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A244251
Numbers k such that if m = (sum of the reverses of the aliquot parts of k) then k = (sum of the reverses of the aliquot parts of m).
1
6, 98, 145, 244, 285, 133857
OFFSET
1,1
COMMENTS
A072228 is a subsequence of this sequence.
EXAMPLE
Aliquot parts of 98 are 1, 2, 7, 14, 49 and the sum of their reverses is 1 + 2 + 7 + 41 + 94 = 145. Aliquot parts of 145 are 1, 5, 29 and the sum of their reverses is 1 + 5 + 92 = 98.
MAPLE
with(numtheory): T:=proc(w) local x, y, z; x:=0; y:=w;
for z from 1 to ilog10(w)+1 do x:=10*x+(y mod 10); y:=trunc(y/10); od; x; end:
P:=proc(q) local a, b, k, n; for n from 1 to q do
a:=sort([op(divisors(n))]); b:=add(T(a[k]), k=1..nops(a)-1); a:=sort([op(divisors(b))]); b:=add(T(a[k]), k=1..nops(a)-1);
if b=n then print(n); fi; od; end: P(10^12);
CROSSREFS
Cf. A072228.
Sequence in context: A288544 A064753 A322739 * A187522 A224615 A138913
KEYWORD
nonn,more,base
AUTHOR
Paolo P. Lava, Dec 09 2014
STATUS
approved