

A199767


Numbers n for which sum_{i=1..k} (1+1/p_i) + product_{i=1..k} (1+1/p_i) is an integer, where p_i are the k prime factors of n (with multiplicity).


6



21, 45, 432, 740, 1728, 3456, 3888, 5616, 12096, 23760, 46656, 52164, 131328, 152064, 186624, 195656, 233280, 311472, 606528, 618192, 746496, 926208, 933120, 979776, 1273536, 1403136, 2985984, 3221456, 3732480, 5178816, 5412096, 5971968, 9704448, 13651200
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OFFSET

1,1


COMMENTS

The numbers of the sequence are the solution of the differential equation n’=(ak)nb, which can also be written as A003415(n)=(ak)*nA003958(n), where k is the number of prime factors of n, and a is the integer Sum(i=1..k)(1+1/p_i) +prod(1=1..k) (1+1/p_i).
The numbers of the sequence satisfy also sum(i=1..k) (11/p_i)  product(i=1..k)(1+1/p_i) = some integer.


LINKS



EXAMPLE

740 has prime factors 2, 2, 5, 37. 1+1/2 +1+1/2 +1+1/5 +1+1/37 =967/185 is the sum over 1+1/p_i. (1+1/2) *(1+1/2) *(1+1/5) *(1+1/37)=513/185 is the product over 1+1/p_i. 967/185 +513/185=8 is an integer.


MAPLE

isA199767 := proc(n)
p := ifactors(n)[2] ;
add(op(2, d)+op(2, d)/op(1, d), d=p) + mul((1+1/op(1, d))^op(2, d), d=p) ;
type(%, 'integer') ;
end proc:
for n from 20 do
if isA199767(n) then
printf("%d, \n", n);
end if;


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



