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A234809 a(n) = |{0 < k < n: p = k + phi(n-k) and 2*(n-p) + 1 are both prime}|, where phi(.) is Euler's totient function. 1
0, 0, 1, 2, 1, 3, 1, 4, 1, 1, 1, 5, 3, 7, 3, 1, 1, 7, 5, 9, 4, 2, 1, 9, 5, 2, 4, 3, 1, 10, 5, 14, 2, 2, 2, 1, 6, 14, 5, 4, 1, 15, 5, 16, 5, 5, 3, 17, 8, 4, 5, 6, 3, 17, 7, 5, 2, 6, 6, 17, 11, 25, 3, 5, 3, 1, 11, 25, 4, 4, 4, 22, 10, 26, 6, 7, 8, 3, 9, 26, 7, 9, 6, 25, 8, 3, 7, 9, 10, 25, 15, 6, 2, 9, 9, 2, 13, 29, 3, 7 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

Conjecture: a(n) > 0 for all n > 2.

Clearly, this implies Lemoine's conjecture which states that any odd number 2*n + 1 > 5 can be written as 2*p + q with p and q both prime.

See also A234808 for a similar conjecture.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

EXAMPLE

a(5) = 1 since 1 + phi(4) = 3 and 2*(5-3) + 1 = 5 are both prime.

a(16) = 1 since 7 + phi(9) = 13 and 2*(16-13) + 1 = 7 are both prime.

a(41) = 1 since 7 +phi(34) = 23 and 2*(41-23) + 1 = 37 are both prime.

a(156) = 1 since 131 + phi(25) = 151 and 2*(156-151) + 1 = 11 are both prime.

MATHEMATICA

f[n_, k_]:=k+EulerPhi[n-k]

p[n_, k_]:=PrimeQ[f[n, k]]&&PrimeQ[2*(n-f[n, k])+1]

a[n_]:=a[n]=Sum[If[p[n, k], 1, 0], {k, 1, n-1}]

Table[a[n], {n, 1, 100}]

CROSSREFS

Cf. A000010, A000040, A046927, A234470, A234475, A234514, A234567, A234615, A234694, A234808

Sequence in context: A213594 A325252 A243334 * A135591 A114897 A210954

Adjacent sequences:  A234806 A234807 A234808 * A234810 A234811 A234812

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 30 2013

STATUS

approved

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Last modified December 9 09:21 EST 2019. Contains 329877 sequences. (Running on oeis4.)