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A304031
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Number of ways to write 2*n as p + 2^k + 5^m with p prime and 2^k + 5^m a product of at most three distinct primes, where k and m are nonnegative integers.
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6
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0, 1, 1, 3, 3, 2, 2, 3, 3, 4, 3, 5, 4, 4, 3, 4, 5, 7, 4, 7, 4, 8, 7, 6, 7, 6, 5, 5, 5, 7, 5, 8, 5, 5, 8, 6, 9, 9, 6, 8, 6, 6, 7, 8, 4, 7, 8, 7, 3, 10, 6, 7, 8, 7, 7, 9, 5, 8, 7, 6, 5, 5, 6, 3, 11, 7, 9, 12, 8, 12, 10, 11, 11, 9, 7, 9, 7, 8, 8, 11, 7, 11, 8, 9, 15, 11, 8, 9, 8, 9
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OFFSET
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1,4
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COMMENTS
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a(n) > 0 for all 1 < n <= 10^10 with the only exception n = 3114603841, and 2*3114603841 = 6219442049 + 2^3 + 5^10 with 6219442049 prime and 2^3 + 5^10 = 3*17*419*457 squarefree.
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LINKS
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Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)
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EXAMPLE
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a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 = 2^1 + 5^0 prime.
a(7) = 2 since 2*7 = 7 + 2^1 + 5^1 with 7 = 2^1 + 5^1 prime, and 2*7 = 11 + 2^1 + 5^0 with 11 and 2^1 + 5^0 both prime.
a(42908) = 2 since 2*42908 = 85751 + 2^6 + 5^0 with 85751 prime and 2^6 + 5^0 = 5*13, and 2*42908 = 69431 + 2^14 + 5^0 with 69431 prime and 2^14 + 5^0 = 5*29*113.
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MATHEMATICA
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qq[n_]:=qq[n]=SquareFreeQ[n]&&Length[FactorInteger[n]]<=3;
tab={}; Do[r=0; Do[If[qq[2^k+5^m]&&PrimeQ[2n-2^k-5^m], r=r+1], {k, 0, Log[2, 2n-1]}, {m, 0, Log[5, 2n-2^k]}]; tab=Append[tab, r], {n, 1, 90}]; Print[tab]
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CROSSREFS
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Cf. A000040, A000079, A000351, A005117, A118955, A156695, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656, A303660, A303702, A303821, A303932, A303934, A303949, A304032, A304081.
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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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