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A178459
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Partial sums of floor(2^n/31).
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2
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0, 0, 0, 0, 1, 3, 7, 15, 31, 64, 130, 262, 526, 1054, 2111, 4225, 8453, 16909, 33821, 67646, 135296, 270596, 541196, 1082396, 2164797, 4329599, 8659203, 17318411, 34636827, 69273660, 138547326, 277094658, 554189322, 1108378650, 2216757307, 4433514621, 8867029249, 17734058505, 35468117017, 70936234042
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OFFSET
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1,6
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COMMENTS
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LINKS
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FORMULA
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a(n) = round((10*2^n - 31*n - 7)/155).
a(n) = floor((10*2^n - 31*n + 22)/155).
a(n) = ceiling((10*2^n - 31*n - 36)/155).
a(n) = round((10*2^n - 31*n - 10)/155).
a(n) = a(n-5) + 2^(n-4) - 1, n > 4,
G.f.: -x^5/((x-1)^2*(2*x-1)*(x^4 + x^3 + x^2 + x + 1)). - Colin Barker, Oct 27 2012
a(n) = Sum_{k=0..n} 2^(n-k) * floor(k/5).
a(n) = floor(2^(n+1)/31) - floor((n+1)/5). (End)
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EXAMPLE
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a(29) = 0 + 0 + 0 + 0 + 1 + 2 + 4 + 8 + 16 + 33 + 66 + 132 + 264 + 528 + 1057 + 2114 + 4228 + 8456 + 16912 + 33825 + 67650 + 135300 + 270600 + 541200 + 1082401 + 2164802 + 4329604 + 8659208 + 17318416 = 34636827.
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MAPLE
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seq(round((10*2^n-31*n-7)/155), n=1..32)
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MATHEMATICA
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PROG
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(Magma) [Round((10*2^n-31*n-7)/155): n in [1..40]]; // Vincenzo Librandi, Jun 21 2011
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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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