Eric Weisstein's World of Mathematics, <a href="httphttps://mathworld.wolfram.com/PrimeNumberTheorem.html">Prime Number Theorem</a>.

proposed

approved

editing

proposed

E. Eric Weisstein 's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeNumberTheorem.html">Prime Number Theorem</a>.

approved

editing

_Cino Hilliard (hillcino368(AT)hotmail.com), _, Aug 28 2008

proposed

approved

editing

proposed

This formula was derived from the x-th root formula 1/(x^(1/x) - 1)+ 1/2 and the well known approximation Pi(x) ~ x/(log(x) - 1). If x = 2^n, the formula can be evaluated by repeated square roots avoiding logs.

For little googol = 2^100 the formula gives 18556039405581571438895944827, while Riemann's R(x) = 18560140176092446446103729058.

the well known approximation Pi(x) ~ The formula is much more accurate than x/(log(x) - and for small x, Legendre's constant 1.08366 can be used for the 1/x term as 1). If 08366/x. This is more accurate for small x. However, for large x = 2^n, , the more noble formula 1/(x^(1/x) - 1/x - 1) is superior.

can be evaluated by repeated square roots avoiding logs.

For little googol = 2^100 the formula gives

18556039405581571438895944827. Riemann's R(x) =

18560140176092446446103729058. The formula is much more accurate than x/log(x)

and for small x, Legendre's constant 1.08366 can be used for the 1/x term as

1.08366/x. This is more accurate for small x. However, for large x, the more

noble formula 1/(x^(1/x) - 1/x - 1) is superior.

approved

editing

A Fibonacci type sequence related to a previously submitted vector Markov sequenceNumber of primes less than 10^n using the x-th root approximation formula 1/(x^(1/x) - 1/x - 1) where x = 10^n.

0, 1, 1, 2, 4, 5, 9, 14, 20, 33, 49, 74, 116, 173, 265, 406, 612, 937, 1425, 2162, 3300, 5013, 7625, 11614, 17652, 26865, 40881, 62170, 94612, 143933, 218953, 333158, 506820, 771065, 1173137, 1784706, 2715268, 4130981, 6284681, 9561518, 14546644

6, 26, 168, 1217, 9511, 78029, 661458, 5740303, 50701541, 454011970, 4110416300, 37550193649, 345618860220, 3201414635780, 29816233849000, 279007258230819, 2621647966812030, 24723998785919975, 233922961602470390

0,4

1,1

a(n+1) gives diagonal sums of Riordan array (1/(1-x),x(1+2x)) and partial sums of A052947. - Paul Barry (pbarry(AT)wit.ie), Jul 18 2005

This formula was derived from the x-th root formula 1/(x^(1/x) - 1)+ 1/2 and

the well known approximation Pi(x) ~ x/(log(x) - 1). If x = 2^n, the formula

can be evaluated by repeated square roots avoiding logs.

For little googol = 2^100 the formula gives

18556039405581571438895944827. Riemann's R(x) =

18560140176092446446103729058. The formula is much more accurate than x/log(x)

and for small x, Legendre's constant 1.08366 can be used for the 1/x term as

1.08366/x. This is more accurate for small x. However, for large x, the more

noble formula 1/(x^(1/x) - 1/x - 1) is superior.

E. Weisstein <a href="http://mathworld.wolfram.com/PrimeNumberTheorem.html">Prime Number Theorem</a>

a(n) = a(n-1)+a(n-2)+a(n-3)-2*a(n-4)

G.f.: x/((1-x)(1-x^2-2x^3)); a(n+1)=sum{k=0..n, sum{j=0..floor(k/2), C(j, k-2j)2^(k-2j)}}; - Paul Barry (pbarry(AT)wit.ie), Jul 18 2005

For x = 10^3 a(3) = 168.

{{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {-2, 1, 1, 1}}.{a[n - 4], a[n - 3], a[n - 2], a[n - 1]} a[0] = 0; a[1] = 1; a[2] = 1; a[3] = 2; a[n_Integer?Positive] := a[n] = a[n - 1] + a[n - 2] + a[n - 3] - 2a[n - 4]; aa = Table[a[n], {n, 0, 200}]

(PARI) /* b = 10 in this sequence */ g(n, b) = for(j=1, n, x=b^j; y=1/(x^(1/x) - 1/x -1); print1(floor(y)", "))

nonn,uned,new

nonn

Roger Bagula Cino Hilliard (rlbagulatftnhillcino368(AT)yahoohotmail.com), Mar 25 2005Aug 28 2008

G.f.: x/((1-x)(1-x^2-2x^3)); a(n+1)=sum{k=0..n, sum{j=0..floor(k/2), C(j, k-2j)2^(k-2j)}}; - Paul Barry (pbarry(AT)wit.ie), Jul 18 2005

nonn,uned,new