A263344 Consider the abundant aliquot parts, in ascending order, of a composite number. Take their sum and repeat the process deleting the minimum number and adding the previous sum. The sequence lists the numbers that after some number of iterations reach a sum equal to themselves.
1700, 5950, 155574, 274550, 300894, 715275, 758625, 1365234, 1404172, 1542500, 1661750, 2095250, 2239750, 2673250, 2962250, 3106750, 3395750, 3829250, 4226625, 4262750, 4407250, 4700619, 5398750, 6371092, 8167635, 8560024, 12305620, 13725855, 15497625, 15586263
Offset: 1
Keywords
Examples
Aliquot parts of 1700 are 1, 2, 4, 5, 10, 17, 20, 25, 34, 50, 68, 85, 100, 170, 340, 425, 850. The abundant numbers are 20, 100, 340. Therefore: 20 + 100 + 340 = 460; 100 + 340 + 460 = 900; 340 + 460 + 900 = 1700.
Crossrefs
Programs
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Maple
with(numtheory):P:=proc(q,h) local a,b,k,t,v; global n; v:=array(1..h); for n from 1 to q do if not isprime(n) then b:=sort([op(divisors(n))]); a:=[]; for k from 1 to nops(b)-1 do if sigma(b[k])>2*b[k] then a:=[op(a),b[k]]; fi; od; a:=sort(a); b:=nops(a); if b>1 then for k from 1 to b do v[k]:=a[k]; od; t:=b+1; v[t]:=add(v[k], k=1..b); while v[t]
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Mathematica
seqQ[n_] := Module[{d = Select[Most[Divisors[n]], DivisorSigma[1, #] > 2 # &]}, Switch[Length[d], ?(# < 1 &), False, ?(# == 1 &), d[[1]] == n, , k = 0; While[k < n, k = Total[d]; d = Rest[AppendTo[d, k]]]; k == n]]; seq = {}; Do[ If[seqQ[n], AppendTo[seq, n]], {n, 2, 10^6}]; seq (* _Amiram Eldar, Mar 20 2019 *)
Extensions
More terms from Amiram Eldar, Mar 20 2019