A359610 Numbers k such that the sum of the 5th powers of the digits of k is prime.

11, 101, 110, 111, 119, 128, 133, 182, 188, 191, 218, 223, 227, 229, 232, 247, 272, 274, 281, 292, 313, 322, 331, 337, 346, 359, 364, 368, 373, 377, 379, 386, 395, 397, 427, 436, 463, 472, 478, 487, 539, 557, 568, 575, 577, 586, 593, 634, 638, 643, 658, 667

Offset: 1

Author

José Hernández, Jan 06 2023

Keywords

• nonn
• base
• easy

Comments

It is easy to establish that the sequence is infinite: if x is in the sequence, so is 10*x.

Alternatively: the sequence is infinite as the sequence contains all numbers consisting of a prime number of 1s and an arbitrary number of 0s. - Charles R Greathouse IV, Jan 06 2023

Examples

			11 is a term since 1^5 + 1^5 = 2 is prime.
		

Crossrefs

A031974 is a subsequence.

Cf. A055014 (sum of the 5th powers of digits).

Cf. A028834, A108662, A225534, A210767, A245358.

Programs

Mathematica

top = 999; (* Find all terms <= top *)
For[t = 11, t <= top, t++, k = IntegerLength[t]; sum = 0;
   For[e = 0, e <= k - 1, e++, sum = sum + NumberDigit[t, e]^5];
      If[PrimeQ[sum] == True, Print[t]]]
Select[Range[670],PrimeQ[Total[IntegerDigits[#]^5]] &] (* Stefano Spezia, Jan 08 2023 *)

PARI

isok(k) = isprime(vecsum(apply(x->x^5, digits(k)))); \\ Michel Marcus, Jan 07 2023