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
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)
Programs
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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));