**If the prime-factorization of n is n = product{p|n} p^b(p,n) (p = distinct primes divisors of n, each b(p,n) is a positive integer), then a(n) is (sum{p|n} p^b(p,n)) taken mod (sum{p|n} p).**

*0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0*

*1 seqfan posts*

Index of A-numbers in seqfan: by ascending order by month by frequency by keyword

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