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A293902
If n = p_1^e_1 * ... * p_k^e_k, p_1, ..., p_k primes, then a(n) = Product c! where c ranges over products of all combinations of exponents e_1, ..., e_k as {e_1, e_1*e_2, e_1*e_3, e_2*e_3, e_1*e_2*e_3, ..., e_1*e_2*...*e_k}.
3
1, 1, 1, 2, 1, 1, 1, 6, 2, 1, 1, 4, 1, 1, 1, 24, 1, 4, 1, 4, 1, 1, 1, 36, 2, 1, 6, 4, 1, 1, 1, 120, 1, 1, 1, 96, 1, 1, 1, 36, 1, 1, 1, 4, 4, 1, 1, 576, 2, 4, 1, 4, 1, 36, 1, 36, 1, 1, 1, 16, 1, 1, 4, 720, 1, 1, 1, 4, 1, 1, 1, 8640, 1, 1, 4, 4, 1, 1, 1, 576, 24, 1, 1, 16, 1, 1, 1, 36, 1, 16, 1, 4, 1, 1, 1, 14400, 1, 4, 4, 96, 1, 1, 1, 36, 1
OFFSET
1,4
COMMENTS
a(1) = 1 (an empty product).
FORMULA
For n = p^k * q * ... * r (with only one of the prime factors occurring multiple times), a(n) = A000142(k)^(2^(A001221(n)-1)).
a(p^n) = A000142(n), for any prime p.
For n > 1, a(n) = A163820(n) / A293900(n).
EXAMPLE
For n = 36 = 2^2 * 3^2 the combinations of the exponents are [], [2] (as exponent of 2), [2] (as exponent of 3) and [2, 2]. Taking products of these multisets we get 1 (as an empty product), 2, 2 and 4. Thus a(36) = 1! * 2! * 2! * 4! = 1*2*2*24 = 96.
For n = 72 = 2^3 * 3^2 the combinations of the exponents are [], [2], [3] and [2, 3]. Taking products of these multisets we get 1, 2, 3 and 6. Thus a(72) = 1! * 2! * 3! * 6! = 1*2*6*720 = 8640.
MATHEMATICA
Array[Apply[Times, Map[Times @@ # &, Subsets@ FactorInteger[#][[All, -1]]]!] &, 105] (* Michael De Vlieger, Oct 23 2017 *)
PROG
(PARI)
A293902(n) = { my(exp_combos=powerset(factor(n)[, 2]), m=1); for(i=1, #exp_combos, m *= vecproduct(exp_combos[i])!); m; };
vecproduct(v) = { my(m=1); for(i=1, #v, m *= v[i]); m; };
powerset(v) = { my(siz=2^length(v), pv=vector(siz)); for(i=0, siz-1, pv[i+1] = choosebybits(v, i)); pv; };
choosebybits(v, m) = { my(s=vector(hammingweight(m)), i=j=1); while(m>0, if(m%2, s[j] = v[i]; j++); i++; m >>= 1); s; };
(Scheme) (define (A293902 n) (if (= 1 n) n (/ (A163820 n) (A293900 n))))
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Oct 22 2017
STATUS
approved