A132022 Decimal expansion of Product_{k>=0} (1 - 1/(2*6^k)).

4, 5, 0, 7, 1, 2, 6, 2, 5, 2, 2, 6, 0, 3, 9, 1, 3, 0, 8, 3, 0, 0, 0, 0, 7, 8, 9, 5, 8, 3, 5, 2, 7, 1, 5, 5, 6, 0, 4, 4, 6, 7, 8, 5, 0, 0, 5, 4, 0, 0, 8, 5, 4, 7, 4, 3, 9, 0, 4, 5, 8, 3, 4, 8, 9, 2, 4, 4, 0, 9, 6, 0, 7, 5, 4, 0, 6, 2, 9, 4, 0, 7, 8, 2, 4, 3, 5, 3, 4, 5, 3, 1, 8, 6, 0, 8, 9, 6, 2, 6, 9, 2, 7

Offset: 0

Author

Hieronymus Fischer, Aug 14 2007

Keywords

• nonn
• cons

Examples

			0.45071262522603913...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..2000

Crossrefs

Cf. A000217, A048651, A098844, A067080, A132019, A132026, A132030, A132034, A167747.

Programs

Mathematica

digits = 103; NProduct[1-1/(2*6^k), {k, 0, Infinity}, NProductFactors -> 200, WorkingPrecision -> digits+5] // N[#, digits+5]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
(1/2)*N[QPochhammer[1/12, 1/6], 200] (* G. C. Greubel, Dec 20 2015 *)

PARI

prodinf(x=0, 1-1/(2*6^x)) \\ Altug Alkan, Dec 20 2015

Formula

Equals lim inf_{n->oo} Product_{k=0..floor(log_6(n))} floor(n/6^k)*6^k/n.

Equals lim inf_{n->oo} A132030(n)/n^(1+floor(log_6(n)))*6^(1/2*(1+floor(log_6(n)))*floor(log_6(n))).

Equals lim inf_{n->oo} A132030(n)/n^(1+floor(log_6(n)))*6^A000217(floor(log_6(n))).

Equals (1/2)*exp(-Sum_{n>0} 6^(-n)*Sum{k|n} 1/(k*2^k)).

Equals lim inf_{n->oo} A132030(n)/A132030(n+1).

Equals (1/2)*(1/12; 1/6){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 20 2015

Equals Product_{n>=1} (1 - 1/A167747(n)). - Amiram Eldar, May 09 2023