OFFSET
1,2
COMMENTS
If p > 3 is a prime, then p^2 | a(p).
Does the above statement follow from Wolstenholme's theorem?
If p is a Wolstenholme prime (A088164), then p^3 | a(p).
However, it should be noted that also 7^3 | a(7).
Conjecture: there are no pseudoprimes m such that m^2 | a(m).
Is 7^2 the only weak pseudoprime (i.e., a composite m such that m | a(m))?
LINKS
Robert Israel, Table of n, a(n) for n = 1..1367
FORMULA
EXAMPLE
a(5) = numerator(-1/5 + 1/1+2/2+4/3+8/4+16/5) = numerator(128/15 - 1/5) = numerator(25/3) = 25.
MAPLE
f:= proc(n) local k; numer(-1/n + add(2^(k-1)/k, k=1..n)) end proc:
map(f, [$1..30]); # Robert Israel, Sep 15 2024
MATHEMATICA
n = 30; Numerator[Accumulate @ Table[(2^(k-1))/k, {k, 1, n}] - 1/Range[n]] (* Amiram Eldar, Feb 24 2020 *)
PROG
(PARI) a(n) = numerator(-1/n + sum(k=1, n, 2^(k-1)/k)); \\ Michel Marcus, Feb 24 2020
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Thomas Ordowski, Feb 24 2020
EXTENSIONS
More terms from Amiram Eldar, Feb 24 2020
STATUS
approved