A255242 -id:A255242 - cp's OEIS frontend

A255243 Number x such that x | A255242(x).

1, 4, 10, 16, 18, 64, 80, 96, 195, 256, 462, 576, 768, 880, 1024, 2560, 3120, 3136, 4096, 6656, 16384, 40704, 53248, 57344, 64000, 65536, 67896, 78864, 80640, 101376, 103680, 120320, 120336, 125440, 126208, 139264, 147968, 195840, 217600, 225280, 250624, 262144

Offset: 1

Paolo P. Lava, Feb 19 2015

nonn

For 4, 10, 195 we have x = A255242(x).

			For a(1) = 1 we have sigma(1) - 1 = 0 and 0 / 1 = 0.
Aliquot parts of a(2) = 4 are 1, 2 and their sum is 3.
Let us repeat the calculation with 1 and 2: 1 => 0; 2 => 1.
Their sum is 1. Finally, 3 + 1 = 4 and 4 / 4 = 1.
Aliquot parts of a(3) = 10 are 1, 2, 5. Their sum is 8.
Let us repeat the calculation with 1, 2 and 5: 1 => 0; 2 => 1; 5 => 1.
Their sum is 2. Finally, 8 + 2 = 10 and 10 / 10 = 1.
Aliquot parts of a(4) = 16 are 1, 2, 4, 8. Their sum is 15.
Let us repeat the calculation with 1, 2, 4 and 8: 1 => 0; 2 => 1; 4 => 1, 2; 8 => 1, 2, 4.
Their sum is 1 + 1 + 2 + 1 + 2 + 4 = 11.
Repeat the calculation with 1, 1, 2, 1, 2, 4: 1 => 0; 1 => 0; 2 => 1; 1 => 0; 2 => 1;  4 => 1, 2.
Their sum is 1 + 1 + 1 + 2 = 5.
Repeat the calculation with 1, 1, 1, 2: 1 => 0; 1 => 0; 1 => 0; 2 => 1; Their sum is 1.
Finally, 15 + 11 + 5 + 1 = 32 and 32 / 16 = 2.

Amiram Eldar, Table of n, a(n) for n = 1..50

Cf. A001065, A255242.

with(numtheory): P:=proc(q) local a,b,c,k,n,t,v;
for n from 1 to q do b:=0; a:=sort([op(divisors(n))]); t:=nops(a)-1;
while add(a[k],k=1..t)>0 do b:=b+add(a[k],k=1..t); v:=[];
for k from 2 to t do c:=sort([op(divisors(a[k]))]); v:=[op(v),op(c[1..nops(c)-1])]; od;
a:=v; t:=nops(a); od; if type(b/n,integer) then print(n); fi; od; end: P(10^9);

f[s_] := Flatten[Most[Divisors[#]] & /@ s]; a[n_] := Total@Flatten[FixedPointList[ f, {n}]] - n; Select[Range[10000], Divisible[a[#], #] &] (* Amiram Eldar, Apr 06 2019 *)

a(32)-a(42) from Amiram Eldar, Apr 06 2019

A330575 a(n) = n + Sum_{d|n and d1; a(1) = 1.

Original entry on oeis.org

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

Michel Marcus, Dec 18 2019

nonn

			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.

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.

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

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

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

a[1] = 1; a[n_] := a[n] = n + DivisorSum[n, a[#] &, # < n &]; Array[a, 65] (* Amiram Eldar, Apr 12 2020 *)

a(n) = if (n==1, 1, n + sumdiv(n, d, if (d

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

A191161 Hypersigma(n), definition 2: sum of the divisors of n plus the recursive sum of the divisors of the proper divisors.

Original entry on oeis.org

1, 4, 5, 12, 7, 22, 9, 32, 19, 30, 13, 72, 15, 38, 37, 80, 19, 90, 21, 96, 47, 54, 25, 208, 39, 62, 65, 120, 31, 178, 33, 192, 67, 78, 65, 316, 39, 86, 77, 272, 43, 222, 45, 168, 147, 102, 49, 560, 67, 174, 97, 192, 55

Offset: 1

Alonso del Arte, May 26 2011

In wanting to ensure the definition was not arbitrary, I initially thought that 1s had to stop the recursion. But as T. D. Noe showed me, this doesn't have to be the case: the 1s can be included in the recursion.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Jon Maiga, Computer-generated formulas for A191161, Sequence Machine.

Cf. A000203, A191150, A202687, A255242, A378211 (Dirichlet inverse).

Sequences that appear in the convolution formulas: A000010, A000203, A007429, A038040, A060640, A067824, A074206, A174725, A253249, A323910, A323912, A330575.

hsTD[n_] := hsTD[n] = Module[{d = Divisors[n]}, Total[d] + Total[hsTD /@ Most[d]]]; Table[hsTD[n], {n, 100}] (* From T. D. Noe *)

a(n)=sumdiv(n,d,if(dCharles R Greathouse IV, Dec 20 2011

a(n) = sigma(n) + sum_{d | n, d < n} a(d). - Charles R Greathouse IV, Dec 20 2011
From Antti Karttunen, Nov 22 2024: (Start)
Following formulas were conjectured by Sequence Machine:
For n > 1, a(n) = A191150(n) + A074206(n).
a(n) = A330575(n) + A255242(n) = 2*A255242(n) + n = 2*A330575(n) - n.
a(n) = Sum_{d|n} A330575(d).
a(n) = Sum_{d|n} d*A067824(n/d).
a(n) = Sum_{d|n} A000203(d)*A074206(n/d).
a(n) = Sum_{d|n} A007429(d)*A174725(n/d).
a(n) = Sum_{d|n} A000010(d)*A253249(n/d).
a(n) = Sum_{d|n} A038040(d)*A323912(n/d).
a(n) = Sum_{d|n} A060640(d)*A323910(n/d).
(End)

A363595 Recursive product of aliquot divisors of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 6, 1, 16, 3, 10, 1, 1728, 1, 14, 15, 2048, 1, 5832, 1, 8000, 21, 22, 1, 4586471424, 5, 26, 81, 21952, 1, 24300000, 1, 67108864, 33, 34, 35, 101559956668416, 1, 38, 39, 163840000000, 1, 130691232, 1, 85184, 91125, 46, 1, 16543163447903718821855232, 7, 125000, 51, 140608, 1, 1338925209984

Offset: 1

Michael De Vlieger, Jul 10 2023

			Define S(n) to be the set of proper divisors of n.
a(2) = 1, since 2 is prime, S(2) = {1} and the product of S(2) is 1.
a(4) = 2, since S(4) = {1, 2}; S(2) = 1, hence we have (1 X 2) X 1 = 2.
a(6) = 6, since S(6) = {1, 2, 3}; 2 and 3 are primes p and both have S(p) = 1,
  hence we have (1 X 2 X 3) X 1 X 1 = 6.
a(8) = 16, since S(8) = {1, 2, 4}; a(2) = 1, a(4) = 2,
  therefore (1 X 2 X 4) X 1 X 2 = 16.
a(9) = 3, since S(9) = {1, 3}, a(3) = 1,
  therefore (1 X 3) X 1 = 3.
a(10) = 10, since S(10) = {1, 2, 5};
  a(2) = a(5) = 1, a(4) = 2,
  therefore (1 X 2 X 5) X 1 X 1 = 10.
a(12) = 1728, since S(12) = {1, 2, 3, 4, 6};
  a(2) = a(3) = 1, a(4) = 2, a(6) = 6,
  therefore (1 X 2 X 3 X 4 X 6) X 1 X 1 X 2 X 6
  = 144 X 12 = 1728.

Michael De Vlieger, Table of n, a(n) for n = 1..719

Cf. A000295, A007955, A007956, analogous to A255242.

f[x_] := f[x] = Times @@ # * Times @@ Map[f, #] &@ Most@ Divisors[x]; Table[f[n], {n, 120}]

ali(n) = setminus(divisors(n), Set(n));
a(n) = my(list = List(), v = [n]); while (#v, my(w = []); for (i=1, #v, my(s=ali(v[i])); for (j=1, #s, w = concat(w, s[j]); listput(list, s[j]));); v = w;); vecprod(Vec(list)); \\ Michel Marcus, Jul 15 2023

a(n) >= A007956(n).
a(p) = 1 for prime p.
a(p^2) = p.
a(p^e) = A000295(e).
a(p*q) = p*q for primes p, q, p < q.
A007947(n) | a(n) for n with omega(n) > 2.

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A255243 Number x such that x | A255242(x).

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A330575 a(n) = n + Sum_{d|n and d1; a(1) = 1.

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A191161 Hypersigma(n), definition 2: sum of the divisors of n plus the recursive sum of the divisors of the proper divisors.

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A363595 Recursive product of aliquot divisors of n.

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