%I
%S 11,51,675,1450,1755,1995,5536,16448,621056,35186255331328,
%T 144115340815630336
%N Numbers equal to the sum of their aliquot parts, each of them increased by 10.
%C Searched up to n = 10^12.
%C a(12) > 10^18.  _Hiroaki Yamanouchi_, Aug 28 2018
%C From _Giovanni Resta_, May 11 2018: (Start)
%C If p = 2^(1+t) + (1+2*t)*k  1 is a prime, for some t > 0 and k even, then x = 2^t*p is in the sequence where k is the value by which the sum of aliquot parts is increased.
%C In this sequence k = 10; for t = 22 we get 35186255331328, which is a term greater than 621056, but this does not exclude the existence of other intermediate terms following a different solution pattern. (End)
%C With the same searching range, no value has been found for the similar sequence with aliquot parts decreased by 10. For n > 10^12, solutions of the kind x = 2^t*p can be found for t = 20, 33, 57, 76, 134, 422, 496, ...: e.g. 2^20*2096741 = 2198592290816.
%C Terms using odd values of k seem very hard to find. Up to n = 10^12, only three such terms are known: 2, 98, and 8450, for k = 1, 5, and 7, respectively.
%e Aliquot part of 11 is 1 and 1+10 = 11.
%e Aliquot parts of 51 are 1, 3, 17 and (1+10) + (3+10) + (17+10) = 51.
%p with(numtheory): P:=proc(q,k) local n;
%p for n from 1 to q do if 2*n=sigma(n)+k*(tau(n)1) then print(n);
%p fi; od; end: P(10^12,10);
%t With[{k = 10}, Select[Range[10^6], DivisorSum[#, # + k &]  (# + k) == # &] ] (* _Michael De Vlieger_, May 14 2018 *)
%o (PARI) isok(n) = sumdiv(n, d, if (d < n, d+10)) == n; \\ _Michel Marcus_, May 14 2018
%Y Cf. A000005, A000203, A000396, A304276, A304277, A304278, A304279, A304281, A304282, A304283, A304284.
%K nonn,hard,more
%O 1,1
%A _Paolo P. Lava_, _Giovanni Resta_, May 11 2018
%E a(10)a(11) from _Hiroaki Yamanouchi_, Aug 28 2018
