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%I #15 Nov 23 2021 15:58:43
%S 11,101,113,131,199,223,311,337,353,373,449,461,463,641,643,661,733,
%T 829,883,919,991,1013,1031,1103,1301,1439,1451,1471,1493,1499,1697,
%U 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
%N 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.
%C 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.
%D Charles W. Trigg, Journal of Recreational Mathematics, Vol. 20(2), 1988.
%H Michael S. Branicky, <a href="/A176179/b176179.txt">Table of n, a(n) for n = 1..10000</a>
%H Mike Mudge, <a href="https://archive.org/details/PersonalComputerWorldMagazine/PCW%20199705%20May%20Created%20From%20PCW%20Cover%20CD/page/n121/mode/1up?view=theater">Morph code</a>, Hands On Numbers Count, Personal Computer World, May 1997, p. 290.
%e For the prime number n =5693 we obtain :
%e 5 + 6 + 9 + 3 = 23 ;
%e 5^2 + 6^2 + 9^2 + 3^2 = 151 ;
%e 5^3 + 6^3 + 9^3 + 3^3 = 1097.
%p 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:
%t 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 *)
%o (Python)
%o from sympy import isprime, primerange
%o def ok(p):
%o return all(isprime(sum(int(d)**k for d in str(p))) for k in [1, 2, 3])
%o def aupto(limit): return [p for p in primerange(1, limit+1) if ok(p)]
%o print(aupto(4409)) # _Michael S. Branicky_, Nov 23 2021
%Y Cf. A109181, A046704, A052034, A091366.
%K nonn,base
%O 1,1
%A _Michel Lagneau_, Apr 10 2010
%E Corrected and extended by _Harvey P. Dale_, Jan 18 2011
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