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A096773
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a(n+2) = 4*a(n) + 1; a(1)=0, a(2) = 3.
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1
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0, 3, 1, 13, 5, 53, 21, 213, 85, 853, 341, 3413, 1365, 13653, 5461, 54613, 21845, 218453, 87381, 873813, 349525, 3495253, 1398101, 13981013, 5592405, 55924053, 22369621, 223696213, 89478485, 894784853, 357913941, 3579139413, 1431655765
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Remainders for classes m of integers n (mod 2^(m+1)). After applying one Collatz (3x+1)-transformation to the so-classified integers the result can be written in two classes (mod 6) only.
This classifying scheme covers all integers >0.
With one 3x+1-transformation T(x;p) := x'= (3x+1)/2^p
all numbers x, described in the form, with the free parameter i>=0, x = i*2^N +a(N) result in x', describable by the two classes with the same parameter i:
x' = i*6 + 1 (for odd N>2), or x' = i*6 + 5 (for even N). Thus
x = 4*i + 3 -> x' = 6*i + 5, x = 8*i + 1 -> x' = 6*i + 1,
x =16*i +13 -> x' = 6*i + 5, x = 32*i + 5 -> x' = 6*i + 1,
x =64*i +53 -> x' = 6*i + 5, x =128*i +21 -> x' = 6*i + 1,
....
all with "i" as a free parameter >=0 covering all natural numbers
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FORMULA
| a(2m) = (5*2^(2m-1) - 1)/3, a(2m-1) = (2^(2m-2)-1)/3.
a(2n+1)=10a(2n)+3. a(n+1)-2a(n)= A001045(n+3), signed. - Paul Curtz (bpcrtz(AT)free.fr), Jul 01 2008
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EXAMPLE
| a(1) = (2^0-1)/3 = 0, a(2) = (5*2^1 - 1) / 3 = 3,
a(3) = (2^2-1)/3 = 1, a(4) = (5*2^3 - 1) / 3 =13,
a(5) = (2^4-1)/3 = 5, a(6) = (5*2^5 - 1) / 3 =53,
a(7) = (2^6-1)/3 =21.
....
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MATHEMATICA
| a[1] = 0; a[2] = 3; a[n_] := a[n] = 4a[n - 2] + 1; Table[ a[n], {n, 35}] (from Robert G. Wilson v Aug 20 2004)
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CROSSREFS
| Bisections are A002450 & A072197.
Sequence in context: A134768 A113139 A143411 * A118384 A133176 A089435
Adjacent sequences: A096770 A096771 A096772 * A096774 A096775 A096776
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KEYWORD
| easy,nonn
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AUTHOR
| Gottfried Helms (helms(AT)uni-kassel.de), Aug 15 2004
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EXTENSIONS
| More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 20 2004
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