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 A225868 Numbers n for which max_{2 <= k <= (n-2)/2} Sum_{d>1, d|n+k, k|n+d} 1 = 3. 5
 6, 9, 12, 13, 16, 19, 24, 31, 32, 48, 53, 83, 89, 107, 113, 131, 139, 149, 167, 179, 191, 199, 227, 233, 251, 263, 409, 431, 449, 467, 479, 503, 587, 599, 631, 659, 683, 719, 769, 827, 983, 1019, 1091, 1259, 1367, 1409, 1439, 1487, 1499, 1511, 1583, 1619, 1979 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Terms >= 53 are primes p such that p + 2 is either prime or semiprime or, maybe, cube of a prime. However, according to calculations of Peter J. C. Moses, up to 4.2*10^13 there are no p in the sequence for which p + 2 is cube of a prime. One can prove that if such a prime p exists, then it is necessary (but not sufficient) that all numbers of the quadruple {r, 2*r - 1, 4*r^2 - 6*r + 3, (2*r - 1)^3 - 2} should be primes, where r == 19 (mod 30) is defined by the equality (2r-1)^3 - 2 = p. The first 3 suitable r are 229, 3109, 17449. But the corresponding p's are not in the sequence. We conjecture that all primes of the sequence are Chen primes, that is, all of them are in A109611. LINKS Peter J. C. Moses, Table of n, a(n) for n = 1..300 MATHEMATICA f[n_] := (m = 0; Do[s = Sum[ Boole[ Divisible[n+d, k]], {d, Divisors[n+k] // Rest}]; If[s > m, m = s], {k, 2, (n-2)/2}]; m); Reap[ For[n = 1, n <= 2000, n = If[n < 53, n+1, NextPrime[n]], If[f[n] == 3, Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jul 09 2013, after Vladimir Shevelev *) CROSSREFS Cf. A225867, A188579, A109611. Sequence in context: A087022 A315951 A081392 * A185177 A185179 A072546 Adjacent sequences:  A225865 A225866 A225867 * A225869 A225870 A225871 KEYWORD nonn AUTHOR Vladimir Shevelev, May 18 2013 STATUS approved

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Last modified April 21 04:56 EDT 2019. Contains 322310 sequences. (Running on oeis4.)