A235061 Number of ways to write n = k*(k+1)/2 + m with k > 0 and m > 0 such that prime(k*(k+1)/2) + phi(m) is prime, where phi(.) is Euler's totient function.

0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 2, 2, 1, 3, 2, 2, 1, 1, 1, 2, 3, 3, 2, 1, 1, 3, 3, 3, 3, 2, 3, 2, 3, 2, 3, 4, 2, 4, 4, 2, 3, 1, 4, 4, 1, 4, 3, 3, 3, 4, 5, 4, 3, 1, 3, 3, 5, 4, 4, 5, 1, 5, 3, 5, 5, 4, 2, 2, 5, 4, 5, 1, 1, 6, 5, 6, 6, 4, 5, 5, 8, 5, 2, 1, 4, 6, 4, 6, 7, 3, 3, 6, 4, 7, 5, 2, 7, 6

Offset

1,13

Comments

Conjecture: (i) a(n) > 0 for all n > 8.

(ii) Any integer n > 100 can be written as k^2 + m with k > 0 and m > 0 such that phi(k^2) + prime(m) is prime.

(iii) Any integer n > 187 can be written as k + m with k > 0 and m > 0 such that prime(k) + phi(m) is a triangular number. Also, each integer n > 45 can be written as k + m with k > 0 and m > 0 such that prime(k) + phi(m)/2 is a triangular number.

(iv) Any integer n > 293 can be written as k + m with k > 0 and m > 0 such that prime(k) + phi(m) is a square. Also, each integer n > 83 can be written as k + m with k > 0 and m > 0 such that prime(k) + phi(m)/2 is a square.

Links

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

Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Example

a(10) = 1 since 10 = 2*(2+1)/2 + 7 = 3 + 7 with prime(3) + phi(7) = 5 + 6 = 11 prime.
a(20) = 1 since 20 = 3*(3+1)/2 + 14 = 6 + 14 with prime(6) + phi(14) = 13 + 6 = 19 prime.
a(86) = 1 since 86 = 12*(12+1)/2 + 8 = 78 + 8 with prime(78) + phi(8) = 397 + 4 = 401 prime.

Mathematica

a[n_]:=Sum[If[PrimeQ[Prime[k(k+1)/2]+EulerPhi[n-k(k+1)/2]], 1, 0], {k, 1, (Sqrt[8n-7]-1)/2}]; Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A000010, A000040, A000217, A234615.

Sequence in context: A363193 A124018 A309288 * A355244 A364730 A283004

Adjacent sequences: A235058 A235059 A235060 * A235062 A235063 A235064

Keyword

nonn

Author

Zhi-Wei Sun, Jan 03 2014

Status

approved