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%I #16 Sep 12 2017 08:59:38
%S 120,1456,121024,2198352216064,576458003527499776
%N Let s0 and s1 be the sums of the reciprocals of the even and odd divisors of n, respectively. The sequence lists the numbers n such that 3*s0 - 2*s1 = 1.
%C Let D0 = {d0(i)}, i = 1..p, the set of the p even divisors of a number n and D1 = {d1(n)}, j = 1..q the set of the q odd divisors of n. Then a(n) is the number such that 3*Sum_{i=1..p} 1/d0(i)- 2*Sum_{j=1..q} 1/d1(j) = 1.
%C Property of the sequence:
%C We observe that a(n) = 2^(k+1)*(2^k-1)*(2^(k+1) - 3) = (2*A000668(m) + 2)*A000668(m)*(2*A000668(m) - 1) where A000668(m) = 2^k - 1 is a Mersenne prime and (2*A000668(m)-1) = 2^(k+1)- 3 is also a prime number.
%C The corresponding values of k are 2, 3, 5, 13, 19, ... and the corresponding values of m are 1, 2, 3, 5, 7, ...
%C Generalization:
%C It is possible to introduce general sequences of numbers such that a*s0 + b*s1 = c with very interesting properties for some integers a, b, c.
%C Example 1: with (a, b, c) = (2, -1, 1) we find the sequence A064591 = 24, 112, 1984, 32512, ... (non-unitary perfect numbers).
%C Example 2: with (a, b, c) = (2, -1, 0) we find the sequence A016825(n) = 2, 6, 10, 14, 18, 22, ...
%C Example 3: with (a, b, c) = (1, 1, 2) we find the sequence A000396(n) = 6, 28, 496, 8128,... (perfect numbers).
%C Example 4: with (a, b, c) = (4, -3, 1) we find the sequence 48, 224, 3968, 65024, ... = 2*A064591(n) = A000668(n)*2^p for some p where A000668 lists the Mersenne primes.
%C Example 5: with (a, b, c) = (6, -5, 1) we find the sequence 240, 2912, 242048, ... which equals twice the sequence obtained with (a, b, c) = (3, -2, 1).
%C Example 6: with (a, b, c) = (7, -6, 1) we find the sequence 2150, 13104, 24800, ...
%e 120 = 2^3*3*5 = (2*A000668(1)+2)* A000668(1)*(2*A000668(1)-1);
%e 1456 = 2^4*7*13 = (2*A000668(2)+2)* A000668(2)*(2*A000668(2)-1);
%e 121024 = 2^6*31*61 =(2*A000668(3)+2)* A000668(3)*(2*A000668(3)-1);
%e 2198352216064 = 2^14*8191*16381= (2*A000668(5)+2)*A000668(5)*(2*A000668(5)-1);
%e 576458003527499776 = 2^20*524287*1048573 = (2*A000668(7)+2)* A000668(7)*(2*A000668(7)-1).
%p with(numtheory):nn:=100000:
%p for n from 2 by 2 to nn do :
%p x:=divisors(n):n0:=nops(x):s:=sum('x[i]', 'i'=1..n0):
%p s0:=0:s1:=0:
%p for k from 1 to n0 do:
%p if irem(x[k],2)=0
%p then
%p s0:=s0+1/x[k]
%p else
%p s1:=s1+1/x[k]:
%p fi:
%p od:
%p if 3*s0-2*s1=1 then print(n):else fi:od:
%t Do[s0=0;s1=0;Do[d=Divisors[n][[i]];If[Mod[d,2]==0,s0=s0+1/d,s1=s1+1/d],{i,1,Length[Divisors[n]]}];If[3*s0-2*s1==1,Print[n]],{n,2,10^9,2}]
%o (PARI) siod(n) = sumdiv(n, d, (d%2)/d);
%o seod(n) = sumdiv(n, d, (1-d%2)/d);
%o isok(n) = 3*seod(n)-2*siod(n) == 1; \\ _Michel Marcus_, May 16 2015
%Y Cf. A000668.
%K nonn
%O 1,1
%A _Michel Lagneau_, May 16 2015
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