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Add square root of sum of terms.
1

%I #27 Feb 24 2021 16:29:55

%S 1,2,3,5,8,12,17,23,31,41,52,65,81,99,119,142,168,197,229,264,303,346,

%T 392,442,497,556,619,687,760,838,921,1009,1103,1203,1308,1419,1537,

%U 1661,1791,1928,2072,2223,2381,2546,2719,2900,3088

%N Add square root of sum of terms.

%C a(0) = 1; a(n) = a(n-1) + floor(sqrt(Sum_{i=0..n-1} a(i))).

%C This appears to give asymptotically a(n) = n^3/36, sum of terms n^4/144, regardless of the starting value a(0).

%H Robert Israel, <a href="/A275580/b275580.txt">Table of n, a(n) for n = 0..10000</a>

%H Robert Israel, <a href="/A275580/a275580.pdf">Formulas for A275580</a>

%H Christian Krause, <a href="https://github.com/ckrause/loda">LODA</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (3,-4,5,-6,5,-4,3,-1).

%F G.f.: (1-x+x^2-x^3+x^4)/((1-x)^3(1+x^2-x^3-x^5)). See link "Formulas for A275580". - _Robert Israel_, Aug 09 2016

%F a(n) = n + 1 + Sum_{i=0..n} floor((floor(i^2 / 3) + i) / 4); derived from corresponding LODA program (see link). - _Hugo van der Sanden_, Feb 24 2021

%e a(3) = a(2) + floor(sqrt(1 + 2)) = 2 + 1 = 3;

%e a(4) = a(3) + floor(sqrt(1 + 2 + 3)) = 3 + 2 = 5.

%p G:= (x^4-x^3+x^2-x+1)/((x^5+x^3-x^2-1)*(x-1)^3):

%p S:= series(G,x,101):

%p seq(coeff(S,x,j),j=0..100); # _Robert Israel_, Aug 09 2016

%t a = {1}; Do[AppendTo[a, a[[k]] + Floor@ Sqrt@ Total@ a], {k, 46}]; a (* _Michael De Vlieger_, Aug 03 2016 *)

%o (PARI) lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = va[n-1] + floor(sqrt(sum(k=1, n-1, va[k])));); va;} \\ _Michel Marcus_, Aug 02 2016

%K easy,nonn

%O 0,2

%A _Hugo van der Sanden_, Aug 02 2016