

A083339


a(n) is the number of distinct prime factors of n that occur in partitions into two primes when n is even and into three primes when n is odd.


2



0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 2, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 2, 1, 2, 0, 0, 1, 2, 0, 0, 0, 0, 0, 2, 1, 0, 0, 1, 0, 2, 0, 0, 0, 2, 0, 2, 1, 0, 0, 0, 1, 2, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 1, 2, 0, 2, 0, 0, 0, 1, 1, 0, 0, 2, 1, 2, 0, 0, 0, 2, 0, 2, 1, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 3
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OFFSET

1,15


COMMENTS

Number of distinct prime factors of n that occur in primepartitions confirming Goldbach's conjectures. (The original name of this sequence.)
Conjecture: Apart from k=2, A070826(k): 1, 3, 15, 105, 1155, 15015, 255255, gives the positions of records (each equal to k1). This follows from the conjectured formula.  Antti Karttunen, Sep 14 2017


LINKS



FORMULA

If n is even, a(n) = A010051(n/2), if n is an odd prime, a(n) = 0, and for odd composites (conjecturally), a(n) = A001221(n).  Antti Karttunen, Sep 14 2017


EXAMPLE

For n = 14 = 2*7 = 3 + 11 = 7 + 7, only one factor of 14 occurs, thus a(14) = 1.
For n = 15 = 3*5 = 2 + 2 + 11 = 3 + 5 + 7 = 5 + 5+ 5, both factors of 15 occur, thus a(15) = 2.
For n = 105 = 3*5*7, with 35 different partitions into three primes, the partition 97 + 5 + 3 contains the prime factors 3 and 5, while the partition 79 + 19 + 7 contains 7, thus all three prime factors of 115 occur and a(115) = 3.
For n = 1155 = 3*5*7*11, among 891 different partitions into three primes, the following four partitions: 1129 + 23 + 3 = 1129 + 19 + 7 = 1109 + 41 + 5 = 1103 + 41 + 11 each have either 3, 5, 7 or 11 as one of their parts, thus a(1155) = 4.


MATHEMATICA

Table[Count[Union@ Flatten@ Select[IntegerPartitions[n, {2 + Boole[OddQ@ n]}], AllTrue[#, PrimeQ] &], p_ /; Divisible[n, p]], {n, 105}] (* Michael De Vlieger, Sep 16 2017 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



