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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
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
OFFSET
0,3
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 = 0..1000 [a(0)=0 prepended by Georg Fischer, Jun 18 2021]
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}, {0, 1, 10, 14, 30, 55, 104, 140, 204, 285, 406}, 50] (* Harvey P. Dale, Jul 01 2020 *)
CROSSREFS
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
Offset adapted to definition by Georg Fischer, Jun 18 2021
STATUS
approved