1+2+3+4+5+6 = 21
1+2+3+4+5+6+7 = 28
1+2+3+4+5+6+7+8 = 36
1+2+3+4+5+6+7+8+9 = 45
Is there a tool, that can do me that job?
How do I get this List {1,3,6,10,15,21,28,36,45}?…
Added by Ahmed Hossam at 2:19pm on September 22, 2013
rated by "<" symbols. Examples: "2<10", "2<4<10", "Pow(2, 1)<5*Sin(3)<10".
The entered text contains 2 or 3 segments separated by two or more consecutive dots. Examples "2..10", "2..4..10", "Pow(2, 1)....5*Sin(3)..10".
If only two segments are provided, then the initial value will be the same as the minimum value. If a bounds number or a default value is written as a simple number, then the number of decimal places will be harvested. I.e. "2..4..10" is not the same as "2..4..10.00" as the former will result in an integer slider and the latter in a slider with two decimal places.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 10:08am on February 15, 2013
13;2} ... 20.{13;12}
21. {21;0}22. {21;1}23. {21;2} ... 41. {21;20}
42. {34;0}43. {34;1}44. {34;2} ... 75. {34;33}
76. {55;0}77. {55;1} ... ....
I want to grab the first 8 [0-7], the next 13[8-20], the next 21[21-42] etc
so i have the (known fibonacci seq) list of numbers on the left here:
C S
8 0
13 8
21 21
34 42
55 76
89 131
144 220
233 364
and i need the list on the right, so that i can select items using a Series (N=1 and S and C from the list above) and a List Item component.
the simple question is:
is there a component that can take a list and accumulate it in this way that I need?
if not, is there anyone that can point me to a simple relevant VB example so i could easily adapt it?
many thanks,
gotjosh…
the paths.
Structure one (paths = 2)
{1;0;0;0;0}(N=10)
{2;0;0;0;0}(N=10)
Structure two (paths = 2)
{1;0;0;0}(N=10)
{2;0;0;0}(N=10)
If i merge the two lists i don't get a structure with 2 paths:
Structure result af merging (2 paths)
{1}(N=20)
{2}(N=20)
as i had expected, but instead a structure with 4 paths because of the difference in amount of zeroes:
Structure result af merging (4 paths)
{1;0;0;0;0}(N=10)
{2;0;0;0;0}(N=10)
{1;0;0;0}(N=10)
{2;0;0;0}(N=10)
The amount of zeroes changes all the time when working on the definition, so what im asking is if there some way to adress paths with * number of zeroes behind.…