A345336 Prime numbers p such that the sum and the product of digits of p^2 are both squares.

2, 3, 101, 103, 257, 283, 347, 401, 463, 491, 499, 509, 571, 599, 653, 661, 743, 751, 797, 1013, 1021, 1031, 1039, 1103, 1201, 1229, 1237, 1301, 1381, 1399, 1427, 1453, 1499, 1553, 1571, 1597, 1667, 1733, 1741, 1759, 1823, 2003, 2011

Offset: 1

Author

Tanya Khovanova, Jun 14 2021

Keywords

• nonn
• base

Comments

Primes in A061868.

Examples

			101^2 = 10201. The sum of the digits is 4, the product is 0: both are squares. Thus, 101 is in the sequence.
		

Links

• Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A061868.

Programs

Maple

filter:= proc(n) local L;
  if not isprime(n) then return false fi;
  L:= convert(n^2,base,10);
  issqr(convert(L,`+`)) and issqr(convert(L,`*`))
end proc:
select(filter, [$1..10000]); # Robert Israel, Jun 17 2021

Mathematica

Select[Range[3000], PrimeQ[#] && IntegerQ[Sqrt[Total[IntegerDigits[#^2]]]] && IntegerQ[Sqrt[Times @@ IntegerDigits[#^2]]] &]
Select[Prime[Range[3500]],With[{c=IntegerDigits[#^2]},AllTrue[{Sqrt[Total[c]],Sqrt[Times@@c]},IntegerQ]]&] (* Harvey P. Dale, Feb 05 2025 *)

PARI

isok(p) =if (isprime(p), my(d=digits(p^2)); issquare(vecsum(d)) && issquare(vecprod(d))); \\ Michel Marcus, Jun 14 2021

Python

from numbthy import isprime
counter = 1
for p in range (2,1090821):
    if isprime(p) and (counter <= 10000):
        pp_product = 1
        pp_sum = 0
        for digit in range (0, len(str(p*p))):
            pp_product *= int(str(p*p)[digit])
            pp_sum += int(str(p*p)[digit])
        if pow(int(pp_product**0.5),2) == pp_product:
            if pow(int(pp_sum**0.5),2) == pp_sum:
                print(counter, p)
                counter += 1; # Karl-Heinz Hofmann, Jun 17 2021