A sequence with the recurrence f(n)=3*f(n1)+f(n2) starts with two 1digit numbers. The sequence contains the 8digit number ABCDAECD. A≠0, and A, B, C, D, and E are not necessarily distinct. Find all possible values of ABCDAECD.
(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 YX=3N. Let XN=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 1digit numbers, M and N are either positive or negative 1digit 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=160051 and 5116=5100+16. Therefore, 1549=511600510051 and 5116=160016154900. The sequence continues like this.
1549=510051+511600
5116=160016154900
16897=30003+46900
55807=7000714200
184318=180018+4300
608761=6100611300
2010601=2010201+400
6640564=6640664100
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 20170118 16:25:20 