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%I #13 Mar 14 2023 10:14:44
%S 1,2,8,80,2240,215040,77414400,61931520000,170930995200000,
%T 1340099002368000000
%N a(0) = 1; for n >= 1, a(n) is the least integer k > a(n-1) such that k / A000005(k) = a(n-1).
%e a(0) = 1.
%e a(1) = 2 because the least k > 1 such that k / A000005(k) = 1 is k = 2.
%e a(2) = 8 because the least k > 2 such that k / A000005(k) = 2 is k = 8.
%e a(3) = 80 because the least k > 8 such that k / A000005(k) = 8 is k = 80.
%e and so on.
%t a[0] = 1; a[n_] := a[n] = Module[{m = a[n - 1], k}, k = 2*m; While[k != m*DivisorSigma[0, k], k += m]; k]; Array[a, 9, 0] (* _Amiram Eldar_, Feb 21 2023 *)
%Y Cf. A000005, A033950.
%K nonn,more
%O 0,2
%A _Ctibor O. Zizka_, Feb 21 2023
%E a(5)-a(9) from _Amiram Eldar_, Feb 21 2023