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A164765 Partial sums of [A080782^2].

%I A164765 #15 Aug 03 2020 12:02:43
%S A164765 1,10,14,30,66,91,140,221,285,385,529,650,819,1044,1240,1496,1820,
%T A164765 2109,2470,2911,3311,3795,4371,4900,5525,6254,6930,7714,8614,9455,
%U A164765 10416,11505,12529,13685,14981,16206,17575,19096,20540,22140,23904,25585
%N A164765 Partial sums of [A080782^2].
%C A164765 Yet another plausible solution to A115603.
%C A164765 The first differences of A115603 are all squares (assuming a prior term of 0), meaning that any sequence beginning 1,3,2,4 is sufficient to account for them; This solution chooses the permutation of integers A080782 = {1,3,2,4,6,5,7,9,8,...}
%C A164765 Ultimately that means this sequence is equal to A000330 for every two out of three consecutive terms, and is greater by 2n+1 where different.
%H A164765 Colin Barker, <a href="/A164765/b164765.txt">Table of n, a(n) for n = 1..1000</a>
%H A164765 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,2,-4,2,-1,2,-1).
%F A164765 a(n) = ( n(n+1) + 6 - 8*sin^2(Pi*(n+1)/3) )*(2n+1)/6.
%F A164765 a(n) = Sum_{k=0..n} A080782(k)^2.
%F A164765 From _Colin Barker_, Aug 03 2020: (Start)
%F A164765 G.f.: x*(1 + 8*x - 5*x^2 + 10*x^3 + 4*x^4 - x^5 + x^6) / ((1 - x)^4*(1 + x + x^2)^2).
%F A164765 a(n) = 2*a(n-1) - a(n-2) + 2*a(n-3) - 4*a(n-4) + 2*a(n-5) - a(n-6) + 2*a(n-7) - a(n-8) for n>8.
%F A164765 (End)
%t A164765 Accumulate[Array[#+Mod[#+1,3]&,70,0]^2] (* _Harvey P. Dale_, Mar 29 2013 *)
%o A164765 (PARI) Vec(x*(1 + 8*x - 5*x^2 + 10*x^3 + 4*x^4 - x^5 + x^6) / ((1 - x)^4*(1 + x + x^2)^2) + O(x^40)) \\ _Colin Barker_, Aug 03 2020
%Y A164765 Original puzzle: A115603; Used in this solution: A080782, A000330; Other solutions: A115391, A116955, A162899
%K A164765 easy,nonn
%O A164765 1,2
%A A164765 _Carl R. White_, Aug 25 2009