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A115391
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a(0)=0; then a(4*k+1)=a(4*k)+(4*k+1)^2, a(4*k+2)=a(4*k+1)+(4*k+3)^2, a(4*k+3)=a(4*k+2)+(4*k+2)^2, a(4*k+4)=a(4*k+3)+(4*k+4)^2.
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4
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0, 1, 10, 14, 30, 55, 104, 140, 204, 285, 406, 506, 650, 819, 1044, 1240, 1496, 1785, 2146, 2470, 2870, 3311, 3840, 4324, 4900, 5525, 6254, 6930, 7714, 8555, 9516, 10416, 11440, 12529, 13754, 14910, 16206, 17575, 19096, 20540, 22140, 23821, 25670, 27434, 29370
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OFFSET
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0,3
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COMMENTS
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Probable answer to the riddle in A115603.
Partial sums of the squares of the terms of A116966.
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (2,-1,0,2,-4,2,0,-1,2,-1).
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FORMULA
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G.f.: x*(4*x^7-3*x^6+8*x^5+7*x^4+12*x^3-5*x^2+8*x+1) / ((x-1)^4*(x+1)^2*(x^2+1)^2). - Colin Barker, Jul 18 2013
a(n) = (2*n+1)*(2*n*(n+1)+3*(1+cos(n*Pi)-2*cos(n*Pi/2)))/12. - Luce ETIENNE, Feb 01 2017
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MATHEMATICA
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LinearRecurrence[{2, -1, 0, 2, -4, 2, 0, -1, 2, -1}, {0, 1, 10, 14, 30, 55, 104, 140, 204, 285, 406}, 50] (* Harvey P. Dale, Jul 01 2020 *)
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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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EXTENSIONS
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STATUS
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approved
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