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A254879
Let us denote 's' the sum of the deficient numbers in the aliquot parts of x. Sequence lists numbers x such that sigma(s)-s is equal to x.
3
6, 28, 54, 284, 496, 1184, 1210, 2924, 5564, 6232, 6368, 8128, 10744, 10856, 14595, 18150, 18416, 66928, 66992, 71145, 76084, 87633, 88730, 123152, 124155, 139815, 153176, 168730, 176336, 180848, 193720, 202444, 203432, 365084, 389924, 399592, 430402, 455344
OFFSET
1,1
COMMENTS
Perfect numbers belong to the sequence.
LINKS
EXAMPLE
Aliquot parts of 28 are 1, 2, 4, 7, 14 and they are all deficient numbers: sigma(1 + 2 + 4 + 7 + 14) = sigma(28) = 56 and 56 - 28 = 28.
Aliquot parts of 18150 are 1, 2, 3, 5, 6, 10, 11, 15, 22, 25, 30, 33, 50, 55, 66, 75, 110, 121, 150, 165, 242, 275, 330, 363, 550, 605, 726, 825, 1210, 1650, 1815, 3025, 3630, 6050, 9075 and the deficient numbers are 1, 2, 3, 5,10, 11, 15, 22, 25, 33, 50, 55, 75, 110, 121, 165, 242, 275, 363, 605, 825, 1210, 1815, 3025, 9075: sigma(1 + 2 + 3 + 5 + 10 + 11 + 15 + 22 + 25 + 33 + 50 + 55 + 75 + 110 + 121 + 165 + 242 + 275 + 363 + 605 + 825 + 1210 + 1815 + 3025 + 9075) = 18138 and sigma(18138) - 18138 = 18150.
MAPLE
with(numtheory); P:=proc(q) local a, b, k, n;
for n from 1 to q do a:=sort([op(divisors(n))]); b:=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)-b=n then print(n); fi; od; end: P(10^9);
CROSSREFS
KEYWORD
nonn
AUTHOR
Paolo P. Lava, Feb 10 2015
STATUS
approved