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A347945
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To get {a(n)}, start with the nonnegative integers sequence f() and, for each y>=0, shift the f(y) to position f(2y) and reset indices.
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0
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0, 2, 3, 1, 6, 7, 5, 10, 11, 8, 14, 15, 4, 18, 19, 13, 22, 23, 16, 26, 27, 12, 30, 31, 21, 34, 35, 24, 38, 39, 17, 42, 43, 29, 46, 47, 32, 50, 51, 9, 54, 55, 37, 58, 59, 40, 62, 63, 28, 66, 67, 45, 70, 71, 48, 74, 75, 33, 78, 79, 53, 82, 83, 56
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OFFSET
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0,2
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COMMENTS
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To get {a(n)}, we use the working sequences f_x(y), where y is the index and x is both, the x-th working sequence and a control variable. x=0,1,2,3,... up to infinity.
f_x(y) = f_(x-1)(x) if y = 2x, f_(x-1)(y+1) if x <= y < 2x, and s_(x-1)(y) otherwise.
Start with the sequence of nonnegative integers
{f(y)} = 0,1,2,3,4,5,6,7,8,9,10,11,12,...
The first index is 0: For f(y=0), nothing is changed, since f(2*0=0), so we still have
{f_0(y)} = 0,1,2,3,4,5,6,7,8,9,10,11,12,...
For x=1: Shift f(1)=1 to f(2*1)=f(2), and all f(1<y<=2) to f(y-1): f(2)=2 to f(1), leaving
{f_1(y)} = 0,2,1,3,4,5,6,7,8,9,10,11,12,...
For x=2: Shift f(2)=1 to f(2*2)=f(4), and all f(1<y<=4) to f(y-1), leaving
{f_2(y)} = 0,2,3,4,1,5,6,7,8,9,10,11,12,...
Iterating the "cyclic shifting" indefinitely produces {a(n)}.
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Visualization of the term index-position shift:
f_1: 0 1 2 3 4 5 6 7 8 9 10 11 12 ...
└+1─^
f_2: 0 2 1 3 4 5 6 7 8 9 10 11 12 ...
└──+2──^
f_3: 0 2 3 4 1 5 6 7 8 9 10 11 12 ...
└───+3────^
f_4: 0 2 3 1 5 6 4 7 8 9 10 11 12 ...
└─────+4─────^
f_5: 0 2 3 1 6 4 7 8 5 9 10 11 12 ...
└───────+5───────^
f_6: 0 2 3 1 6 7 8 5 9 10 4 11 12 ...
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In the Formula section, formulas (2) and (3) (which build pairs of terms with a certain repeating spacing and difference, unlike formula (1)) give the terms with the single most f(x) to f(2x) shifts.
Some terms f(y) only get shifted to f(y-1), decreasing their final index n; other terms f(y) change position to f(2y) multiple times.
The value "t" in formulas (1), (2) and (3) gives the number of f(2y) shifts "s" of the resulting terms: For (1): s = t, for (2),(3): s = t-1.
E.g., the terms f(y) for the formulas (2) and (3) with t=1 (e.g., 2,3,6,7,10,11,...) only get shifted to f(y-1), e.g., f(y)=14=a(10), decreasing its index n from 14 to 10.
Other terms f(y) for formulas (2) and (3) with t>1 change position to f(2y) one or multiple times; e.g., for formula (2) t=3, k=1: f(y)=28=a(48) increases its position 2 times from its original y=28 up to n=48.
The number of times that the terms f(y) for formula (1) change position to f(2y) is t. E.g., t=5: f(y)=20=a(120) changes positions 5 times, up to n=120.
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LINKS
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FORMULA
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To obtain all terms there are three formulas.
For any t=1,2,3,... and k=0,1,2,3,... constellation:
(1) a((3^t - 3)/2) = (2^(t+2) + (-1)^(t-1) - 9)/6.
(2) a(k*3^t + (5*3^(t-1)-3)/2) = k*2^(t+1) + 2^t + (2^(t+2) + (-1)^(t-1) - 9)/6.
(3) a(k*3^t + (7*3^(t-1)-3)/2) = k*2^(t+1) + 2^t + (2^(t+3) - (-1)^(t-1) - 9)/6.
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EXAMPLE
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Formula (1) has no control value "k" and produces small values (terms) for large index numbers n, compared to formulas (2) and (3).
E.g.:
For formula (1) t=5: a(363)=41.
For formula (2) t=1, k=100: a(301)=402.
Formula (1) produces the "Generalized Jacobsthal numbers" as a subsequence "s": s(A029858(t))=A084639(t), and the differences between those terms are the "Jacobsthal numbers" A001045.
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PROG
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(PARI) shiftv(v, n) = {my(w = v); for (i=1, n-1, w[i] = v[i]; ); for (i=n, 2*n-1, w[i] = v[i+1]; ); w[2*n] = v[n]; w; }
lista(nn) = {my(v = [1..nn], va); for (n=1, nn\2, va = shiftv(v, n); v = va; ); concat(0, vector(#v\2, k, v[k])); } \\ Michel Marcus, Sep 21 2021
(PARI) a(n) = n++; my(c=1, r); while([n, r]=divrem(n, 3); r==1, c++); n<<(c+1) + (r<<1+1)<<c\3 - 1; \\ Kevin Ryde, Oct 09 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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