A372869 Decimal expansion of the number whose continued fraction coefficients are given in A084580.

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

Offset: 0

Author

Jwalin Bhatt, Jul 04 2024

Keywords

• cons
• nonn

Examples

			0.5815803358828329856145006072280655247763056696200923013621215551576710...
		

Crossrefs

Cf. A084580 (continued fraction).

Programs

Python

# Using `sample_gauss_kuzmin_distribution` function from A084580.
from mpmath import mp, iv
def decimal_from_cf(coeffs):
    num = iv.mpf([coeffs[-1], coeffs[-1]+1])
    for coeff in coeffs[-2::-1]:
        num = coeff + 1/iv.mpf(num)
    return 1/num
def get_matching_digits(interval_a, interval_b):
    match_index = 0
    for i, j in zip(interval_a, interval_b):
        if i != j:  break
        match_index += 1
    return interval_a[:match_index]
def compute_kuzmin_digits(prec, num_coeffs):
    assert prec > num_coeffs
    mp.dps = iv.dps = prec
    coeffs = sample_gauss_kuzmin_distribution(num_coeffs)
    x = decimal_from_cf(coeffs)
    a = mp.nstr(mp.mpf(x.a), n=prec, strip_zeros=False)
    b = mp.nstr(mp.mpf(x.b), n=prec, strip_zeros=False)
    return get_matching_digits(a, b)
num = compute_kuzmin_digits(prec=200, num_coeffs=180)
A372869 = [int(d) for d in num[1:] if d != '.']