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Aliquot parts of 924 are 1, 2, 3, 4, 6, 7, 11, 12, 14, 21, 22, 28, 33, 42, 44, 66, 77, 84, 132, 154, 231, 308, 462. Abundant numbers are 12, 42, 66, 84, 132, 308 and 462. Then sigma(12+42+66+84+132+308+462) = sigma(1106) = 1920 = usigma(924). Etc.
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Amiram Eldar, <a href="/A258135/b258135.txt">Table of n, a(n) for n = 1..1000</a>
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Let 's' denote the sum of the abundant numbers in the aliquot parts of x. Sequence lists numbers x such that sigma(s) = usigma(x), where usigma(x) is the sum of the unitary divisors of x (A034448).
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Let us 's' denote “s” as the sum of the abundant numbers in the aliquot parts of x. Sequence lists numbers x such that sigma(s) = usigma(x), where usigma(x) is the sum of the unitary divisors of x (A034448).
allocated for Paolo PLet us denote “s” as the sum of the abundant numbers in the aliquot parts of x. Sequence lists numbers x such that sigma(s) = usigma(x), where usigma(x) is the sum of the unitary divisors of x (A034448). Lava
760, 918, 924, 1540, 4648, 6204, 8260, 15210, 20070, 21450, 27450, 30114, 41052, 47344, 50464, 55952, 60040, 60534, 61088, 63080, 77024, 77994, 81320, 99084, 117572, 132210, 136068, 150750, 169480, 215325, 215740, 226422, 309160, 476196, 495444, 505720, 530292
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Aliquot parts of 760 are 1, 2, 4, 5, 8, 10, 19, 20, 38, 40, 76, 95, 152, 190, 380. Abundant numbers are 20, 40 and 380. Then sigma(20+40+380) = sigma(440) = 1080 = usigma(760).
Aliquot parts of 918 are 1, 2, 3, 6, 9, 17, 18, 27, 34, 51, 54, 102, 153, 306, 459. Abundant numbers are 18, 54, 102 and 306. Then sigma(18+54+102+306) = sigma(480) = 1512 = usigma(918).
Aliquot parts of 924 are 1, 2, 3, 4, 6, 7, 11, 12, 14, 21, 22, 28, 33, 42, 44, 66, 77, 84, 132, 154, 231, 308, 462. Abundant numbers are 12, 42, 66, 84, 132, 308 and 462. Then sigma(12+42+66+84+132+308+462) = sigma(1106) = 1920 = usigma(924). Etc.
with(numtheory); P:=proc(q) local a, b, d, k, n; for n from 1 to q do
a:=sort([op(divisors(n))]); b:=0; d:=0;
for k from 1 to nops(a)-1 do if sigma(a[k])>2*a[k] then b:=b+a[k]; fi; od;
for k from 1 to nops(a) do if gcd(a[k], n/a[k])=1 then d:=d+a[k]; fi; od;
if sigma(b)=d then print(n); fi; od; end: P(10^9);
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Paolo P. Lava, May 21 2015
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