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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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
The numbers of the sequence are the solution of the differential equation n’=(a-k)n-b, which can also be written as A003415(n)=(a-k)*n-A003958(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) (1-1/p_i) - product(i=1..k)(1+1/p_i) = some integer.
LINKS
R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
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;
end do: # R. J. Mathar, Nov 23 2011
CROSSREFS
Sequence in context: A372290 A102603 A147222 * A044098 A044479 A139581
KEYWORD
nonn,easy
AUTHOR
Paolo P. Lava, Nov 22 2011
STATUS
approved

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Last modified June 22 23:47 EDT 2024. Contains 373629 sequences. (Running on oeis4.)