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URL: https://oeis.org/A390321

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A390321
Limit of A390254(n)/(4/3)^n.
5
2, 0, 7, 6, 3, 9, 5, 7, 3, 0, 7, 9, 9, 6, 6, 6, 7, 1, 3, 8, 1, 4, 0, 7, 1, 7, 6, 5, 3, 6, 2, 7, 5, 4, 6, 0, 6, 7, 1, 4, 2, 5, 1, 7, 9, 1, 4, 7, 5, 0, 9, 8, 4, 3, 1, 7, 3, 8, 2, 8, 8, 3, 0, 3, 7, 4, 4, 5, 0, 7, 3, 6, 6, 5, 0, 9, 1, 8, 8, 4, 4, 5, 1, 0, 0, 7, 6, 9, 7, 7, 4, 9, 9, 5, 7, 1, 2, 9, 3, 7, 9
OFFSET
1,1
COMMENTS
In general, let {f_n}_{n>=0} be the sequence defined by f_{n+1} = alpha*f_n + e_n, where alpha > 1, a <= e_n <= b, then f_n = (f_0 + e_0/alpha + ... + e_{n-1}/alpha^n)*alpha^n = c*alpha^n - (e_n/alpha + e_{n+1}/alpha^2 + ...), where c = f_0 + Sum_{n>=0} e_n/alpha^{n+1}. We conclude that -(b/(alpha - 1))/alpha^n <= f_n/alpha^n - c <= -(a/(alpha - 1))/alpha^n.
Here we have alpha = 4/3, a = -1/3, and b = 1/3, so we have -(3/4)^n < A390254(n)/(4/3)^n - c < (3/4)^n, since we have neither e_n = -1/3 for all sufficiently large n nor e_n = 1/3 for all sufficiently large n.
EXAMPLE
2.0763957307996667138140717653627546067142...
PROG
(PARI) A390321(prec) = my(N = ceil(log(2)/log(4/3) + prec*log(10)/log(4/3)), A390254_N = 2); for(n=1, N, A390254_N = round(4*A390254_N/3)); [(A390254_N-1)/(4/3.)^N, (A390254_N+1)/(4/3.)^N] \\ outputs a lower bound and an upper bound that differ by at most 10^(-prec), given sufficiently large realprecision
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Nov 01 2025
STATUS
approved