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A245510
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Records in A245509: smallest m > 1 such that the first odd number greater than m^k is prime for every 0 < k < n, but not for k = n.
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7
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OFFSET
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1,1
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COMMENTS
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For more comments and a program, see A245509. a(9), if it exists, certainly exceeds 1050000000. It is not clear whether this sequence is infinite, nor whether a(n) is defined for every n.
For n > 3, a(n) is always odd, because A245509(i) can exceed 3 only when i is odd. Therefore to find more terms, it suffices to find odd bases m such that m+2, m^2+2, m^3+2, m^4+2, ..., m^N+2 is a long list of primes. - Jeppe Stig Nielsen, Sep 09 2022
For any term m beyond a(8) that exists, each of the following holds:
m = p - 2, where p is a prime (so m is odd);
m == 0 (mod 3);
m == {-1, 0, 1} (mod 5);
m == {-1, 0, 1} (mod 11);
consequently, m mod 330 is one of 9 values: {21, 45, 99, 111, 165, 219, 231, 285, 309}.
(End)
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LINKS
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EXAMPLE
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a(4) = 105 because 105 is the smallest m such that the first odd numbers after m^k are prime for k = 1,2,3, but composite for k = 4.
909+2, 909^2+2, 909^3+2, 909^4+2 and 909^5+2 are five primes, but 909^6+2 is composite, and 909 is minimal with this property. Therefore, a(6)=909 (and A245509(909)=6). - Jeppe Stig Nielsen, Sep 09 2022
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MATHEMATICA
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f[n_] := Block[{d = If[ OddQ@ n, 2, 1], m = 1, t}, While[t = n^m + d; EvenQ@ t || PrimeQ@ t, m++]; m]; t = Table[0, {25}]; k = 2; While[k < 29000000, a = f@ k; If[ t[[a]] == 0, t[[a]] = k; Print[{a, k}]]; k++]; t (* Robert G. Wilson v, Aug 04 2014 *)
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PROG
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(PARI) a(n) = for(k=1, oo, c=0; for(i=1, n-1, if(isprime(k^i+(k%2)+1), c++)); if(c==n-1&&!isprime(k^n+(k%2)+1), return(k)))
n=1; while(n<10, print1(a(n), ", "); n++) \\ Derek Orr, Jul 27 2014
(PARI) upto(n)=v=vector(n); forstep(m=3, +oo, 2, k=1; while(ispseudoprime(m^k+2), k++); if(k<=n&&v[k]==0, v[k]=m-(k==3)*7; print(v); vecprod(v)!=0&&return(v))) \\ Jeppe Stig Nielsen, Sep 09 2022
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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a(4) and example corrected by Derek Orr, Jul 27 2014
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STATUS
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approved
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