

A234200


a(n) = {0 < k < n/2: k*phi(nk)  1 and k*phi(nk) + 1 are both prime}, where phi(.) is Euler's totient function.


8



0, 0, 0, 0, 1, 2, 1, 3, 3, 2, 3, 2, 2, 3, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 4, 2, 2, 4, 4, 3, 2, 4, 4, 3, 2, 3, 7, 2, 4, 4, 3, 7, 3, 6, 5, 3, 6, 5, 4, 3, 4, 3, 7, 4, 6, 3, 3, 4, 6, 7, 3, 7, 4, 6, 8, 2, 4, 6, 7, 8, 5, 2, 2, 10, 6, 3, 7, 7, 3, 7, 6, 2, 7, 4, 2, 6, 7, 9, 8, 4, 1, 3, 2, 4, 5, 8, 10, 4, 10, 7
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OFFSET

1,6


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 4.
(ii) If n > 3 is different from 9 and 29, then k*sigma(nk)  1 and k*sigma(nk) + 1 are both prime for some 0 < k < n.
Obviously, either of the two parts implies the twin prime conjecture. We have verified part (i) for n up to 10^8.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, New representation problems involving Euler's totient function, a message to Number Theory List, Dec. 18, 2013.


EXAMPLE

a(5) = 1 since 2*phi(3)  1 = 3 and 2*phi(3) + 1 = 5 are both prime.
a(7) = 1 since 3*phi(4)  1 = 5 and 3*phi(4) + 1 = 7 are both prime.
a(18) = 1 since 5*phi(13)  1 = 59 and 5*phi(13) + 1 = 61 are both prime.
a(91) = 1 since 13*phi(78)  1 = 311 and 13*phi(78) + 1 = 313 are both prime.
a(101) = 1 since 6*phi(95)  1 = 431 and 6*phi(95) + 1 = 433 are both prime.


MATHEMATICA

TQ[n_]:=PrimeQ[n1]&&PrimeQ[n+1]
a[n_]:=Sum[If[TQ[k*EulerPhi[nk]], 1, 0], {k, 1, (n1)/2}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A000010, A000040, A001359, A006512, A014574, A233542, A233544, A233547, A233566, A233567, A233867, A233918.
Sequence in context: A046819 A159945 A089216 * A102746 A287618 A123143
Adjacent sequences: A234197 A234198 A234199 * A234201 A234202 A234203


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Dec 21 2013


STATUS

approved



