OFFSET
1,1
COMMENTS
For odd numbers {5,7,9,11,13,.,k,..} we define the product (k-1)/(k+1) if k == 1 (mod 4), otherwise (k+1)/(k-1). Then (2/3)*(4/3)*(4/5)*(6/5)*(6/7)*(8/7)*... = Pi/4 = 0.78539816339744... < 1, based on the Wallis product. This sequence answers the question, what terms must be deleted such that the product of remaining terms equals 1?.
Since Pi/4 < 1 the terms that must be deleted to make the product equal 1 are all 1 (mod 4). These are the least possible, and apparently their product is Pi/4. The remaining sequence which has product 1 contains more terms that are 3 (mod 4) than 1 (mod 4).
PROG
(PARI) {
pp=1.0;
forstep(n=5, 10^10, 4,
p=(n-1)/(n+1);
n2=n+2; p2=(n2+1)/(n2-1);
pp*=p2;
if(pp>1, pp*=p, print1(n", "))
)
}
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Dimitris Valianatos, Jan 03 2019
STATUS
approved