cp's OEIS Frontend

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

%I A366397 #11 Dec 12 2023 19:33:39
%S A366397 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,
%T A366397 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,
%U A366397 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
%N A366397 Decimal expansion of the number whose continued fraction terms are one larger than those of Pi.
%H A366397 Sean A. Irvine, <a href="https://github.com/archmageirvine/joeis/blob/master/src/irvine/oeis/a366/A366397.java">Java program</a> (github)
%e A366397 4.12406010228786539167585... = 4 + 1/(8 + 1/(16 + 1/(2 + 1/(293 + ...)))).
%e A366397 Pi = 3.141592653589793238... = 3 + 1/(7 + 1/(15 + 1/(1 + 1/(292 + ...)))).
%o A366397 (PARI)
%o A366397 N = 25;
%o A366397 cf(v) = my(m=contfracpnqn(v)); m[1, 1]/m[2, 1];
%o A366397 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));
%o A366397 pi1 = contfrac(1/2^6*sum(k=0,N,summand(k)));
%o A366397 pi2 = contfrac(1/2^6*sum(k=0,N+1,summand(k)));
%o A366397 n = 0; while(pi1[1..n+1] == pi2[1..n+1], n++);
%o A366397 ap1 = cf(apply(x->x+1, pi1[1..n-1]));
%o A366397 ap2 = cf(apply(x->x+1, pi1[1..n]));
%o A366397 n = 0; while(digits(floor(10^(n+1)*ap1)) == digits(floor(10^(n+1)*ap2)), n++);
%o A366397 A366397 = digits(floor(10^n*ap1));
%Y A366397 Cf. A000796, A001203, A059833.
%K A366397 cons,nonn
%O A366397 1,1
%A A366397 _Rok Cestnik_, Oct 08 2023