A258270
Consider the unitary 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 iterations reach a sum equal to the reverse of themselves.
Original entry on oeis.org
6, 75, 133, 1005, 1603, 4258, 5299, 84292, 89944, 170568, 192901, 303003, 695364, 1633303
Offset: 1
Unitary aliquot parts of 6 are 1, 2, 3. We have: 1 + 2 + 3 = 6 that is equal to its reverse.
Unitary aliquot parts of 75 are 1, 3, 25. We have: 1 + 3 + 25 = 29; 3 + 25 + 29 = 57 that is the reverse of 75.
Unitary aliquot parts of 84292 are 1, 4, 13, 52, 1621, 6484, 21073. We have: 1 + 4 + 13 + 52 + 1621 + 6484 + 21073 = 29248 that is the reverse of 84292.
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with(numtheory): R:=proc(w) local x, y; x:=w; y:=0;while x>0 do
y:=10*y+(x mod 10); x:=trunc(x/10); od: y; end:
P:=proc(q, h) local a,b,c,k,n,t,v; v:=array(1..h);
for n from 1 to q do if not isprime(n) then a:=sort([op(divisors(n))]);
b:=[]; c:=ilog10(n)+1; for k from 1 to nops(a)-1 do if gcd(a[k],n/a[k])=1
then b:=[op(b),a[k]]; fi; od; if nops(b)>1 then
for k from 1 to nops(b) do v[k]:=b[k]; od; t:=nops(b)+1; v[t]:=add(v[k],k=1..nops(b)); if R(v[t])=n then print(n); else
while ilog10(v[t])+1<=c do t:=t+1; v[t]:=add(v[k], k=t-nops(b)..t-1);
if R(v[t])=n then print(n); break; fi; od; fi; fi; fi; od;
end: P(10^9,1000);
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.
Original entry on oeis.org
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
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.
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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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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 *)
A307859
Consider the non-unitary 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 iterations reach a sum equal to themselves.
Original entry on oeis.org
24, 112, 189, 578, 1984, 2125, 3993, 5043, 9583, 19197, 32512, 126445, 149565, 175689, 225578, 236883, 1589949, 1862935, 1928125, 3171174, 5860526, 6149405, 11442047, 16731741, 60634549, 75062535, 134201344, 177816209, 1162143369, 4474779517, 10369035821
Offset: 1
Divisors of 578 are 1, 2, 17, 34, 289, 578. Non-unitary aliquot parts are 17 and 34.
We have:
17 + 34 = 51;
34 + 51 = 85;
51 + 85 = 136;
85 + 136 = 221;
136 + 221 = 357;
221 + 357 = 578.
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with(numtheory):P:=proc(q,h) local a,b,c,k,n,t,v; 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 2 to nops(b)-1 do if gcd(b[k],n/b[k])>1 then
a:=[op(a),b[k]]; fi; od; b:=nops(a); if b>1 then c:=0;
for k from 1 to b do v[k]:=a[k]; c:=c+a[k]: od;
t:=b+1; v[t]:=c; while v[t]
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aQ[n_] := CompositeQ[n] && Module[{s = Select[Divisors[n], GCD[#, n/#] != 1 &]}, If[Length[s] < 2, False, While[Total[s] < n, AppendTo[s, Total[s]]; s = Rest[s]]; Total[s] == n]]; Select[Range[10^4], aQ] (* Amiram Eldar, May 07 2019 *)
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