A276459 Nested radical expansion of Pi: Pi = sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + sqrt(a(4) + ...)))), with a(1) = 7 and 2 <= a(n) <= 6 for n>1.

7, 6, 2, 6, 6, 5, 5, 2, 4, 6, 3, 4, 2, 4, 6, 3, 6, 3, 3, 5, 4, 3, 6, 3, 3, 3, 4, 3, 6, 6, 4, 3, 3, 4, 5, 5, 2, 6, 2, 5, 4, 3, 4, 6, 6, 2, 3, 5, 2, 3, 5, 4, 2, 3, 2, 4, 2, 6, 4, 6, 3, 3, 4, 3, 4, 6, 3, 4, 6, 5, 2, 2, 2, 3, 4, 5, 5, 5, 2, 4, 3, 6, 4, 3, 6, 3, 2, 6, 2, 4, 5, 6, 2, 3, 2, 5, 2, 3, 2, 3, 3, 5, 4, 4, 6, 4, 2, 4, 5, 4, 6, 5, 3

Offset: 1

Author

Yuriy Sibirmovsky, Sep 03 2016

Keywords

• nonn

Comments

Similar to Bolyai expansion. Uses the fact that for 0

Examples

			Pi^2=7+2+p1, thus a(1)=7;
(2+p1)^2=6+2+p2, thus a(2)=6;
(2+p2)^2=2+2+p3, thus a(3)=2; ... 0<pn<1.
		

Links

• Yuriy Sibirmovsky, Table of n, a(n) for n = 1..250
• Eric Weisstein's World of Mathematics, Bolyai Expansion.

Crossrefs

Cf. A000796 (digits), A001203 (continued fraction).

Programs

Mathematica

Nm=100;
A=Table[1,{j,1,Nm}];
V=Table[1,{j,1,Nm}];
P=Pi;
p0=P;
Do[p1=Floor[p0^2]-2;
A[[j]]=p1;
p0=N[2+p0^2-Floor[p0^2],300],{j,1,Nm}];
Do[v0=Sqrt[A[[n]]];
Do[v1=A[[n-j]]+v0;
v0=Sqrt[v1],{j,1,n-1}];
V[[n]]=v0,{n,1,Nm}];
A

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