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Pyramid angle = arccos(1/sqrt(e)) in radians.
0

%I #16 May 15 2024 01:22:02

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

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

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

%N Pyramid angle = arccos(1/sqrt(e)) in radians.

%F arccos(1/sqrt(e)) = arccos(1/sqrt(A001113)) = arccos(1/A019774) = arccos(A092605).

%e = 0.91910665729358842204136171489674270575563073402035114...

%e = 52°.660932385299557203750879574284228798519442622806897...

%e = 52°39'.655943117973432225052774457053727911166557368413...

%e = 52°39'39.3565870784059335031664674232236746699934421048..."

%t RealDigits[ ArcCos[ 1/Sqrt[ E]], 10, 111][[1]]

%o (PARI) acos(exp(-1/2)) \\ _Charles R Greathouse IV_, Mar 05, 2012

%Y Cf. A001113, A019774, A092605.

%K nonn,easy,cons

%O 0,1

%A _Robert G. Wilson v_, Mar 04 2012