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 A322485 The sum of the semi-unitary divisors of n. 5

%I

%S 1,3,4,5,6,12,8,11,10,18,12,20,14,24,24,19,18,30,20,30,32,36,24,44,26,

%T 42,31,40,30,72,32,39,48,54,48,50,38,60,56,66,42,96,44,60,60,72,48,76,

%U 50,78,72,70,54,93,72,88,80,90,60,120,62,96,80,71,84,144

%N The sum of the semi-unitary divisors of n.

%C A semi-unitary divisor of n is defined as the largest divisor d of n such that the largest divisor of d that is a unitary divisor of n/d is 1 (see A322483).

%D J. Chidambaraswamy, Sum functions of unitary and semi-unitary divisors, J. Indian Math. Soc., Vol. 31 (1967), pp. 117-126.

%H Amiram Eldar, <a href="/A322485/b322485.txt">Table of n, a(n) for n = 1..10000</a>

%H Pentti Haukkanen, <a href="https://www.researchgate.net/profile/Pentti_Haukkanen/publication/266363785_Basic_properties_of_the_bi-unitary_convolution_and_the_semi-unitary_convolution/links/5469efe50cf2397f782f4e3b/Basic-properties-of-the-bi-unitary-convolution-and-the-semi-unitary-convolution.pdf">Basic properties of the bi-unitary convolution and the semi-unitary convolution</a>, Indian J. Math, Vol. 40 (1998), pp. 305-315.

%H D. Suryanarayana and V. Siva Rama Prasad, <a href="https://doi.org/10.1017/S1446788700012908">Sum functions of k-ary and semi-k-ary divisors</a>, Journal of the Australian Mathematical Society, Vol. 15, No. 2 (1973), pp. 148-162.

%H Laszlo Tóth, <a href="http://annalesm.elte.hu/annales41-1998/Annales_1998_T-XLI.pdf#page=167">Sum functions of certain generalized divisors</a>, Ann. Univ. Sci. Budap. Rolando Eötvös, Sect. Math., Vol. 41 (1998), pp. 165-180.

%F Multiplicative with a(p^e) = sigma(p^floor((e-1)/2)) + p^e = (p^floor((e+1)/2) - 1)/(p-1) + p^e.

%F In particular a(p) = p + 1, a(p^2) = p^2 + 1, a(p^3) = p^3 + p + 1.

%F a(n) <= A000203(n) with equality if and only if n is squarefree (A005117).

%e The semi-unitary divisors of 8 are 1, 2, 8 (4 is not semi-unitary divisor since the largest divisor of 4 that is a unitary divisor of 8/4 = 2 is 2 > 1), and their sum is 11, thus a(8) = 11.

%t f[p_, e_] := (p^Floor[(e+1)/2] - 1)/(p-1) + p^e; susigma[n_] := If[n==1, 1, Times @@ (f @@@ FactorInteger[n])]; Array[susigma, 100]

%o (PARI) a(n) = {my(f = factor(n)); for (k=1, #f~, my(p=f[k,1], e=f[k,2]); f[k,1] = (p^((e+1)\2) - 1)/(p-1) + p^e; f[k,2] = 1;); factorback(f);} \\ _Michel Marcus_, Dec 14 2018

%Y Cf. A000203, A005117, A034448, A188999, A322483.

%K nonn,mult

%O 1,2

%A _Amiram Eldar_, Dec 11 2018

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