A114833 Each term is previous term plus ceiling of root mean square of two previous terms.

0, 1, 2, 4, 8, 15, 28, 51, 93, 168, 304, 550, 995, 1799, 3253, 5882, 10635, 19229, 34767, 62861, 113656, 205497, 371550, 671782, 1214618, 2196094, 3970654, 7179153, 12980288, 23469047, 42433278, 76721609, 138716724, 250807167, 453472612, 819902445, 1482426947, 2680306255, 4846135343

Offset: 0

Author

Jonathan Vos Post, Feb 19 2006

Keywords

• easy
• nonn

Examples

			a(3) = 2 + ceiling[sqrt[(1^2 + 2^2)/2]] = 2 + ceiling[Sqrt[5/2]] = 2 + 2 = 4.
a(4) = 4 + ceiling[sqrt[(2^2 + 4^2)/2]] = 4 + ceiling[Sqrt[20/2]] = 4 + 4 = 8.
a(5) = 8 + ceiling[sqrt[(4^2 + 8^2)/2]] = 8 + ceiling[Sqrt[80/2]] = 8 + 7 = 15.
a(6) = 15 + ceiling[sqrt[(8^2 + 15^2)/2]] = 15 + ceiling[Sqrt[289/2]] = 15 + 13 = 28.
a(7) = 28 + ceiling[sqrt[(15^2 + 28^2)/2]] = 28 + ceiling[Sqrt[1009/2]] = 28 + 23 = 51.
a(8) = 51 + ceiling[sqrt[(28^2 + 51^2)/2]] = 51 + ceiling[Sqrt[3385/2]] = 51 + 42 = 93.
a(9) = 93 + ceiling[sqrt[(51^2 + 93^2)/2]] = 93 + ceiling[Sqrt[11250/2]] = 93 + 75 = 168 [the 75 is an exact value].
a(10) = 168 + ceiling[sqrt[(93^2 + 168^2)/2]] = 168 + ceiling[Sqrt[36873/2]] = 168 + 136 = 304.
a(11) = 304 + ceiling[sqrt[(168^2 + 304^2)/2]] = 304 + ceiling[Sqrt[120640/2]] = 304 + 246 = 550.
a(12) = 550 + ceiling[sqrt[(304^2 + 550^2)/2]] = 550 + ceiling[Sqrt[394916/2]] = 550 + 445 = 995.
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 0..1000
• Eric Weisstein's World of Mathematics, Root-Mean-Square.
• Eric Weisstein's World of Mathematics, Mean.

Crossrefs

Cf. A065094, A065095.

Programs

Mathematica

a[n_] := a[n] = a[n - 1] + Ceiling[ Sqrt[(a[n - 1]^2 + a[n - 2]^2)/2]]; a[0] = 0; a[1] = 1; Array[a, 39, 0] (* Robert G. Wilson v, Jun 22 2014 *)
nxt[{a_,b_}]:={b,b+Ceiling[Sqrt[(a^2+b^2)/2]]}; Transpose[NestList[nxt,{0,1},40]][[1]] (* Harvey P. Dale, May 12 2015 *)

Formula

a(1) = 1, a(2) = 2, for n>2: a(n+1) = a(n) + ceiling(RMS[a(n),a(n-1)]). a(n+1) = a(n) + ceiling[Sqrt[[a(n)^2]+[a(n-1)^2]/2]].

It can easily be proved via induction that a(n)<=2^n. On the other hand we can derive a lower bound: We derive another sequence of the form b(n) = a*c^n, where "a" and "c" are real numbers. If b(1)<=a(1) and b(2)<=a(2) and a(n+1) = a(n)+Ceiling(Sqrt((a(n)^2+a(n-1)^2)/2)) >= b(n)+Sqrt((b(n)^2+b(n-1)^2)/2) >= b(n+1) then, via induction we can safely conclude that a(n)>=b(n). With this method we can derive that a(n) >= 1.80805^(n-1) (where 1.80... is the positive solution of x^2 = x+Sqrt((x^2+1)/2)). Hence we have 1.80805 < a(n)^(1/n) < 2. Can these bounds be improved? - Stefan Steinerberger, Feb 21 2006

Extensions

a(0) and a(13) onward from Robert G. Wilson v, Jun 22 2014