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proposed
(PARI) a(n) = sum(k = 0, n, 2^omega(binomial(n, k))); \\ Amiram Eldar, Jul 22 2024
Sum of the numbers of unitary divisors of the binomial coefficients C[(n,k], ), k=0..n.
a(3) = 6 because the divisors of 1,3,3,1 are {1},{1,3},{1,3},{1}, respectively, all of which are unitary, and 1 + 2 + 2 + 1 = 6.
a[n_] := Sum[2^PrimeNu[Binomial[n, k]], {k, 0, n}]; Array[a, 50, 0] (* Amiram Eldar, Jul 22 2024 *)
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editing
_Emeric Deutsch (deutsch(AT)duke.poly.edu), _, Feb 06 2005
Sum of the numbers of unitary divisors of the binomial coefficients C[n,k], k=0..n.
1, 2, 4, 6, 10, 14, 22, 22, 30, 46, 74, 94, 90, 102, 130, 170, 198, 222, 290, 350, 474, 650, 730, 734, 746, 838, 962, 1214, 2138, 2582, 1890, 1830, 2526, 3498, 4746, 6842, 5098, 6358, 8178, 10634, 8650, 9782, 13634, 14438, 17178, 20202, 22170, 21422, 16298
0,2
Row sums of the triangle A103444.
a(3)=6 because the divisors of 1,3,3,1 are {1},{1,3},{1,3},{1}, respectively, all of which are unitary.
with(numtheory):unitdiv:=proc(n) local A, k: A:={}: for k from 1 to tau(n) do if gcd(divisors(n)[k], n/divisors(n)[k])=1 then A:=A union {divisors(n)[k]} else A:=A fi od end: T:=proc(n, k) if k<=n then nops(unitdiv(binomial(n, k))) else 0 fi end: for n from 0 to 50 do b[n]:=[seq(T(n, k), k=0..n)] od: seq(sum(b[n][j], j=1..n+1), n=0..50);
Cf. A103444.
nonn,new
Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 06 2005
approved