A317986 Odd numbers k > 1 such that k == 1 (mod 4), Product_{n>=1} (a(n)-1)/(a(n)+1) = Pi/4, and Limit_{n->oo} a(n+1)/a(n) = 3, where a(1) = 13 (see comments).

13, 37, 89, 277, 821, 2465, 7389, 22161, 66469, 199389, 598165, 1794469, 5383413, 16150225, 48450673, 145352013, 436056017, 1308168049, 3924504145

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).

Links

Table of n, a(n) for n=1..19.

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

Cf. A016813.

Sequence in context: A124706 A145990 A089528 * A043190 A043970 A183339

Adjacent sequences: A317983 A317984 A317985 * A317987 A317988 A317989

Keyword

nonn,more

Author

Dimitris Valianatos, Jan 03 2019

Status

approved