OFFSET

0,1

COMMENTS

Consider the "add the square root" sequence {b(n)} starting with 1, i.e., b(1)=1, b(n+1) = b(n) + sqrt(b(n)) for n > 0: at each step, the next term is obtained simply by adding the current term to its square root, so the sequence begins {1, 2, 2 + sqrt(2), 2 + sqrt(2) + sqrt(2 + sqrt(2)), ...}, i.e., {1, 2, 3.414213..., 5.261972..., ...}.

The 1st, 10th, 100th, and 1000th terms are

b(10^0) = 1.0

b(10^1) = 25.956...

b(10^2) = 2452.850...

b(10^3) = 248949.869...

b(10^4) = 24983729.385...

b(10^5) = 2499779700.562...

Since the difference between successive terms is b(n+1) - b(n) = sqrt(b(n)) and the similar differential equation (d/dx)f(x) = sqrt(f(x)) is satisified by the function f(x) = (1/4)*x^2, it is perhaps not surprising that lim_{n->infinity} b(n)/n^2 = 1/4. More precisely, it can be shown that, as n increases, b(n) approaches

(1/4)*n^2

+ u*n

+ u^2 - u/2

+ (-(1/2)*u^2 + u/4 - 1/96)/n

+ ((1/3)*u^3 + 0*u^2 + (-5/48)*u + 7/576)/n^2

+ ...

where u = -(1/4)*log(n) + c

and c = 0.675177442458557139813285625075...

It follows that lim_{n->infinity} (b(n)/n - (n - log(n))/4) = c.

If we were to define {b(n)} instead as b(1)=1, b(n+1) = b(n) + 1/b(n) for n > 0 (i.e., "add the reciprocal" rather than "add the square root"), we would obtain the rational-valued sequence {b(n)} = {1, 2, 5/2, 29/10, 941/290, ...} (see A073833 and, for a constant arising from that sequence, A233770).

EXAMPLE

0.67517744245855713981328562507582763368407315898905...

CROSSREFS

KEYWORD

nonn,cons

AUTHOR

Jon E. Schoenfield, Feb 10 2020

STATUS

approved