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A110676 Number of prime factors with multiplicity of 1 + (n^(n+1)). 1
1, 2, 2, 3, 3, 4, 3, 6, 3, 5, 4, 5, 5, 9, 3, 4, 9, 3, 6, 10, 6, 7, 6, 11, 5, 11, 10, 5, 10, 9, 3, 12, 6, 10, 9, 5, 6, 13, 9, 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, 15, 15, 7, 12, 7, 8, 19, 8, 13, 14, 5, 6, 20, 6, 10, 13, 12, 7, 9, 9, 6, 21 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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 previously appear in OEIS. For instance, a(n) = 3 for n = 4, 5, 7, 9, 15, 18, 31, ... Is each such derived sequence finite?

LINKS

Table of n, a(n) for n=1..89.

FORMULA

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

EXAMPLE

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.

Sequence in context: A182921 A242767 A027833 * A117171 A084054 A106747

Adjacent sequences:  A110673 A110674 A110675 * A110677 A110678 A110679

KEYWORD

easy,nonn

AUTHOR

Jonathan Vos Post, Sep 14 2005

STATUS

approved

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Last modified October 1 12:17 EDT 2014. Contains 247510 sequences.