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A141000
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Numbers k for which there is a solution to the Jumping Grasshopper game.
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6
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0, 1, 4, 9, 13, 16, 20, 21, 24, 25, 28, 29, 32, 33, 36, 37, 40, 41, 44, 45, 48, 49, 52, 53, 56, 57, 60, 61, 64, 65, 68, 69, 72, 73, 76, 77, 80, 81, 84, 85, 88, 89, 92, 93, 96, 97, 100, 101, 104, 105, 108, 109, 112, 113, 116, 117, 120, 121, 124, 125, 128, 129, 132, 133
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OFFSET
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1,3
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COMMENTS
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That is, numbers k such that there is a choice of signs s_1, s_2, ..., s_k (each +1 or -1) so that (i) 0 <= Sum_{i = 1..j } i*s_i <= k for all 1 <= j <= k-1 and (ii) Sum_{i = 1..k } i*s_i = k. (This forces s_1 = s_2 = s_k = +1.)
It has been shown by Dick Hess and Benji Fisher that a number k >= 20 is in the sequence iff k == 0 or 1 (mod 4). (For a proof see the Applegate link.) It is easy to see that k == 0 or 1 (mod 4) is a necessary condition.
An obvious greedy algorithm (working backwards) does the following: For j = k, k-1, ..., 1, let target_j = k - Sum_{i = j+1..k} i * s_i and set s_j = +1 if target_j >= j and s_j = -1 otherwise. This works unless we hit one of five exceptions, in which we must set s_j = -1 instead of +1.
The five exceptions are when (j, target_j) is (5,5), (6,9), (7,14), (8,8), or (9,13). The algorithm also works for the more general case when the target total target_k is different from k, with the additional exception of (8,20). (End)
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REFERENCES
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Ivan Moscovich, "MATH - Isn't It Beautiful!", 2009.
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LINKS
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FORMULA
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a(n) = (11 - (-1)^n + 4*n)/2 for n > 6.
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 9.
G.f.: -x^2*(x^7+2*x^6+2*x^4-x^3-4*x^2-3*x-1) / ((x-1)^2*(x+1)). (End)
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EXAMPLE
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4 is a member because we can take s_1 = s_2 = s_4 = +1, s_3 = -1. Note in particular that 1 + 2 -3 + 4 = 4. (See the illustration.)
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MATHEMATICA
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{0, 1, 4, 9, 13, 16}~Join~LinearRecurrence[{1, 1, -1}, {20, 21, 24}, 58] (* Jean-François Alcover, Nov 20 2019 *)
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PROG
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(Tcl) # See the notes by D. Applegate above.
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CROSSREFS
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KEYWORD
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nonn,nice,easy
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AUTHOR
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Ivan Moscovich (i.moscovich2(AT)chello.nl), Jul 07 2008
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STATUS
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approved
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