)
%Y A362677 Cf. A007488, A167217.
%K A362677 base,nonn,easy
%O A362677 1,1
%A A362677 _Zhining Yang_, Jul 03 2023
%E A362677 a(18)-a(30) from _Jon E. Schoenfield_, Jul 03 2023