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

1,1

Links

Table of n, a(n) for n=1..104.

Sean A. Irvine, Java program (github)

Example

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

Prog

(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));

Crossrefs

Cf. A000796, A001203, A059833.

Sequence in context: A127812 A072033 A100353 * A080508 A178141 A063987

Adjacent sequences: A366394 A366395 A366396 * A366398 A366399 A366400

Keyword

cons,nonn

Author

Rok Cestnik, Oct 08 2023

Status

approved