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

1,1

Comments

Primes in A061868.

Links

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

Example

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

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]]] &]

Prog

(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

Crossrefs

Cf. A061868.

Sequence in context: A371270 A249409 A256114 * A062657 A041589 A371122

Adjacent sequences: A345333 A345334 A345335 * A345337 A345338 A345339

Keyword

nonn,base

Author

Tanya Khovanova, Jun 14 2021

Status

approved