OFFSET
1,1
COMMENTS
Minimal number of digits in p is 5. n such that sum of even digits equals sum of odd digits in A036301.
To find terms of this sequence, one could look at zerofree positive integers having the criterion on sum of cubes of digits. Then permute the digits to see which are prime. Using those digits with 0 and permuting then only needs the check on primality. - David A. Corneth, Dec 11 2018
LINKS
Marius A. Burtea, Table of n, a(n) for n = 1..5869
EXAMPLE
16447 is OK because 1^3 + 7^3 = 6^3 + 4^3 + 4^3.
14467 has digits in nondecreasing order (is zerofree). Of the 60 permutations, 16447, 41467, 41647, 44617, 46147, 46471, 76441 are prime. - David A. Corneth, Dec 11 2018
MATHEMATICA
oeQ[n_]:=Module[{idn = IntegerDigits[n]}, Total[Select[idn, OddQ]^3] == Total[ Select[idn, EvenQ]^3]]; Select[Range[100000], PrimeQ[#] && oeQ[#] &] (* Amiram Eldar, Dec 10 2018 after Harvey P. Dale at A076165 *)
PROG
(PARI) isok(p) = isprime(p) && (d=digits(p)) && (sum(i=1, #d, d[i]^3*if(d[i]%2, 1, -1))==0); \\ Michel Marcus, Dec 13 2018
CROSSREFS
KEYWORD
nonn,base,less
AUTHOR
Zak Seidov, Nov 01 2002
STATUS
approved