James E Davis - cp's OEIS frontend

User: James E Davis

James E Davis has authored 2 sequences.

A322332 'Geobonnaci' sequence: a(1)=a(2)=1, thereafter a(n) = round( 2 * sqrt(a(n-1) * a(n-2)) ).

1, 1, 2, 3, 5, 8, 13, 20, 32, 51, 81, 129, 204, 324, 514, 816, 1295, 2056, 3263, 5180, 8222, 13052, 20718, 32888, 52206, 82872, 131551, 208824, 331488, 526204, 835297, 1325951, 2104816, 3341187, 5303804, 8419264, 13364749, 21215216, 33677057, 53458995

Offset: 1

James E Davis, Dec 03 2018

nonn

Named because each term is the nearest integer to twice the geometric mean of the previous two terms. This is similar to the Fibonacci sequence, where each term is twice the arithmetic mean of the previous two terms. In fact, the early terms mirror the Fibonacci sequence.

			For n=6, a(5) is 5 and a(4) is 3. 3 * 5 is 15 and twice the square root of 15 is just above 7.745. This rounds to 8, so a(6) is 8.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

public static int G(int n) {
if(n==1) {return 1;}
if(n==2) {return 1;}
return Math.round(2*Math.sqrt(G(n-1)*G(n-2))); //Recursive definition
} \\ James E Davis, Dec 03 2018

a:=proc(n) option remember: `if`(n<3,1,round(2*sqrt(a(n-1)*a(n-2)))) end: seq(a(n),n=1..50); # Muniru A Asiru, Dec 20 2018

a[1] =1 ; a[2] = 1; a[n_] := a[n] = Round[2 * Sqrt[a[n-1] * a[n-2]]]; Array[a, 40] (* Amiram Eldar, Dec 04 2018 *)
nxt[{a_,b_}]:={b,Round[2*Sqrt[a*b]]}; NestList[nxt,{1,1},40][[;;,1]] (* Harvey P. Dale, Dec 04 2024 *)

seq(n)={my(v=vector(n)); v[1]=v[2]=1; for(n=3, n, v[n]=round(2*sqrt(v[n-1]*v[n-2]))); v} \\ Andrew Howroyd, Dec 03 2018

a(n) ~ c * 2^(2*n/3), where c = 0.50182724761947676453167569419757096890286053854137516239835895319268638286015... - Vaclav Kotesovec, Dec 20 2018

A116543 Number of terms in greedy representation of n in terms of the Lucas numbers.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 3, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 4, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 4, 2, 3, 3, 3, 3, 4, 4, 3, 4, 4, 4, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 4, 2, 3, 3, 3, 3, 4, 4, 3, 4, 4, 4, 2

Offset: 0

James E Davis, Mar 28 2006, Jun 07 2006

I have been studying A007895 and similar sequences and created this sequence as an analog of A007895 for the Lucas sequence (A000032).

			a(12)=2 because 12=11+1.

Clark Kimberling, Table of n, a(n) for n = 0..10000
Ron Knott, Using the Fibonacci numbers to represent whole numbers.

Cf. A000032, A007895, A007953, A130310, A131343.

s = Reverse[Sort[Table[LucasL[n - 1], {n, 1, 22}]]];
t = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2,1]], # > 0 &]] &, Range[1000]] (* Peter J. C. Moses, Oct 18 2012 *)

Let L(n) = max(Lucas numbers < n). Then a(0) = 0, a(n) = 1 + a(n-L(n)).

a(n) = A007953(A130310(n)). - Amiram Eldar, Feb 17 2022

Edited by N. J. A. Sloane, Aug 10 2007

a(0) added by Amiram Eldar, Feb 17 2022

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User: James E Davis

A322332 'Geobonnaci' sequence: a(1)=a(2)=1, thereafter a(n) = round( 2 * sqrt(a(n-1) * a(n-2)) ).

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