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a(n) = n + Sum_{d|n and d<n} a(d) for n>1; a(1) = 1.
11

%I #41 Nov 24 2024 04:30:01

%S 1,3,4,8,6,14,8,20,14,20,12,42,14,26,26,48,18,54,20,58,34,38,24,116,

%T 32,44,46,74,30,104,32,112,50,56,50,176,38,62,58,156,42,132,44,106,96,

%U 74,48,304,58,112,74,122,54,190,74,196,82,92,60,346,62,98,124,256,86

%N a(n) = n + Sum_{d|n and d<n} a(d) for n>1; a(1) = 1.

%H Robert Israel, <a href="/A330575/b330575.txt">Table of n, a(n) for n = 1..10000</a>

%H Thomas Fink, <a href="https://arxiv.org/abs/1912.07979">Recursively divisible numbers</a>, arXiv:1912.07979 [math.NT], 2019. See Table 2 p. 11.

%H Thomas Fink, <a href="https://arxiv.org/abs/2008.10398">Recursively abundant and recursively perfect numbers</a>, arXiv:2008.10398 [math.NT], 2020.

%H T. M. A. Fink, <a href="https://arxiv.org/abs/2307.09140">Properties of the recursive divisor function and the number of ordered factorizations</a>, arXiv:2307.09140 [math.NT], 2023.

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

%F a(n) = n + A255242(n). - _Rémy Sigrist_, Dec 18 2019

%F G.f. A(x) satisfies: A(x) = x/(1 - x)^2 + Sum_{k>=2} A(x^k). - _Ilya Gutkovskiy_, Dec 18 2019

%F a(n) = Sum_{d|n} A074206(d) * n/d. - _David A. Corneth_, Apr 13 2020

%e a(2) = 2 + a(1) = 2 + 1 = 3, since the only proper divisors of 2 is 1.

%e a(4) = 4 + a(1) + a(2) = 4 + 1 + 3 = 8, since the proper divisors of 4 are 1 and 2.

%e a(6) = 6 + a(1) + a(2) + a(3) = 6 + 1 + 3 + 4 = 14, since the proper divisors of 6 are 1, 2 and 3.

%p f:= proc(n) option remember;

%p n + add(procname(d), d = numtheory:-divisors(n) minus {n})

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Dec 19 2019

%t a[1] = 1; a[n_] := a[n] = n + DivisorSum[n, a[#] &, # < n &]; Array[a, 65] (* _Amiram Eldar_, Apr 12 2020 *)

%o (PARI) a(n) = if (n==1, 1, n + sumdiv(n, d, if (d<n, a(d))));

%o (Magma) a:=[1]; for n in [2..65] do Append(~a,(n+&+[a[d]:d in Set(Divisors(n)) diff {n}])); end for; a; // _Marius A. Burtea_, Dec 18 2019

%Y Cf. A067824, A074206, A191161, A255242, A378217 (Dirichlet inverse).

%K nonn

%O 1,2

%A _Michel Marcus_, Dec 18 2019