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A304688
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Primes p > 5 such that there is a polygonal number P_s(k) (with s >= 3, k >= 5) equal to p-1.
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1
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29, 37, 67, 71, 79, 97, 101, 113, 127, 137, 149, 157, 191, 197, 211, 233, 239, 257, 277, 281, 307, 317, 331, 337, 367, 373, 379, 397, 401, 409, 449, 457, 461, 487, 491, 541, 547, 569, 577, 607, 617, 631, 641, 653, 659, 673, 677, 701, 709, 727, 739, 743, 751, 757, 787, 821, 827, 853, 857
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OFFSET
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1,1
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COMMENTS
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For all primes p > 5, at least one polygonal number exists with P_s(k) = p - 1 when k = 3 or 4, dependent on p mod 6; this is why the sequence is defined for k >= 5.
For s = {14,17} no such P_s(k) exists, since P_14(k) + 1 and P_17(k) + 1 are composites.
For k = 4*m + 1, m > 0 all the numbers P_s(k) + 1 are even and > 2, so they cannot be prime.
For s = 2*m, m >= 2, k = 2*j + 1, j >= 1 all the numbers P_s(k) + 1 are even and > 2, so they cannot be prime.
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LINKS
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EXAMPLE
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a(1)-1 = 29-1 = 28 = P_3(7);
a(2)-1 = 37-1 = 36 = P_3(8) = P_4(6).
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MATHEMATICA
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lst = {}; Do[
If[Resolve[
Exists[{s, k},
Prime[m] == 1/2 k (4 + k (-2 + s) - s) + 1 && s >= 3 && k >= 5],
Integers], lst = Union[lst, {Prime[m]}]], {m, 4, 150}]; lst
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PROG
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(PARI) isok(p) = {if ((p > 5) && isprime(p), for (s=3, p, if (ispolygonal(p-1, s, &k) && (k>=5), return (1)); ); ); return (0); } \\ Michel Marcus, May 18 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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