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A203555 a(n) is the index m that maximizes R_m / p_(2m). 0
2, 2, 4, 4, 5, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 20, 21, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 43, 45, 45, 46, 47, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68, 68 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

The R_n is A104272 and p_n is A000040. The function eta_max allow one to take a value of eta(x) from A191228 and finds a(n). Then one can use A179196 to find the prime index value.

a(n) records on the value of the index, m, when the max value, Max(m>=n,R_m/p_(2*m)) for each n.

The function eta_max(n) can be defined in the following way: With a(n) = m so that eta_max(n) := Max(m>=n,R_m/p_(2*m)) for each n.

Because the ratio R_a(n)/p_{2a(n)} approaches 1 as n approaches infinity, and the absolute max is at a(2), we see that this ratio has local maximums at increasing values of n. This sequence removes the "dips" between the local maximums. The values of a(n) implies, for all i >= n, R_(i+1) < R_a(n)/p_(2*a(n))* R_i. Because of these implications, this sequence can be made into a program finding values of a(n) to a limit L.

LINKS

Table of n, a(n) for n=1..66.

EXAMPLE

For m=98, max n>98 (R_m/p_(2*m)) = R_98/p_196 = 1.196144174 < 6/5 = 1.2

For m=99, max n>99 (R_m/p_(2*m)) = R_107/p_214 = 1549/1307 = 1.178070899 < 2^(0.25)

CROSSREFS

Cf. A104272, A000040, A182873, A191228, A179196.

Sequence in context: A341464 A277758 A240027 * A096197 A097264 A339810

Adjacent sequences:  A203552 A203553 A203554 * A203556 A203557 A203558

KEYWORD

nonn

AUTHOR

John W. Nicholson, Jan 02 2012

STATUS

approved

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Last modified June 16 15:37 EDT 2021. Contains 345063 sequences. (Running on oeis4.)