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Mille and Remainder Resolution (Posted on 2016-02-01) Difficulty: 3 of 5
The sequence {S(n)} is defined as follows:
S(1) = S(2) = S(3) = 1, and:
S(k+3) = S(k+2) + S(k+1) + S(k), for k=1,2,.....n

It is known that:
S(28) = 6090307, S(29)= 11201821, and:
S(30) = 20603361.

Determine the remainder when Σk=1 to 28 S(k) is divided by 1000.

No Solution Yet Submitted by K Sengupta    
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Some Thoughts re: spoiler solution; can it clue toward the intended method? Comment 3 of 3 |
(In reply to spoiler solution; can it clue toward the intended method? by Charlie)

Add one more sequence adding consecutive pairs of Sigma sums, call it T(n):

n | T(n) = [Sigma(k=1 to n-1) S(k)] + [Sigma(k=1 to n) S(k)]
5 | 17
6 | 31
7 | 57
8 | 105
9 | 193
etc.

This is S(n) shifted a few places, specifically T(n) = S(n+2)

Also, any sequence defined with the recursion S(k+3) = S(k+2) + S(k+1) + S(k) will have lim S(k+1)/S(k) = 1.8393..., which is the positive real root of x^3-x^2-x-1 = 0.  This is analogous to the Fibonacci sequence limit of 1.618..., the positive real root of x^2-x-1 = 0.

OEIS has S(k) as A000213 and the sums are A001590, which actually has the same recursive generator with different starting values.

  Posted by Brian Smith on 2016-02-02 10:45:39
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