|
|
A254878
|
|
Let 's' denote the sum of the deficient numbers in the aliquot parts of x. Sequence lists numbers x such that sigma(s) is equal to x.
|
|
2
|
|
|
4, 8, 32, 128, 168, 224, 756, 8192, 131072, 524288, 2147483648
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
All numbers of the form 2^A000043(n) belong to the sequence.
Terms that are not of this form begin: 168, 224, 756, ... - Amiram Eldar, Mar 24 2019
|
|
LINKS
|
|
|
EXAMPLE
|
Aliquot parts of 8 are 1, 2, 4 that are all deficient numbers: sigma(1 + 2 + 4) = sigma(7) = 8.
Aliquot parts of 168 are 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84 and the deficient numbers are 1, 2, 3, 4, 7, 8, 14, 21: sigma(1 + 2 + 3 + 4 + 7 + 8 + 14 + 21) = sigma(60) = 168.
|
|
MAPLE
|
with(numtheory); P:=proc(q) local a, b, c, k, n;
for n from 1 to q do a:=sort([op(divisors(n))]); b:=0; c:=0;
for k from 1 to nops(a)-1 do if sigma(a[k])<2*a[k] then b:=b+a[k]; fi; od;
if sigma(b)=n then print(n); fi; od; end: P(10^9);
|
|
MATHEMATICA
|
seqQ[n_] := Module[{s = DivisorSum[n, # &, #<n && DivisorSigma[1, #] < 2# &]}, s>0 && DivisorSigma[1, s] == n]; Select[Range[10000], seqQ] (* Amiram Eldar, Mar 24 2019 *)
|
|
PROG
|
(PARI) isok(n) = my (s = sumdiv(n, d, d*((d!=n) && (sigma(d)/d < 2)))); s && (sigma(s) == n); \\ Michel Marcus, Feb 19 2015
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,more
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|