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

%I A114833 #15 Feb 16 2025 08:33:00
%S A114833 0,1,2,4,8,15,28,51,93,168,304,550,995,1799,3253,5882,10635,19229,
%T A114833 34767,62861,113656,205497,371550,671782,1214618,2196094,3970654,
%U A114833 7179153,12980288,23469047,42433278,76721609,138716724,250807167,453472612,819902445,1482426947,2680306255,4846135343
%N A114833 Each term is previous term plus ceiling of root mean square of two previous terms.
%H A114833 Robert G. Wilson v, <a href="/A114833/b114833.txt">Table of n, a(n) for n = 0..1000</a>
%H A114833 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Root-Mean-Square.html">Root-Mean-Square.</a>
%H A114833 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Mean.html">Mean.</a>
%F A114833 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]].
%F A114833 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
%e A114833 a(3) = 2 + ceiling[sqrt[(1^2 + 2^2)/2]] = 2 + ceiling[Sqrt[5/2]] = 2 + 2 = 4.
%e A114833 a(4) = 4 + ceiling[sqrt[(2^2 + 4^2)/2]] = 4 + ceiling[Sqrt[20/2]] = 4 + 4 = 8.
%e A114833 a(5) = 8 + ceiling[sqrt[(4^2 + 8^2)/2]] = 8 + ceiling[Sqrt[80/2]] = 8 + 7 = 15.
%e A114833 a(6) = 15 + ceiling[sqrt[(8^2 + 15^2)/2]] = 15 + ceiling[Sqrt[289/2]] = 15 + 13 = 28.
%e A114833 a(7) = 28 + ceiling[sqrt[(15^2 + 28^2)/2]] = 28 + ceiling[Sqrt[1009/2]] = 28 + 23 = 51.
%e A114833 a(8) = 51 + ceiling[sqrt[(28^2 + 51^2)/2]] = 51 + ceiling[Sqrt[3385/2]] = 51 + 42 = 93.
%e A114833 a(9) = 93 + ceiling[sqrt[(51^2 + 93^2)/2]] = 93 + ceiling[Sqrt[11250/2]] = 93 + 75 = 168 [the 75 is an exact value].
%e A114833 a(10) = 168 + ceiling[sqrt[(93^2 + 168^2)/2]] = 168 + ceiling[Sqrt[36873/2]] = 168 + 136 = 304.
%e A114833 a(11) = 304 + ceiling[sqrt[(168^2 + 304^2)/2]] = 304 + ceiling[Sqrt[120640/2]] = 304 + 246 = 550.
%e A114833 a(12) = 550 + ceiling[sqrt[(304^2 + 550^2)/2]] = 550 + ceiling[Sqrt[394916/2]] = 550 + 445 = 995.
%t A114833 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 *)
%t A114833 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 *)
%Y A114833 Cf. A065094, A065095.
%K A114833 easy,nonn
%O A114833 0,3
%A A114833 _Jonathan Vos Post_, Feb 19 2006
%E A114833 a(0) and a(13) onward from _Robert G. Wilson v_, Jun 22 2014