A234246 a(n) = |{0 < k < n: k*phi(n-k) + 1 is a square}|, where phi(.) is Euler's totient function.

0, 0, 0, 1, 1, 0, 2, 1, 1, 3, 2, 1, 1, 2, 3, 4, 5, 4, 2, 2, 2, 5, 4, 1, 5, 4, 4, 3, 2, 8, 5, 2, 1, 3, 9, 5, 9, 4, 4, 6, 2, 4, 9, 5, 5, 7, 9, 3, 1, 10, 6, 8, 3, 6, 4, 5, 7, 8, 3, 5, 5, 4, 6, 6, 10, 14, 8, 3, 3, 6, 9, 5, 7, 7, 9, 2, 8, 8, 9, 5, 6, 6, 6, 8, 9, 7, 9, 4, 5, 9, 10, 8, 8, 7, 14, 9, 5, 7, 6, 10

Offset

1,7

Comments

Conjecture: (i) a(n) > 0 if n is not a divisor of 6. The only values of n with a(n) = 1 are 4, 5, 8, 9, 12, 13, 24, 33, 49.

(ii) If n >= 60, then k + phi(n-k) is a square for some 0 < k < n. If n > 60, then sigma(k) + phi(n-k) is a square for some 0 < k < n, where sigma(k) is the sum of all positive divisors of k.

(iii) If n > 7 is not equal to 10 or 20, then phi(k)*phi(n-k) + 1 is a square for some 0 < k < n.

(iv) If n > 7 is not equal to 10 or 19, then (phi(k) + phi(n-k))/2 is a triangular number for some 0 < k < n.

Note that (n - 1)*phi(1) + 1 = n. So a(n) > 0 if n is a square.

Links

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

Example

a(4) = 1 since 3*phi(1) + 1 = 2^2.
a(5) = 1 since 3*phi(2) + 1 = 2^2.
a(8) = 1 since 4*phi(4) + 1 = 3^2.
a(9) = 1 since 8*phi(1) + 1 = 3^2.
a(12) = 1 since 2*phi(10) + 1 = 3^2.
a(13) = 1 since 4*phi(9) + 1 = 5^2.
a(14) = 2 since 2*phi(12) + 1 = 3^2 and 6*phi(8) + 1 = 5^2.
a(24) = 1 since 12*phi(12) + 1 = 7^2.
a(33) = 1 since 3*phi(30) + 1 = 5^2.
a(49) = 1 since 48*phi(1) + 1 = 7^2.

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
a[n_]:=Sum[If[SQ[k*EulerPhi[n-k]+1], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A000010, A000290, A233542, A233544, A233547, A233566, A233567, A233867, A233918, A234200

Sequence in context: A296774 A066099 A254111 * A006375 A327520 A184441

Adjacent sequences: A234243 A234244 A234245 * A234247 A234248 A234249

Keyword

nonn

Author

Zhi-Wei Sun, Dec 21 2013

Status

approved