# Find the Fibonacci Patterns

Engineer Toast 05/15/2018. 5 answers, 135 views

You are probably familiar with the Fibonacci sequence where the first two terms are 0, 1 (or sometimes 1, 1) and every term after that is the sum of the previous two. It starts like this:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

Sometimes, the sequence contains numbers that have a particular pattern that I find interesting: the difference between any pair of adjacent digits is the same as any other pair. For instance, in the sequence starting with 0, 1, the 18th term is 987. 9-8=1 and 8-7=1. I am mildly satisfied.

Challenge

Given two initial values F(0) and F(1), output every number in the sequence generated by F(n) = F(n-1) + F(n-2) that meets the following criteria:

• The difference between any pair of adjacent digits is the same as any other pair
• It is as least three digits long (1 and 2 digit numbers are not interesting for this pattern)

Input

• Two non-negative integers less than 10**10 (10 billion)

Output

• All integers that are less than 10**10 and meet the criteria in the Challenge section
• It is acceptable to output digits greater than 10**10 but it is not a requirement
• Given that repeated digits meet the pattern (e.g. 777), it is possible that there are infinite numbers that meet the criteria but your program is not required to output forever
• If no such integers exist, output whatever you want so long as it's not a number (nothing, null, empty array, error message, sad face, etc.)
• If a number matching the pattern appears more than once in the sequence, you can output it once or as many times as it occurs
• If any input meets the criteria, it should be included in the output

Rules

Examples / Test Cases

Input , Output
[1,10] , []

[0,1] , [987]
[2,1] , [123]
[2,3] , [987]

[61,86] , [147]
[75,90] , [420]
[34,74] , [1234]
[59,81] , [2468]
[84,85] , [7531]

[19,46] , [111]
[60,81] , [222]
[41,42] , [333]
[13,81] , [444]
[31,50] , [555]
[15,42] , [666]
[94,99] , [777]
[72,66] , [888]

[15,3] , [159,258]
[22,51] , [321,1357]
[74,85] , [159,4444]
[27,31] , [147,11111]

[123,0] , [123,123,246,369]
[111,0] , [111,111,222,333,555,888]
[111,222] , [111,222,333,555,888]

[3894621507,5981921703] , [9876543210]
[765432099,111111111] , [111111111,876543210,987654321]   

### 5 Answers

Stewie Griffin 05/15/2018.

# Octave, 145 bytes

@(a,x=0)eval('do++x;t{x}=(a*[1,1;1,0]^x)(1);until t{x}>1e11,for i=1:x,if t{i}>99,if diff(k=+num2str(t{i}))==diff(k(1:2)),disp(t{i}),end,end,end')

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Good night! Explanation tomorrow! :)

Chas Brown 05/15/2018.

# Python 2, 102 98 bytes

f=lambda x,y:x<1e10and[x]*(len(set(int(a)-int(b)for a,b in zip(x,x[1:])))<2<99<x)+f(y,x+y)or[]

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Thx for 4 bytes from ovs

dylnan 05/15/2018.

# Jelly, 20 19 bytes

>ȷ2ȧDIEƊ
;+ƝṪȷ¡ÇƇ

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Luis Mendo 05/15/2018.

# MATL, 13 bytes

wy+ttVdd~?tD

Infinite loop that outputs the numbers as they are found.

Try it online! Note that in the online interpreter the results are displayed after the 1-minute time-out.

Let F(n) and F(n+1) denote two generic consecutive terms of the Fibonacci sequence.

### Explanation

         % Do...while
w       %   Swap. Takes the two inputs (implicitly) the first time
%   STACK: F(n+1), F(n)
y       %   Duplicate from below
%   STACK: F(n+1), F(n), F(n+1)
+       %   Add. Note that F(n)+F(n+1) is F(n+2)
%   STACK: F(n+1), F(n+2)
tt      %   Duplicate twice
%   STACK: F(n+1), F(n+2), F(n+2), F(n+2)
V       %   Convert to string. Let the digits of F(n+2) be '3579' for example
%   STACK: F(n+1), F(n+2), F(n+2), '3579'
d       %   Consecutive differences (of ASCII codes)
%   STACK: F(n+1), F(n+2), F(n+2), [2 2 2]
d       %   Consecutive differences
%   STACK: F(n+1), F(n+2), F(n+2), [0 0]
~       %   Logical negate, element-wise
%   STACK: F(n+1), F(n+2), F(n+2), [1 1]
?       %   If top of the stack is non-empty and only contains non-zero entries
%   (this is the case for digits '3579', but not for '3578' or '33')
t     %     Duplicate
%     STACK: F(n+1), F(n+2), F(n+2), F(n+2)
D     %     Display immediately. This prints the top-most copy of F(n+2)
%     STACK: F(n+1), F(n+2), F(n+2)
%   End (implicit)
% End (implicit). The top of the stack is consumed as loop condition.
% Since it is non-zero, and thus truthy, a new iteration will begin,
% with the stack containing F(n+1), F(n+2)

Arnauld 05/15/2018.

# JavaScript (ES6), 85 bytes

f=(p,q,a=[])=>p<1e10?f(q,p+q,p<100|[...p+''].some(x=d=n=>d-(d=x-(x=n)))?a:[...a,p]):a

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