A366397 Decimal expansion of the number whose continued fraction terms are one larger than those of Pi.

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

Offset: 1

Author

Rok Cestnik, Oct 08 2023

Keywords

• cons
• nonn

Examples

			4.12406010228786539167585... = 4 + 1/(8 + 1/(16 + 1/(2 + 1/(293 + ...)))).
Pi = 3.141592653589793238... = 3 + 1/(7 + 1/(15 + 1/(1 + 1/(292 + ...)))).
		

Links

• Sean A. Irvine, Java program (github)

Crossrefs

Cf. A000796, A001203, A059833.

Programs

PARI

N = 25;
cf(v) = my(m=contfracpnqn(v)); m[1, 1]/m[2, 1];
summand(k) = (-1)^k/2^(10*k)*(-2^5/(4*k+1)-1/(4*k+3)+2^8/(10*k+1)-2^6/(10*k+3)-2^2/(10*k+5)-2^2/(10*k+7)+1/(10*k+9));
pi1 = contfrac(1/2^6*sum(k=0,N,summand(k)));
pi2 = contfrac(1/2^6*sum(k=0,N+1,summand(k)));
n = 0; while(pi1[1..n+1] == pi2[1..n+1], n++);
ap1 = cf(apply(x->x+1, pi1[1..n-1]));
ap2 = cf(apply(x->x+1, pi1[1..n]));
n = 0; while(digits(floor(10^(n+1)*ap1)) == digits(floor(10^(n+1)*ap2)), n++);
A366397 = digits(floor(10^n*ap1));