cp's OEIS Frontend

A254009 Numbers that divide the sum of the reverse of their aliquot parts (A069250).

%I A254009 #20 Nov 09 2018 02:47:36
%S A254009 1,6,244,285,944,1242,3738,22644,37686,58950,85512,124944,130410,
%T A254009 133857,235644,3202101,5367582,5663697,45165231,141635817,214939686,
%U A254009 736140702,2395863144,4992033177,28406362140,30364415451
%N A254009 Numbers that divide the sum of the reverse of their aliquot parts (A069250).
%C A254009 A072228 is a subsequence. - _Paolo P. Lava_, Nov 09 2018
%e A254009 Aliquot parts of 944 are 1, 2, 4, 8, 16, 59, 118, 236, 472 and the sum of their reverse is 1 + 2 + 4 + 8 +61 + 95 + 811 + 632 + 274 = 1888. Finally, 1888 / 944 = 2.
%p A254009 with(numtheory):
%p A254009 T:=proc(w) local x,y,z; x:=w; y:=0;
%p A254009 for z from 1 to ilog10(x)+1 do y:=10*y+(x mod 10); x:=trunc(x/10); od; y; end:
%p A254009 P:=proc(q) local a,b,k; global n;
%p A254009 for n from 1 to q do a:=sort([op(divisors(n))]); b:=add(T(a[k]),k=1..nops(a)-1);
%p A254009 if type(b/n,integer) then print(n); fi; od; end: P(10^9);
%o A254009 (PARI) isok(n) = (sumdiv(n, d, (d != n)* eval(concat(Vecrev(Str(d))))) % n) == 0; \\ _Michel Marcus_, Feb 27 2015
%Y A254009 Cf. A069250, A072228, A007691, A254008.
%K A254009 nonn,base
%O A254009 1,2
%A A254009 _Paolo P. Lava_, Jan 22 2015
%E A254009 a(16)-a(26) from _Lars Blomberg_, Feb 27 2015