A330575
a(n) = n + Sum_{d|n and d1; a(1) = 1.
1, 3, 4, 8, 6, 14, 8, 20, 14, 20, 12, 42, 14, 26, 26, 48, 18, 54, 20, 58, 34, 38, 24, 116, 32, 44, 46, 74, 30, 104, 32, 112, 50, 56, 50, 176, 38, 62, 58, 156, 42, 132, 44, 106, 96, 74, 48, 304, 58, 112, 74, 122, 54, 190, 74, 196, 82, 92, 60, 346, 62, 98, 124, 256, 86
Offset: 1
Keywords
Examples
a(2) = 2 + a(1) = 2 + 1 = 3, since the only proper divisors of 2 is 1. a(4) = 4 + a(1) + a(2) = 4 + 1 + 3 = 8, since the proper divisors of 4 are 1 and 2. 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.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Thomas Fink, Recursively divisible numbers, arXiv:1912.07979 [math.NT], 2019. See Table 2 p. 11.
- Thomas Fink, Recursively abundant and recursively perfect numbers, arXiv:2008.10398 [math.NT], 2020.
- T. M. A. Fink, Properties of the recursive divisor function and the number of ordered factorizations, arXiv:2307.09140 [math.NT], 2023.
Programs
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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 -
Maple
f:= proc(n) option remember; n + add(procname(d), d = numtheory:-divisors(n) minus {n}) end proc: map(f, [$1..100]); # Robert Israel, Dec 19 2019 -
Mathematica
a[1] = 1; a[n_] := a[n] = n + DivisorSum[n, a[#] &, # < n &]; Array[a, 65] (* Amiram Eldar, Apr 12 2020 *)
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PARI
a(n) = if (n==1, 1, n + sumdiv(n, d, if (d
Formula
a(p) = p+1 for p prime.
a(n) = n + A255242(n). - Rémy Sigrist, Dec 18 2019
G.f. A(x) satisfies: A(x) = x/(1 - x)^2 + Sum_{k>=2} A(x^k). - Ilya Gutkovskiy, Dec 18 2019
a(n) = Sum_{d|n} A074206(d) * n/d. - David A. Corneth, Apr 13 2020