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A127330
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Begin with the empty sequence and a starting number s = 0. At step k (k >= 1) append the k consecutive numbers s to s+k-1 and change the starting number (for the next step) to 2s+2.
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5
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0, 2, 3, 6, 7, 8, 14, 15, 16, 17, 30, 31, 32, 33, 34, 62, 63, 64, 65, 66, 67, 126, 127, 128, 129, 130, 131, 132, 254, 255, 256, 257, 258, 259, 260, 261, 510, 511, 512, 513, 514, 515, 516, 517, 518, 1022, 1023, 1024, 1025, 1026, 1027, 1028, 1029, 1030, 1031, 2046
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OFFSET
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0,2
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COMMENTS
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From a TV show.
A129142 and A129143 are similar, slightly more natural, but for a puzzle perhaps too transparent sequences.
Can be seen as a triangle (row by step) read by rows: T(n,k) = T(n-1,k) + 2^n for k < n and T(n,n) = T(n-1,n-1) + 2^n + 1. - Reinhard Zumkeller, Nov 16 2013
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LINKS
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EXAMPLE
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In step 1 starting number 0 is appended to the empty sequence and the next starting number is 2*0 + 2 = 2. In step 2 the two numbers 2, 3 are appended and the starting number is changed to 2*2 + 2 = 6.
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MATHEMATICA
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Join[{0}, Flatten[With[{nn=10}, Range[#[[1]], Total[#]]&/@Thread[ {Accumulate[ 2^Range[nn]], Range[nn]}]]]] (* Harvey P. Dale, Nov 05 2017 *)
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PROG
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(PARI) {v=[]; s=0; for(k=1, 11, w=vector(k, j, j+s-1); s=2*s+2; v=concat(v, w)); for(n=1, #v, print1(v[n], ", "))} \\ Klaus Brockhaus, Mar 31 2007
(Magma) &cat[ [2^k-2..2^k+k-3]: k in [1..11] ]; // Klaus Brockhaus, Mar 31 2007
(Haskell)
a127330 n k = a127330_tabl !! n !! k
a127330_row n = a127330_tabl !! n
a127330_tabl = step 0 1 where
step s k = [s .. s + k - 1] : step (2 * s + 2) (k + 1)
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CROSSREFS
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KEYWORD
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AUTHOR
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Steven Cartier (steven.cartier(AT)rogers.com), Mar 30 2007
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EXTENSIONS
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STATUS
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approved
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