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a(n) = Sum_{d|n, d^2>=n} 1+d+n/d.
0

%I #7 Nov 06 2021 02:51:30

%S 3,4,5,11,7,14,9,17,18,20,13,31,15,26,26,38,19,42,21,45,34,38,25,64,

%T 38,44,42,59,31,76,33,66,50,56,50,102,39,62,58,94,43,100,45,87,81,74,

%U 49,129,66,96,74,101,55,124,74,124,82,92,61,174,63,98,107,139,86,148

%N a(n) = Sum_{d|n, d^2>=n} 1+d+n/d.

%H Roland Bacher, <a href="https://arxiv.org/abs/2111.02788">Yet another Proof of an old Hat</a>, arXiv:2111.02788 [math.HO], 2021.

%F a(n) = A060866(n) + A038548(n).

%t a[n_] := DivisorSum[n, 1 + # + n/# &, #^2 >= n &]; Array[a, 100] (* _Amiram Eldar_, Nov 05 2021 *)

%o (PARI) a(n) = sumdiv(n, d, if (d^2>=n, d+1+n/d));

%Y Cf. A038548, A060866.

%K nonn

%O 1,1

%A _Michel Marcus_, Nov 05 2021