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A104418 Number of prime factors, with multiplicity, of the nonzero 9-acci numbers. 0
0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 2, 1, 6, 3, 5, 7, 9, 9, 11, 9, 3, 2, 2, 8, 7, 7, 7, 10, 11, 10, 3, 2, 7, 8, 11, 7, 12, 13, 15, 11, 3, 2, 6, 7, 7, 10, 9, 12, 12, 13, 5, 2, 5, 8, 8, 7, 13, 12, 10, 12, 6, 3, 3, 6, 12, 11, 12, 10, 12, 12, 2, 6, 12, 8, 11, 9, 14, 13, 13, 13, 7, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

Prime 9-acci numbers: b(3) = 2, b(12) = 1021, ... Semiprime 9-acci numbers: b(4) = 4 = 2^2, b(11) = 511 = 7 * 73, b(22) = 1035269 = 47 * 22027, b(23) = 2068498 = 2 * 1034249, b(32) = 1049716729 = 1051 * 998779 b(42) = 1064366053385 = 5 * 212873210677, b(52) = 1079219816432629 = 28669 * 37644138841, b(71) = 555323195719171835391 = 3 * 185107731906390611797, b(82) = 1125036467745713090813969 = 37 * 30406391020154407859837.

LINKS

Table of n, a(n) for n=1..82.

FORMULA

a(n) = A001222(A104144(n+7)).

EXAMPLE

a(1)=a(2)=0 because the first two nonzero 9-acci numbers are both 1, which has zero prime divisors.

a(3)=1 because the 3rd nonzero 9-acci number is 2, a prime, with only one prime divisor.

a(4)=2 because the 4th nonzero 9-acci number is 4 = 2^2 which has (with multiplicity) 2 prime divisors (which happen to be equal).

a(5)=3 because the 5th nonzero 9-acci number is 8 = 2^3.

a(13) = 6 because b(13) = 2040 = 2^3 * 3 * 5 * 17 so has 6 prime factors (2 with multiplicity 3 and 3, 5 and 17 once each).

CROSSREFS

Cf. A001222, A104411, A104412, A104413, A104414, A104415.

Sequence in context: A040115 A080465 A151950 * A173528 A245345 A043268

Adjacent sequences:  A104415 A104416 A104417 * A104419 A104420 A104421

KEYWORD

easy,nonn

AUTHOR

Jonathan Vos Post, Mar 06 2005

STATUS

approved

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Last modified December 18 23:46 EST 2018. Contains 318245 sequences. (Running on oeis4.)