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
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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.
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Programs
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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 *)
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