OFFSET
2,1
COMMENTS
Least k such that Product_{p|k} (p+1)/(p-1) = (n+1)/(n-1). As a result, all terms are squarefree. - Charles R Greathouse IV, Sep 06 2017
a(102) > 100000000. - Robert G. Wilson v, Sep 08 2017
a(102) = 8759437837. - Giovanni Resta, Sep 11 2017
a(108) > 2550000000. - Robert G. Wilson v, Sep 20 2017
LINKS
Robert G. Wilson v, Table of n, a(n) for n = 2..107
FORMULA
a(p) = p for all primes p.
EXAMPLE
a(4) = 95 = 5*19 because (psi(5*19) + phi(5*19)) / (psi(5*19) - phi(5*19)) = (6*20 + 4*18) / (6*20 - 4*18) = 4 and 95 is the least number with this property.
MAPLE
N:= 10^7: # to get all terms before the first with a(n) > N
M:= nextprime(N):
A:= Vector(M):
R:= proc(n) mul((i[1]+1)/(i[1]-1), i=ifactors(n)[2]) end proc:
for k from 2 to N do
r:= R(k);
n:= (r+1)/(r-1);
if n::integer and n <= M and A[n] = 0 then
A[n]:= k;
fi
od:
m:=min(select(t -> A[t]=0, [$2..M]))-1:
seq(A[i], i=2..m); # Robert Israel, Sep 06 2017
MATHEMATICA
psi[n_] := If[n < 1, 0, n Sum[ MoebiusMu[d]^2/d, {d, Divisors@ n}]]; f[n_] := Block[{k = 1}, While[(n + 1)*EulerPhi[k] != (n - 1)*psi[k], k++]; k]; Array[f, 52, 2] (* Robert G. Wilson v, Sep 06 2017 *)
PROG
(PARI) a(n)=my(target=2/(n-1)+1, start=n, end=10*n, f); while(1, forfactored(k=start, end, f=k[2][, 1]; if(vecmax(k[2][, 2])==1 && prod(i=1, #f, 2/(f[i]-1)+1)==target, return(k[1]))); start=end+1; end*=2) \\ Charles R Greathouse IV, Sep 06 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Altug Alkan, Sep 06 2017
STATUS
approved