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.

6, 75, 133, 1005, 1603, 4258, 5299, 84292, 89944, 170568, 192901, 303003, 695364, 1633303

Offset

1,1

Links

Table of n, a(n) for n=1..14.

Eric Weisstein's World of Mathematics, Unitary Divisor

Eric Weisstein's World of Mathematics, Unitary Divisor Function

Eric Weisstein's World of Mathematics, Unitary Perfect Number

Wikipedia, Unitary divisor

Example

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.

Maple

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);

Crossrefs

Cf. A246544, A247012, A258142.

Sequence in context: A269334 A069852 A266574 * A281797 A049235 A129031

Adjacent sequences: A258267 A258268 A258269 * A258271 A258272 A258273

Keyword

nonn,more,base

Author

Paolo P. Lava, May 25 2015

Status

approved