# Investigating Minimal Recursive Growth.  we choose starting terms a 1 … a k and then determine each later term  a 1 always equals zero  a 2 is any.

## Presentation on theme: "Investigating Minimal Recursive Growth.  we choose starting terms a 1 … a k and then determine each later term  a 1 always equals zero  a 2 is any."— Presentation transcript:

Investigating Minimal Recursive Growth

 we choose starting terms a 1 … a k and then determine each later term  a 1 always equals zero  a 2 is any number, or zero  a 3 you pick any number that is greater than or equal to a 1 + a 2 = a 2  For a 2 to a k, must obey the rule that a n greater than or equal to max {a i + a n-i } for i=1…n-1  Then for n > k we define a n = max {a i + a n-i } for i=1…n-1  Is there a shortcut?

 For example if the starting points are 0, 1, 2, 4, and 9 then what are the next five terms? a10 a21 a32 a44 a59 a69 a710 a811 a913 a1018

This is the Mathematica program to get the 1 st 25 terms clear[a] a[1]=0; a[2]=1; a[3]=2; a[4]=4; a[5]=9; a[n_]:=a[n]=Max[Table[a@i+a@(n-i),{i,1,n- 1}]];Table[a@n,{n,1,25}]

a10 a21 a32 a44 a59 a10 a25 a310 a415 a520 a10 a23 a36 a48 a59 a10 a22 a32 a44 a56

a10 a21 a32 a44 a59 a69 a710 a811 a913 a1018 a1118 a1219 a1320 a1422 a1527 a1627 a1728 a1829 a1931 a2036 a2136 a2237 a2338 a2441 a2545

a10 a23 a36 a48 a59 a612 a714 a816 a918 a1020 a1122 a1224 a1326 a1428 a1530 a1632 a1734 a1836 a1938 a2040 a2142 a2244 a2346 a2448 a2550

a10 a25 a310 a415 a520 a625 a730 a835 a940 a1045 a1150 a1255 a1360 a1465 a1570 a1675 a1780 a1885 a1990 a2095 a21100 a22105 a23110 a24115 a25120

a10 a22 a32 a44 a56 a68 a78 a810 a912 a1012 a1114 a1214 a1316 a1418 a1518 a1620 a1720 a1822 a1924 a2024 a2126 a2226 a2328 a2430 a2530

 If the starting points are 0, 1,2, 4, and 9 then the next terms can be calculated by adding the last given term to the first to give the a 6 term then add the last term to second term to get a 7 and add last term to third term to get a 8 and add last term to fourth term to get a 9 and add last term to itself to get a 10  We found 0,1,2,4,9 by evaluating f(x) at the integers, for f(x) continuous, differentiable, increasing, and concave up on (1, 5).

 By concave up or flat we mean f’’(x) ≥ 0  for any two a i, a j where 1 ≤ i < j ≤ k  a h for i ≤ h ≤ j lies below or on the line connecting a i and a j  The shortcut says that a n = pa k + a q where n=pk+q, 0 ≤q<k

 We conjecture that to have the shortcut the starting points should be found by evaluating f at the integers, for f continuous, differentiable, increasing, and concave up on (1, k).  If for a i, a j where 1 ≤ i < j ≤ k, we have that a h for i ≤ h ≤ j lies on or below the line connecting a i and a j  Then a n = pa k + a q where n=pk+q, 0 ≤q<k

Our sequences are examples of Operads. Illustrate the Vincent Ferreiro, Jack F. Douglas, James Warren, and Alamgir Karim Measurements they took in 2002 as certain crystals formed in solution. Their first study was Graph of sinusoidal model of Crystal growth is very similar to one of our zigzag series of graphs C 0,0,0,0,0,1,2,3,4,5,6 Dr. STEFAN FORCEY’s N-fold operads research Graph of experimental Crystal growth is very similar to one of our graphs

Download ppt "Investigating Minimal Recursive Growth.  we choose starting terms a 1 … a k and then determine each later term  a 1 always equals zero  a 2 is any."

Similar presentations