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Number of divisors of n-th term of sequence a(n+1) = a(n)*(a(0) + ... + a(n)) (A001697).
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%I #17 Feb 21 2024 01:19:32

%S 1,1,2,4,12,40,160,792,9408,783360,55987200,35610624000,

%T 269007298560000

%N Number of divisors of n-th term of sequence a(n+1) = a(n)*(a(0) + ... + a(n)) (A001697).

%F a(n) = A000005(A001697(n)). - _Amiram Eldar_, Feb 17 2019

%t a[0] = 1; a[1] = 1; a[n_] := a[n] = a[n - 1]^2*(1 + 1/a[n - 2]); Table[DivisorSigma[0, a[n]], {n, 0, 10}] (* _Amiram Eldar_, Feb 17 2019 after Jean-François Alcover at A001697 *)

%o (PARI) a(n)=if(n<2,n >= 0,a(n-1)^2*(1+1/a(n-2)));

%o for(n=0,11,print1(numdiv(a(n)), ", "))

%Y Cf. A000005, A001697.

%K nonn,hard,more

%O 0,3

%A _Jason Earls_, Jul 22 2001

%E a(11) from _Amiram Eldar_, Feb 17 2019

%E a(12) from _Max Alekseyev_, Feb 20 2024