OFFSET
1,2
COMMENTS
If there is some n > 47 such that a(n) < 0, then there is some k^2 > 47 such that a(k^2) < 0.
If n > 1 is a square number, then a(n) = a(n-1) - n^tau(n).
If n > 1 is a nonsquare number, then a(n) = a(n-1) + n^tau(n).
If n > 1 is a prime, then a(n) = a(n-1) + n^2.
LINKS
Simon Jensen, Table of n, a(n) for n = 1..10000
Simon Jensen, On an extended divisor product summatory function
EXAMPLE
a(4) = a(1) + a(2) + a(3) + (-4)^tau(4) = (-1) + 3 + 12 + (-64) = -52.
MATHEMATICA
Accumulate@ Table[(-n)^DivisorSigma[0, n], {n, 28}] (* Michael De Vlieger, Mar 18 2016 *)
PROG
(PARI) a(n) = sum(k=1, n, (-k)^numdiv(k)); \\ Michel Marcus, Mar 18 2016
CROSSREFS
KEYWORD
sign
AUTHOR
Simon Jensen, Apr 19 2013
STATUS
approved