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A350129
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a(0) = 0; thereafter a(n) = a(n-1)/2 + n if a(n-1) is even, otherwise a(n) = a(n-1) + a(n-2).
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2
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0, 1, 1, 2, 5, 7, 12, 13, 25, 38, 29, 67, 96, 61, 157, 218, 125, 343, 468, 253, 721, 974, 509, 1483, 1992, 1021, 3013, 4034, 2045, 6079, 8124, 4093, 12217, 16310, 8189, 24499, 32688, 16381, 49069, 65450, 32765, 98215, 130980, 65533, 196513, 262046, 131069, 393115, 524184
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OFFSET
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0,4
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COMMENTS
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Terms are even if and only if n is a multiple of 3. This can be shown by induction. The predictable pattern means the sequence is given by a linear recurrence with constant coefficients. - Andrew Howroyd, Dec 15 2021
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LINKS
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FORMULA
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G.f.: x*(1 + 2*x^3)*(1 + x + 2*x^2 - x^3 + x^4)/((1 - x)^2*(1 + x + x^2)^2*(1 - 2*x^3)). - Andrew Howroyd, Dec 15 2021
a(n) = 2^(n/3+3)-2n-8 if n == 0 (mod 3);
a(n) = 2^((n+5)/3)-3 if n == 1 (mod 3);
a(n) = 3*2^((n+4)/3)-2n-7 if n == 2 (mod 3).
(End)
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EXAMPLE
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Start at a(0)=0.
0 is even, so to get a(1), divide by 2 and add n: a(1) = 0/2 + 1 = 1.
1 is odd, so to get a(2), add the previous term: a(2) = a(1) + a(0) = 1 + 1 = 1.
Continuing, we get
n a(n)
- ----
0 0
1 0/2 = 0 + 1 = 1
2 1 + 0 = 1
3 1 + 1 = 2
4 2/2 = 1 + 4 = 5
5 5 + 2 = 7
6 7 + 5 = 12
7 12/2 = 6 + 7 = 13
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MATHEMATICA
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a[0] = 0; a[n_] := a[n] = If[EvenQ[a[n - 1]], a[n - 1]/2 + n, a[n - 1] + a[n - 2]]; Array[a, 50, 0] (* Amiram Eldar, Dec 15 2021 *)
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PROG
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(Python)
a = [0]
[a.append(a[-1] + a[-2] if a[-1]%2 else a[-1]//2 + n) for n in range(1, 49)]
(PARI) seq(n)={my(a=vector(n)); a[1]=0; for(n=2, #a, a[n] = if(a[n-1]%2==0, a[n-1]/2+(n-1), a[n-1]+a[n-2])); a} \\ Andrew Howroyd, Dec 15 2021
(PARI) concat([0], Vec((1 + 2*x^3)*(1 + x + 2*x^2 - x^3 + x^4)/((1 - x)^2*(1 + x + x^2)^2*(1 - 2*x^3)) + O(x^30))) \\ Andrew Howroyd, Dec 15 2021
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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