

A304280


Numbers equal to the sum of their aliquot parts, each of them increased by 10.


8



11, 51, 675, 1450, 1755, 1995, 5536, 16448, 621056, 35186255331328, 144115340815630336
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OFFSET

1,1


COMMENTS

Searched up to n = 10^12.
a(12) > 10^18.  Hiroaki Yamanouchi, Aug 28 2018
From Giovanni Resta, May 11 2018: (Start)
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.
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)
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.
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.


LINKS

Table of n, a(n) for n=1..11.


EXAMPLE

Aliquot part of 11 is 1 and 1+10 = 11.
Aliquot parts of 51 are 1, 3, 17 and (1+10) + (3+10) + (17+10) = 51.


MAPLE

with(numtheory): P:=proc(q, k) local n;
for n from 1 to q do if 2*n=sigma(n)+k*(tau(n)1) then print(n);
fi; od; end: P(10^12, 10);


MATHEMATICA

With[{k = 10}, Select[Range[10^6], DivisorSum[#, # + k &]  (# + k) == # &] ] (* Michael De Vlieger, May 14 2018 *)


PROG

(PARI) isok(n) = sumdiv(n, d, if (d < n, d+10)) == n; \\ Michel Marcus, May 14 2018


CROSSREFS

Cf. A000005, A000203, A000396, A304276, A304277, A304278, A304279, A304281, A304282, A304283, A304284.
Sequence in context: A321421 A317021 A199895 * A231386 A191099 A325727
Adjacent sequences: A304277 A304278 A304279 * A304281 A304282 A304283


KEYWORD

nonn,hard,more


AUTHOR

Paolo P. Lava, Giovanni Resta, May 11 2018


EXTENSIONS

a(10)a(11) from Hiroaki Yamanouchi, Aug 28 2018


STATUS

approved



