A342661 a(n) = n * sigma(A064989(n)), where A064989 is multiplicative with a(2^e) = 1 and a(p^e) = prevprime(p)^e for odd primes p, and sigma gives the sum of the divisors of its argument.
1, 2, 9, 4, 20, 18, 42, 8, 63, 40, 88, 36, 156, 84, 180, 16, 238, 126, 342, 80, 378, 176, 460, 72, 325, 312, 405, 168, 696, 360, 930, 32, 792, 476, 840, 252, 1184, 684, 1404, 160, 1558, 756, 1806, 352, 1260, 920, 2068, 144, 1519, 650, 2142, 624, 2544, 810, 1760, 336, 3078, 1392, 3186, 720, 3660, 1860, 2646, 64, 3120
Offset: 1
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Programs
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Mathematica
f[p_, e_] := If[p == 2, 2^e, Module[{q = NextPrime[p, -1]}, p^e*(q^(e + 1) - 1)/(q - 1)]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Dec 24 2022 *)
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PARI
A064989(n) = { my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f) }; A326041(n) = sigma(A064989(n)); A342661(n) = (n*A326041(n));
Formula
Multiplicative with a(p^e) = (p^e) * (q^(e+1)-1)/(q-1), where q = 1 for p = 2, and for odd primes p, q = A151799(p), i.e., the previous prime.
Sum_{k=1..n} a(k) ~ c * n^3, where c = (2/9) * Product_{p prime > 2} (p^3/((p+1)*(p^2-prevprime(p)))) = 0.1815217..., where prevprime is A151799. - Amiram Eldar, Dec 24 2022