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Old 2017-05-16, 16:52   #1
MattcAnderson
 
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"Matthew Anderson"
Dec 2010
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Default This is a sequence with integers

Hi Mersenneforum,

Maple software is so overwhelming. It can algebraically solve almost anything. I don't type Pari much. But I try to shine a light on some math / computer topics.

Request for comments.

Regards,
Matt
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File Type: pdf fibonacci with coefficients again 4.pdf (123.6 KB, 156 views)
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Old 2017-05-17, 00:00   #2
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Quote:
Originally Posted by MattcAnderson View Post
Hi Mersenneforum,

Maple software is so overwhelming. It can algebraically solve almost anything. I don't type Pari much. But I try to shine a light on some math / computer topics.

Request for comments.

Regards,
Matt
you can reduce both the start values by a factor of two as well as the recurrence relationship and get the odd part of the values given.

x1=1
x2=3
x[a]=6*x[a-1]+5*x[a-2]

x[3]=18+5 =23
x[4]=138+15 =153

etc. edit:okay maybe not but you do eliminate a constant factor of two and get another recurrence that is always odd.

Last fiddled with by science_man_88 on 2017-05-17 at 00:02
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Old 2017-05-17, 13:11   #3
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more generally there are 4 ways for the first values to be setup within the integers mod 2

1 0
1 1
0 0
0 1

and 4 ways for the coefficients ( same as above) these combinations result in:


Code:
(10:02) gp > my(a=[0,0]);for(x=1,10,a=concat(a,1*a[#a]+0*a[#a-1]));a
%1 = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(10:03) gp > my(a=[0,1]);for(x=1,10,a=concat(a,1*a[#a]+0*a[#a-1]));a
%2 = [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
(10:03) gp > my(a=[1,1]);for(x=1,10,a=concat(a,1*a[#a]+0*a[#a-1]));a
%3 = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
(10:03) gp > my(a=[1,0]);for(x=1,10,a=concat(a,1*a[#a]+0*a[#a-1]));a
%4 = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(10:03) gp > my(a=[1,0]);for(x=1,10,a=concat(a,0*a[#a]+0*a[#a-1]));a
%5 = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(10:04) gp > my(a=[1,1]);for(x=1,10,a=concat(a,0*a[#a]+0*a[#a-1]));a
%6 = [1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(10:04) gp > my(a=[0,1]);for(x=1,10,a=concat(a,0*a[#a]+0*a[#a-1]));a
%7 = [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(10:04) gp > my(a=[0,0]);for(x=1,10,a=concat(a,0*a[#a]+0*a[#a-1]));a
%8 = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(10:04) gp > my(a=[0,0]);for(x=1,10,a=concat(a,0*a[#a]+1*a[#a-1]));a
%9 = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(10:05) gp > my(a=[1,0]);for(x=1,10,a=concat(a,0*a[#a]+1*a[#a-1]));a
%10 = [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0]
(10:05) gp > my(a=[0,1]);for(x=1,10,a=concat(a,0*a[#a]+1*a[#a-1]));a
%11 = [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1]
(10:05) gp > my(a=[1,1]);for(x=1,10,a=concat(a,0*a[#a]+1*a[#a-1]));a
%12 = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
(10:05) gp > my(a=[1,1]);for(x=1,10,a=concat(a,1*a[#a]+1*a[#a-1]));a
%13 = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144]
(10:06) gp > my(a=[0,1]);for(x=1,10,a=concat(a,1*a[#a]+1*a[#a-1]));a
%14 = [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
(10:06) gp > my(a=[0,0]);for(x=1,10,a=concat(a,1*a[#a]+1*a[#a-1]));a
%15 = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(10:06) gp > my(a=[1,0]);for(x=1,10,a=concat(a,1*a[#a]+1*a[#a-1]));a
%16 = [1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
with the ones that don't result in 0 and 1 equating to 1,1,0,1,1,0,1,1,0,1,1,0 and 0,1,1,0,1,1,0,1,1,0,1,1 and 1,0,1,1,0,1,1,0,1,1,0 when done mod 2.
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Old 2017-05-17, 13:31   #4
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Question

Draw one line on this equation to make it correct
5+5+5+5= 555

Last fiddled with by Harrywill on 2017-05-17 at 13:35
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Old 2017-05-17, 14:09   #5
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Quote:
Originally Posted by Harrywill View Post
Draw one line on this equation to make it correct
5+5+5+5\neq 555
now how does it relate to the topic ?
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Old 2017-05-17, 15:49   #6
LaurV
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Quote:
Originally Posted by Harrywill View Post
Draw one line on this equation to make it correct
5+5+5+5= 555
repeat after me: B, A, N
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Old 2017-05-18, 08:33   #7
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Quote:
Originally Posted by science_man_88 View Post
now how does it relate to the topic ?
[offtopc]Connect the upper and left line end points of one plus sign.[/offtopc]

Last fiddled with by kar_bon on 2017-05-18 at 08:34
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Old 2017-05-23, 21:38   #8
MattcAnderson
 
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Hi Mersenneforum,

So here is another Fibonacci like sequence with different starting values.

Regards,
Matt
Attached Files
File Type: pdf Fibonacci like sequence starting with 2 and 2.pdf (83.6 KB, 179 views)
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Old 2017-05-23, 21:59   #9
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Quote:
Originally Posted by MattcAnderson View Post
Hi Mersenneforum,

So here is another Fibonacci like sequence with different starting values.

Regards,
Matt
you'll note not that other than 2 no other entries in this sequence are prime. They are in fact twice the normal Fibonacci numbers.
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Old 2017-05-27, 01:39   #10
MattcAnderson
 
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"Matthew Anderson"
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Hi Mersenneforum and others,

New sequence, new integers

This one is c(r) = 8*c(r-1)+9*c(r-2).

I am not saying that there is any special significance in this one, but check it out.

Regards,

Matt
Attached Files
File Type: pdf Fibonacci with 8 and 9.pdf (67.9 KB, 143 views)
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