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A233296
a(n) = |{0 < k < n: k*prime(n-k) + 1 is prime}|
5
0, 1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 4, 2, 2, 5, 4, 1, 6, 1, 2, 5, 4, 4, 4, 3, 3, 2, 4, 6, 5, 4, 5, 6, 7, 7, 6, 5, 8, 6, 4, 5, 7, 8, 4, 6, 7, 6, 10, 7, 4, 8, 11, 9, 11, 6, 5, 5, 8, 8, 10, 5, 7, 10, 10, 11, 7, 6, 6, 12, 6, 10, 11, 6, 11, 7, 11, 8, 12, 7, 7, 9, 13, 9, 10, 14, 7, 13, 8, 10, 11, 9, 14, 10, 14, 17, 14, 13, 8, 12, 12
OFFSET
1,4
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 1. Similarly, for any integer n > 2, k*prime(n-k) - 1 (or k^2*prime(n-k) - 1) is prime for some 0 < k < n.
(ii) Let n > 3 be an integer. Then k + prime(n-k) is prime for some 0 < k < n. Also, if n is not equal to 13, then k^2 + prime(n-k)^2 is prime for some 0 < k < n.
LINKS
Z.-W. Sun, On a^n+ bn modulo m, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014-2016.
EXAMPLE
a(17) = 1 since 17 = 14 + 3 with 14*prime(3) + 1 = 14*5 + 1 = 71 prime.
a(19) = 1 since 19 = 18 + 1 with 18*prime(1) + 1 = 18*2 + 1 = 37 prime.
MATHEMATICA
a[n_]:=Sum[If[PrimeQ[k*Prime[n-k]+1], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 07 2013
STATUS
approved