a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - 3*a(n-5), where a(0) = 1, a(1) = 3, a(2) = 6, a(3) = 11, a(4) = 20, a(5) = 33.
G.f.: (3 + 3 x - x^2 + 3 x^3 - 6 x^4)/(1 - x - 2 x^2 + 2 x^3 - 3 x^4 +
3 x^5)
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a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - 3*a(n-5), where a(0) = 1, a(1) = 3, a(2) = 6, a(3) = 11, a(4) = 20, a(5) = 33.
G.f.: (3 + 3 x - x^2 + 3 x^3 - 6 x^4)/(1 - x - 2 x^2 + 2 x^3 - 3 x^4 +
3 x^5)
allocated for Clark Kimberling
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - 3*a(n-5), where a(0) = 1, a(1) = 3, a(2) = 6, a(3) = 11, a(4) = 20, a(5) = 33.
1, 3, 6, 11, 20, 33, 60, 101, 182, 303, 546, 911, 1640, 2733, 4920, 8201, 14762, 24603, 44286, 73811, 132860, 221433, 398580, 664301, 1195742, 1992903, 3587226, 5978711, 10761680, 17936133, 32285040, 53808401, 96855122, 161425203, 290565366, 484275611
0,2
Conjecture: a(n) = least positive whose base-3 total variation is n; see A297440.
Clark Kimberling, <a href="/A297443/b297443.txt">Table of n, a(n) for n = 0..1000</a>
<a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,3,-3)
If n = (1 mod 4) or n = (1 mod 4), then a(n) = 3*a(n-2); otherwise, a(n) = 3*a(n-2) + 2.
Join[{1}, LinearRecurrence[{1, 2, -2, 3, -3}, {3, 6, 11, 20, 33}, 40]]
Cf. A297440.
allocated
nonn,easy
Clark Kimberling, Jan 21 2018
approved
editing
allocated for Clark Kimberling
allocated
approved