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A366397
Decimal expansion of the number whose continued fraction terms are one larger than those of Pi.
1
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
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
KEYWORD
cons,nonn
AUTHOR
Rok Cestnik, Oct 08 2023
STATUS
approved