A176838 Primes p such that p^3 = q//3 for a prime q, where "//" denotes concatenation.
17, 157, 257, 277, 397, 677, 877, 997, 1217, 1697, 1997, 2417, 2777, 3257, 3517, 3697, 4157, 4177, 5077, 5197, 5897, 6277, 7417, 7517, 8377, 9397, 9497, 9677, 9857, 11197, 11597, 12157, 12457, 12697, 13397, 13477, 13877, 14057, 14197, 15017, 16477, 17597, 18097
Offset: 1
Examples
17^3 = 4913 = prime(94)//3, 17 = prime(7) is the first term. 157^3 = 3869893 = prime(32838)//3, 157 = prime(37) is the second term.
References
- J.-P. Allouche and J. Shallit, Automatic Sequences, Theory, Applications, Generalizations, Cambridge University Press, 2003.
- G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers (Fifth edition), Oxford University Press, 1980.
- F. Padberg, Zahlentheorie und Arithmetik, Spektrum Akademie Verlag, Heidelberg - Berlin 1999.
Links
- Harvey P. Dale, Table of n, a(n) for n = 1..1000
Programs
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Maple
q:= n-> isprime(iquo(n^3, 10, 'd')) and d=3: select(q, [ithprime(i)$i=1..3000])[]; # Alois P. Heinz, Sep 03 2021
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Mathematica
Select[Range[7,20000,10],PrimeQ[#]&&PrimeQ[FromDigits[Most[IntegerDigits[ #^3]]]]&] (* Harvey P. Dale, Oct 03 2013 *)
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PARI
isok(p) = if (isprime(p), my(v=divrem(p^3,10)); isprime(v[1]) && (v[2] == 3)); \\ Michel Marcus, Sep 03 2021
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Python
from sympy import isprime, primerange def ok(p): q, r = divmod(p**3, 10); return r == 3 and isprime(q) print(list(filter(ok, primerange(2, 18200)))) # Michael S. Branicky, Aug 31 2021
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