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An interesting recurrence (Posted on 2017-01-17) Difficulty: 3 of 5
A sequence with the recurrence f(n)=3*f(n-1)+f(n-2) starts with two 1-digit numbers. The sequence contains the 8-digit number ABCDAECD. A≠0, and A, B, C, D, and E are not necessarily distinct. Find all possible values of ABCDAECD.

No Solution Yet Submitted by Math Man    
Rating: 5.0000 (1 votes)

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Why there are many solutions | Comment 4 of 5 |
(In reply to Unreasonable by Jer)

Here is why there are many solutions. Let X=f(1) and Y=f(2). Notice that in all of the solutions, X and Y differ by a multiple of 3. Let Y-X=3N. Let X-N=M. Then, X=M+N and Y=M+4N. The sequence continues like this.


f(1)=M+N
f(2)=M+4N
f(3)=4M+13N
f(4)=13M+43N
f(5)=43M+142N
f(6)=142M+469N
f(7)=469M+1549N
f(8)=1549M+5116N
f(9)=5116M+16897N
f(10)=16897M+55807N
f(11)=55807M+184318N
f(12)=184318M+608761N
f(13)=608761M+2010601N
f(14)=2010601M+6640564N
f(15)=6640564M+21932293N
f(16)=21932293M+72437443N

The coefficients of M and N are numbers in the sequence starting with 1 and 1. This sequence is 1, 1, 4, 13, 43, 142, 469, 1549, 5116, 16897, 55807, 184318, 608761, 2010601, 6640564, 21932293, 72437443... It is A003688 in OEIS. Since X and Y are 1-digit numbers, M and N are either positive or negative 1-digit numbers. Since 2010601, 6640564, 21932293, and 72437443 are all of the form ABCDAECD, then 2010601M+6640564N, 6640564M+21932293N, and 21932293M+72437443N are likely to be of the form ABCDAECD. That is why there are a lot of solutions.

Now, the big question is why the sequence starting with 1 and 1 has numbers of the form ABCDAECD. Well, 1549=1600-51 and 5116=5100+16. Therefore, 1549=511600-510051 and 5116=160016-154900. The sequence continues like this.

1549=-510051+511600
5116=160016-154900
16897=-30003+46900
55807=70007-14200
184318=180018+4300
608761=610061-1300
2010601=2010201+400
6640564=6640664-100
21932293=21932193+100
72437443=72437243+200

The sequence is the sum of the sequence starting with -510051 and 160016 and the sequence starting with 511600 and -154900. The sequence starting with -510051 and 160016 has multiples of 10001. The sequence starting with 511600 and -154900 leads to -100 and 100. Therefore, the sequence starting with 1549 and 5116 has sums of multiples of 10001 and small multiples of 100. That is why the sequence starting with 1 and 1 has numbers of the form ABCDAECD.


  Posted by Math Man on 2017-01-18 16:25:20
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