|
| |
|
|
A285735
|
|
a(1) = 1, and for n > 1, a(n) = the least squarefree number x such that x > n-x, and n-x is also squarefree.
|
|
9
|
|
|
|
1, 1, 2, 2, 3, 3, 5, 5, 6, 5, 6, 6, 7, 7, 10, 10, 10, 11, 13, 10, 11, 11, 13, 13, 14, 13, 14, 14, 15, 15, 17, 17, 19, 17, 21, 19, 22, 19, 22, 21, 22, 21, 22, 22, 23, 23, 26, 26, 26, 29, 29, 26, 30, 31, 29, 30, 31, 29, 30, 30, 31, 31, 33, 33, 34, 33, 34, 34, 35, 35, 37, 37, 38, 37, 38, 38, 39, 39, 41, 41, 42, 41, 42, 42, 43, 43, 46, 46, 46, 47, 53, 46, 47, 47
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
For n > 1, a(n) = the least squarefree number x >= n/2 for which n-x is also squarefree.
For any n > 1 there is at least one decomposition of n as a sum of two squarefree numbers (cf. A071068 and the Mathematics Stack Exchange link). Of all pairs (x,y) of squarefree numbers for which x <= y and x+y = n, sequences A285734 and A285735 give the unique pair for which the difference y-x is the least possible.
|
|
|
LINKS
|
|
|
|
FORMULA
|
|
|
|
PROG
|
(Python)
from sympy.ntheory.factor_ import core
def issquarefree(n): return core(n) == n
def a285734(n):
if n==1: return 0
j=n//2
while True:
if issquarefree(j) and issquarefree(n - j): return j
else: j-=1
def a285735(n): return n - a285734(n)
print([a285735(n) for n in range(1, 101)]) # Indranil Ghosh, May 02 2017
(PARI) a(n)=for(x=(n+1)\2, n, if(issquarefree(x) && issquarefree(n-x), return(x))); 1 \\ Charles R Greathouse IV, Nov 05 2017
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
