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A228116
a(n) = A006879(n) - A228115(n).
6
1, 0, 0, 0, -3, -26, 200, 2154, 11484, 19600, -477397, -8219901, -91253055, -827443165, -6390673975, -40675147794, -175537475858, 224340865430, 16557635792557, 240512852610684, 2400398259375610, 16146663225893061, 5309635516930146, -2257043208658957597, -52738581235904454897
OFFSET
1,5
COMMENTS
Difference between the number of primes with n digits (A006879) and its estimate by squares of odd-indexed Fibonacci polynomials (A228115).
The sequence (A228115) provides exactly the values of pi(10^n)- pi(10^(n-1)) for n=2 to 4 and yields an average relative difference in absolute value, i.e. Average(Abs(A228116(n))/ (A006879(n)) = 1.01656…x10^-2 for 1<=n<=25, better than when using the ((10^n)/log(10^n)) function (Average(Abs(A228066(n))/ (A006879(n)) = 4.69094…x10^-2 (see A228066)), or the Logarithm integral (Li(10^n)-Li(2)) function (Average(Abs(A228068(n))/ (A006879(n)) = 1.75492…x10^-2 (see A228068)), or the Riemann(Riemann (10^n)) function (Average(Abs(A228114(n))/ (A006879(n)) = 1.03936…x10^-2) for 1<=n<=25.
Furthermore, if the first value for n=1 is skipped, the average relative difference in absolute value is improved by nearly two orders of magnitude, i.e. Average(Abs(A228116(n))/ (A006879(n)) = 1.72564…x10^-4 for 2<=n<=25, better than when using the ((10^n)/log(10^n)) function (Average(Abs(A228066(n))/ (A006879(n)) = 4.88640…x10^-2 (see A228066)), or the Logarithm integral (Li(10^n)-Li(2)) function (Average(Abs(A228068(n))/ (A006879(n)) = 7.86383…x10^-3 (see A228068)), or the Riemann(Riemann (10^n)) function (Average(Abs(A228114(n))/ (A006879(n)) = 4.10042…x10^-4), or the product of x and Fibonacci polynomials of multiple of 4 indices F[4n](x) (Average(Abs(A228064(n))/ (A006879(n)) = 3.90981…x10^-3 (see A228112)) for 2<=n<=25.
LINKS
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
Eric Weisstein's World of Mathematics, Prime Counting Function.
FORMULA
a(n) = A006879(n) - A228115(n).
KEYWORD
sign,less,base
AUTHOR
Vladimir Pletser, Aug 10 2013
STATUS
approved