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A362677
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Primes whose reversal + 1 is a cube.
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0
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7, 421, 827, 4733, 32831, 57571, 228301, 364751, 892079, 1932677, 2256713, 3684211, 4213591, 6751853, 7218259, 7887707, 8497033, 15720487, 19925251, 21055813, 28756943, 29547961, 47369149, 51881849, 55033973, 57954643, 59677001, 63062963, 74415157, 88535987
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OFFSET
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1,1
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COMMENTS
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Equivalently, primes whose reversal is one less than the cube of a positive integer whose last digit is not a 1.
Since no prime starts or ends with a 0, reversing the prime will not change the number of digits, and since no prime consists only of 9's, adding 1 to its reversal will not change the number of digits, either, so the cube will have the same number of digits as the prime. Since the prime cannot begin with a 0, its reversal cannot end in 0, so the cube cannot end in 1 (and a cube ends in 1 if and only if its cube root ends in 1). Since cubes are less dense than primes, a reasonably efficient but simple way to search for all terms having at most D digits is to test each positive integer r < 10^(D/3) such that r mod 10 != 1: if the reversal of r^3 - 1 is a prime, then that prime is a term of the sequence. (End)
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LINKS
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EXAMPLE
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421 is prime and reversal(421) + 1 = 124 + 1 = 125 = 5^3.
364751 is in the sequence because it is prime and reversal(364751) + 1 = 157463 + 1 = 157464 = 54^3.
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MATHEMATICA
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Select[Prime@Range@1000000, IntegerQ@CubeRoot@(FromDigits@Reverse@IntegerDigits@#+1) &]
r = Select[Range@300, Mod[#, 3] != 1 && Mod[#, 10] != 1 &];
s = Sort@Select[FromDigits /@ Reverse /@ IntegerDigits@(r^3 - 1), PrimeQ]
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PROG
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(Python)
from sympy import isprime
s=[int(str(k**3-1)[::-1]) for k in range(1, 301)if k%10!=1 and k%3!=1]
t=[p for p in s if isprime(p)]
t.sort
print(t)
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CROSSREFS
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KEYWORD
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base,nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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