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A171769
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Partial sums of A042964 (numbers congruent to 2 or 3 mod 4).
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3
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2, 5, 11, 18, 28, 39, 53, 68, 86, 105, 127, 150, 176, 203, 233, 264, 298, 333, 371, 410, 452, 495, 541, 588, 638, 689, 743, 798, 856, 915, 977, 1040, 1106, 1173, 1243, 1314, 1388, 1463, 1541, 1620, 1702, 1785, 1871, 1958, 2048, 2139, 2233, 2328, 2426, 2525
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refs;
listen;
history;
text;
internal format)
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OFFSET
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1,1
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COMMENTS
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If we insert an initial 0, and alternate the signs: 0,2,-5,11,-18,28,..., we get a sequence where the average of the first n terms is an integer, with no repeats: specifically A001057(n-1). The sum of the first n terms is (-1)^(n-1)*A093353(n-1). - Franklin T. Adams-Watters, May 20 2010
Suppose that n cards have the numbers 1..2n written on them randomly, one number to a side, and are set out on a table randomly. You have the task of maximizing the sum of the visible numbers by flipping cards. If you have no information other than the numbers on the upper faces, and may not flip any particular card more than once, a(n) is the largest sum you can guarantee in the worst case. - Andrew Woods, Jun 06 2013
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LINKS
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FORMULA
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a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - R. H. Hardin, Nov 13 2011
G.f.: x*(2+x+x^2) / ((1-x)^3*(x+1)). - Colin Barker, Jun 04 2014
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MATHEMATICA
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LinearRecurrence[{2, 0, -2, 1}, {2, 5, 11, 18}, 60] (* G. C. Greubel, Jul 02 2019 *)
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PROG
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(PARI) Vec(x*(x^2+x+2)/((1-x)^3*(x+1)) + O(x^60)) \\ Colin Barker, Jun 04 2014
(Sage) [ceiling(n*(1+2*n)/2) for n in (1..60)] # G. C. Greubel, Jul 02 2019
(GAP) a:=[2, 5, 11, 18];; for n in [5..60] do a[n]:=2*a[n-1]-2*a[n-3] + a[n-4]; od; a; # G. C. Greubel, Jul 02 2019
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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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