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A276935 Number of distinct prime factors prime(k) of n such that prime(k)^k, but not prime(k)^(k+1) is a divisor of n. 3
0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,18

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..10000

FORMULA

a(1) = 0, for n > 1, a(n) = a(A028234(n)) + [A067029(n) = A055396(n)], where [] is Iverson bracket, giving 1 as its result when the stated equivalence is true and 0 otherwise.

EXAMPLE

For n = 12 = 2*2*3 = prime(1)^2 * prime(2)^1, neither of the prime factors satisfies the condition, thus a(12) = 0.

For n = 18 = 2*3*3 = prime(1)^1 * prime(2)^2, both prime factors satisfy the condition, thus a(18) = 1+1 = 2.

For n = 750 = 2*3*5*5*5 = prime(1)^1 * prime(2)^1 * prime(3)^3, only the prime factors 2 and 5 satisfy the condition, thus a(750) = 1+1 = 2.

PROG

(Scheme, with Antti Karttunen's IntSeq-library)

(definec (A276935 n) (if (= 1 n) 0 (+ (A276935 (A028234 n)) (if (= (A067029 n) (A055396 n)) 1 0))))

CROSSREFS

Cf. A276077, A276936.

Sequence in context: A219488 A129251 A276077 * A235127 A258059 A093956

Adjacent sequences:  A276932 A276933 A276934 * A276936 A276937 A276938

KEYWORD

nonn

AUTHOR

Antti Karttunen, Sep 24 2016

STATUS

approved

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Last modified January 18 13:59 EST 2019. Contains 319271 sequences. (Running on oeis4.)