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 A115391 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. 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Probable answer to the riddle in A115603. Partial sums of the squares of the terms of A116966. LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (2,-1,0,2,-4,2,0,-1,2,-1). FORMULA 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 MATHEMATICA LinearRecurrence[{2, -1, 0, 2, -4, 2, 0, -1, 2, -1}, {1, 10, 14, 30, 55, 104, 140, 204, 285, 406}, 50] (* Harvey P. Dale, Jul 01 2020 *) CROSSREFS Cf A000330, A004770 Sequence in context: A115603 A074778 A162899 * A241162 A164765 A116955 Adjacent sequences:  A115388 A115389 A115390 * A115392 A115393 A115394 KEYWORD nonn,easy AUTHOR Pierre CAMI, Mar 15 2006 EXTENSIONS More terms from Stefan Steinerberger, Mar 31 2006 Entry revised by Don Reble, Apr 06 2006 More terms from Colin Barker, Jul 18 2013 STATUS approved

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Last modified September 24 23:13 EDT 2020. Contains 337325 sequences. (Running on oeis4.)