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%I #9 Jul 22 2024 15:24:11
%S 1,1,1,1,2,1,1,2,2,1,1,2,4,2,1,1,2,4,4,2,1,1,4,4,4,4,4,1,1,2,4,4,4,4,
%T 2,1,1,2,4,4,8,4,4,2,1,1,2,4,8,8,8,8,4,2,1,1,4,4,8,16,8,16,8,4,4,1,1,
%U 2,4,8,16,16,16,16,8,4,2,1,1,4,8,8,8,8,16,8,8,8,8,4,1,1,2,8,8,8,8,16,16,8,8
%N Triangle read by rows: T(n,k) is number of unitary divisors of C(n,k), 0<=k<=n.
%C Row n contains n+1 terms. Row sums yield A103445.
%e T(6,3) = 4 because the divisors of C(6,3) = 20 are 1,2,4,5,10,20 of which 1,4,5,20 are unitary (i.e. d|20 such that gcd(d,20/d) = 1).
%e Triangle begins:
%e 1;
%e 1,1;
%e 1,2,1;
%e 1,2,2,1;
%e 1,2,4,2,1;
%e 1,2,4,4,2,1;
%p 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 13 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
%t Table[2^PrimeNu[Binomial[n, k]], {n, 0, 10}, {k, 0, n}] // Flatten (* _Amiram Eldar_, Jul 22 2024 *)
%o (PARI) T(n, k) = 2^omega(binomial(n, k)); \\ _Amiram Eldar_, Jul 22 2024
%Y Cf. A007318, A034444, A103445.
%K nonn,tabl,changed
%O 0,5
%A _Emeric Deutsch_, Feb 06 2005
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