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A176179
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Primes such that the sum of digits, the sum of the squares of digits and the sum of 3rd powers of their digits is also a prime.
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2
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11, 101, 113, 131, 199, 223, 311, 337, 353, 373, 449, 461, 463, 641, 643, 661, 733, 829, 883, 919, 991, 1013, 1031, 1103, 1301, 1439, 1451, 1471, 1493, 1499, 1697, 1741, 1949, 2089, 2111, 2203, 2333, 2441, 2557, 3011, 3037, 3307, 3323, 3347, 3491, 3583, 3637, 3659, 3673, 3853, 4049, 4111, 4139, 4241, 4337, 4373, 4391, 4409
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OFFSET
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1,1
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COMMENTS
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See A091365 for the exceptions for the case where the sum of the digits of p is not prime, but the sum of the cubes of the digits of p is prime.
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REFERENCES
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Charles W. Trigg, Journal of Recreational Mathematics, Vol. 20(2), 1988.
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LINKS
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Mike Mudge, Morph code, Hands On Numbers Count, Personal Computer World, May 1997, p. 290.
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EXAMPLE
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For the prime number n =5693 we obtain :
5 + 6 + 9 + 3 = 23 ;
5^2 + 6^2 + 9^2 + 3^2 = 151 ;
5^3 + 6^3 + 9^3 + 3^3 = 1097.
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MAPLE
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with(numtheory):for n from 2 to 10000 do:l:=evalf(floor(ilog10(n))+1):n0:=n:s1:=0:s2:=0:s3:=0:for m from 1 to l do:q:=n0:u:=irem(q, 10):v:=iquo(q, 10):n0:=v :s1:=s1+u:s2:=s2+u^2:s3:=s3+u^3:od:if type(n, prime)=true and type(s1, prime)=true and type(s2, prime)=true and type(s3, prime)=true then print(n):else fi:od:
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MATHEMATICA
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okQ[n_]:=Module[{idn=IntegerDigits[n]}, And@@PrimeQ[Total/@{idn, idn^2, idn^3}]]; Select[Prime[Range[600]], okQ] (* Harvey P. Dale, Jan 18 2011 *)
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PROG
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(Python)
from sympy import isprime, primerange
def ok(p):
return all(isprime(sum(int(d)**k for d in str(p))) for k in [1, 2, 3])
def aupto(limit): return [p for p in primerange(1, limit+1) if ok(p)]
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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