login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A236573 Number of ordered ways to write n = k + m (k > 0, m > 0) such that p = 2*k + phi(m) - 1, prime(p + 2) + 2 and 2*n - p are all prime, where phi(.) is Euler's totient function. 1
0, 0, 0, 1, 2, 2, 1, 2, 3, 0, 2, 3, 0, 1, 0, 0, 2, 1, 2, 0, 2, 3, 1, 4, 3, 3, 8, 3, 2, 5, 5, 4, 3, 1, 2, 7, 6, 0, 8, 4, 2, 8, 4, 4, 7, 4, 4, 3, 6, 3, 5, 3, 1, 4, 6, 4, 9, 2, 4, 11, 2, 1, 5, 2, 4, 4, 1, 2, 9, 4, 0, 3, 2, 2, 5, 2, 4, 4, 1, 4, 1, 1, 1, 4, 0, 0, 3, 2, 5, 5, 0, 1, 2, 2, 1, 2, 1, 2, 2, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,5

COMMENTS

Conjecture: a(n) > 0 for all n > 712.

This implies the conjecture in A236566.

LINKS

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

EXAMPLE

a(100) = 1 since 100 = 10 + 90 with 2*10 + phi(90) - 1 = 20 + 24 - 1 = 43, prime(43 + 2) + 2 = 197 + 2 = 199 and 2*100 - 43 = 157 all prime.

a(1727) = 1 since 1727 = 956 + 771 with 2*956 + phi(771) - 1 = 1912 + 512 - 1 = 2423, prime(2423 + 2) + 2 = 21599 + 2 = 21601 and 2*1727 - 2423 = 1031 all prime.

MATHEMATICA

p[n_]:=PrimeQ[n]&&PrimeQ[Prime[n+2]+2]

f[n_, k_]:=2k+EulerPhi[n-k]-1

a[n_]:=Sum[If[p[f[n, k]]&&PrimeQ[2n-f[n, k]], 1, 0], {k, 1, n-1}]

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

CROSSREFS

Cf. A000010, A000040, A001359, A002372, A002375, A006512, A234808, A236531, A236566, A236568, A236569.

Sequence in context: A075445 A216612 A194447 * A293375 A232174 A077766

Adjacent sequences:  A236570 A236571 A236572 * A236574 A236575 A236576

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Jan 29 2014

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified February 15 20:07 EST 2019. Contains 320138 sequences. (Running on oeis4.)