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a(n) = Sum_{k=1 to d(n)} C(d(n)-1, k-1) d_k, where d(n) is the number of divisors of n and d_k is the k-th divisor of n.
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%I #28 Mar 13 2020 17:43:47

%S 1,3,4,9,6,22,8,27,16,32,12,123,14,42,40,81,18,164,20,171,52,62,24,

%T 704,36,72,64,219,30,808,32,243,76,92,72,1765,38,102,88,944,42,1016,

%U 44,315,276,122,48,4075,64,336,112,363,54,1224,104,1170,124,152,60,17815,62

%N a(n) = Sum_{k=1 to d(n)} C(d(n)-1, k-1) d_k, where d(n) is the number of divisors of n and d_k is the k-th divisor of n.

%H Rémy Sigrist, <a href="/A132065/b132065.txt">Table of n, a(n) for n = 1..10000</a>

%F a(p) = p+1, for p prime. - _Michel Marcus_, Sep 13 2014

%e Since the divisors of 12 are 1,2,3,4,6,12 and since row (d(12)-1) of Pascal's triangle is 1,5,10,10,5,1, a(12) = 1*1 + 5*2 + 10*3 + 10*4 + 5*6 + 1*12 = 123.

%e From _Peter Luschny_, May 18 2016: (Starts)

%e Also the lower vertex of the accumulation triangle of the divisors of n.

%e For instance a(39) = 88 because the lower vertex of ATD(39) = 88. ATD(39) is:

%e [ 39 13 3 1]

%e [ 52 16 4]

%e [ 68 20]

%e [ 88]

%e (End)

%t f[n_] := Block[{d, l, k},d = Divisors[n];l = Length[d];Sum[ Binomial[l - 1, k - 1]*d[[k]], {k, l}]];Array[f, 100] (* _Ray Chandler_, Oct 31 2007 *)

%t Table[Sum[Binomial[Length[Divisors[n]] - 1, k - 1]*Divisors[n][[k]], {k, 1, Length[Divisors[n]]}], {n, 1, 70}] (* _Stefan Steinerberger_, Oct 31 2007 *)

%o (PARI) a(n) = {d = divisors(n); sum(i=1, #d, d[i]*binomial(#d-1, i-1));} \\ _Michel Marcus_, Sep 13 2014

%o (Sage)

%o def A132065(n):

%o D = divisors(n)[::-1]

%o T = matrix(ZZ, len(D))

%o for (m, d) in enumerate(D):

%o T[0, m] = d

%o for k in range(m-1, -1, -1) :

%o T[m-k, k] = T[m-k-1, k+1] + T[m-k-1, k]

%o return T[len(D)-1,0]

%o print([A132065(n) for n in range(1,62)]) # _Peter Luschny_, May 18 2016

%Y Cf. A007318 (Pascal's triangle), A027750 (divisors of n).

%K nonn

%O 1,2

%A _Leroy Quet_, Oct 30 2007

%E Extended by _Ray Chandler_ and _Stefan Steinerberger_, Nov 01 2007