A110676 Number of prime factors with multiplicity of 1 + n^(n+1).

1, 2, 2, 3, 3, 4, 3, 6, 3, 5, 4, 5, 4, 9, 3, 4, 9, 3, 6, 10, 6, 7, 6, 11, 5, 11, 10, 5, 10, 8, 3, 12, 6, 10, 8, 5, 6, 13, 8, 6, 11, 6, 10, 16, 4, 4, 6, 9, 6, 11, 8, 4, 10, 10, 5, 13, 10, 7, 11, 6, 6, 21, 4, 23, 8, 6, 8, 16, 15, 7, 12, 7, 8, 19, 8, 13, 14, 5, 6, 20, 6, 10, 13, 12, 7, 9, 9, 6, 21

Offset: 1

Author

Jonathan Vos Post, Sep 14 2005

Keywords

• easy
• nonn

Comments

As also noticed by T. D. Noe, for odd n: 2 | a(n), for even n: (n+1)^2 | a(n). Coincidentally, a(74) includes 13 multidigit prime factors all of which end with the digit 1. There is no upper limit to this sequence, which rapidly becomes slow to compute. The derived sequences of n such that a(n) = k for any constant k > 2 do not yet appear in the OEIS. For instance, a(n) = 3 for n = 4, 5, 7, 9, 15, 18, 31, ... Is each such derived sequence finite?

Examples

			a(1) = 1 because 1+1^2 = 2 is prime (and the only such prime).
a(2) = 2 because 1 + 2^3 = 9 = 3^2 which has (with multiplicity) two prime factors.
a(3) = 2 because 1 + 3^4 = 82 = 2 * 41 (the last such semiprime?).
a(4) = 3 because 1 + 4^5 = 1025 = 5^2 * 41 which has (with multiplicity) 3 prime factors.
a(8) = 6 because 1 + 8^9 = 134217729 = 3^4 * 19 * 87211.
a(14) = 9 because 1 + 14^15 = 155568095557812225 = 3^2 * 5^2 * 61 * 71 * 101 * 811 * 1948981.
a(1000) > 52.
		

Crossrefs

Cf. A001222, A007778, A110567.

Programs

PARI

a(n) = bigomega(1+(n^(n+1))) \\ Georg Fischer, Jun 21 2024

Formula

a(n) = A001222(A110567(n)) = A001222(1 + A007778(n)) = A001222(1 + (n^(n+1))).

Trivially a(n) << n log n. At most n^(n+1) + 1 is of the form 2*3^k. - Charles R Greathouse IV, Jun 24 2024

Extensions

a(13) and 3 other terms corrected by Georg Fischer, Jun 21 2024