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a(n) = sum_(d|n) product_(d_x|n, d_x<=d) d_x.
1

%I #13 Jan 01 2020 21:57:36

%S 1,3,4,11,6,45,8,75,31,113,12,1905,14,213,244,1099,18,6201,20,8451,

%T 466,509,24,346929,131,705,760,22803,30,839019,32,33867,1126,1193,

%U 1266,10374657,38,1485,1564,2627571,42,3189609,44,87219,93304,2165,48,260348721

%N a(n) = sum_(d|n) product_(d_x|n, d_x<=d) d_x.

%C Conjecture: sequence is injective (all terms of this sequence occur only once).

%C The conjecture is true for the first 100000 terms. - _Harvey P. Dale_, Dec 02 2017

%H Jaroslav Krizek, <a href="/A220848/b220848.txt">Table of n, a(n) for n = 1..100</a>

%F a(p) = p + 1 for prime p.

%e The divisors of 6 are 1, 2, 3, 6. a(n) = 1 + (1*2) + (1*2*3) + (1*2*3*6) = 1 + 2 + 6 + 36 = 45.

%t f[x_]:=Module[{d=Divisors[x]},Total[Table[Times@@Take[d,n],{n,Length[ d]}]]]; Array[f,50] (* _Harvey P. Dale_, Dec 02 2017 *)

%K nonn,easy

%O 1,2

%A _Jaroslav Krizek_, Dec 22 2012