A261079 Sum of index differences between prime factors of n, summed over all unordered pairs of primes present (with multiplicity) in the prime factorization of n.

0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 2, 0, 3, 1, 0, 0, 2, 0, 4, 2, 4, 0, 3, 0, 5, 0, 6, 0, 4, 0, 0, 3, 6, 1, 4, 0, 7, 4, 6, 0, 6, 0, 8, 2, 8, 0, 4, 0, 4, 5, 10, 0, 3, 2, 9, 6, 9, 0, 7, 0, 10, 4, 0, 3, 8, 0, 12, 7, 6, 0, 6, 0, 11, 2, 14, 1, 10, 0, 8, 0, 12, 0, 10, 4, 13, 8, 12, 0, 6, 2, 16, 9, 14, 5, 5, 0, 6, 6, 8, 0, 12, 0, 15, 4, 15, 0, 6, 0, 8, 10, 12, 0, 14, 6, 18, 8, 16, 3, 10

Offset: 1

Author

Antti Karttunen, Sep 23 2015

Keywords

• nonn

Examples

			For n = 1 the prime factorization is empty, thus there is nothing to sum, so a(1) = 0.
For n = 6 = 2*3 = prime(1) * prime(2), a(6) = 1 because the (absolute value of) difference between prime indices of 2 and 3 is 1.
For n = 10 = 2*5 = prime(1) * prime(3), a(10) = 2 because the difference between prime indices of 2 and 5 is 2.
For n = 12 = 2*2*3 = prime(1) * prime(1) * prime(2), a(12) = 2 because the difference between prime indices of 2 and 3 is 1, and the pair (2,3) occurs twice as one can pick either one of the two 2's present in the prime factorization to be a pair of a single 3. Note that the index difference between 2 and 2 is 0, thus the pair (2,2) of prime divisors does not contribute to the sum.
For n = 36 = 2*2*3*3, a(36) = 4 because the index difference between 2 and 3 is 1, and the prime factor pair (2,3) occurs 2^2 = four times in total. As the index difference is zero between 2 and 2 as well as between 3 and 3, the pairs (2,2) and (3,3) do not contribute to the sum.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A000720.

Cf. A000961 (positions of zeros), A006094 (positions of ones).

Cf. A243055, A242411.

Cf. also A260737.

A055396 gives minimum prime index, maximum A061395.

A112798 list prime indices, length A001222, sum A056239.

A304818 adds up partial sums of reversed prime indices, row sums of A359361.

A318283 adds up partial sums of prime indices, row sums of A358136.

Cf. A029931, A316413, A325362, A326836, A326844, A355536, A358137, A359362.

Programs

Mathematica

Table[Function[p, Total@ Map[Function[b, Times @@ {First@ Differences@ PrimePi@ b, Count[Subsets[p, {2}], c_ /; SameQ[c, b]]}], Subsets[Union@ p, {2}]]][Flatten@ Replace[FactorInteger@ n, {p_, e_} :> ConstantArray[p, e], 2]], {n, 120}] (* Michael De Vlieger, Mar 08 2017 *)

Formula

a(n) = A304818(n) - A318283(n). - Gus Wiseman, Jan 09 2023

a(n) = 2*A304818(n) - A359362(n). - Gus Wiseman, Jan 09 2023