%I #75 Sep 22 2021 19:36:20
%S 24,76,160,276,424,604,816,1060,1336,1644,1984,2356,2760,3196,3664,
%T 4164,4696,5260,5856,6484,7144,7836,8560,9316,10104,10924,11776,12660,
%U 13576,14524,15504,16516,17560,18636,19744,20884,22056,23260,24496,25764,27064,28396
%N Sum of the corners of a 2n+1 X 2n+1 square spiral.
%C The 3 X 3 and 5 X 5 spirals are
%C .
%C 7---8---9
%C |
%C 6 1---2
%C | |
%C 5---4---3
%C .
%C with corners 7 + 9 + 5 + 3 = 24
%C and
%C .
%C 21--22--23--24--25
%C |
%C 20 7---8---9--10
%C | | |
%C 19 6 1---2 11
%C | | | |
%C 18 5---4---3 12
%C | |
%C 17--16--15--14--13
%C .
%C with corners 21 + 25 + 17 + 13 = 76.
%C An issue arises when considering a 1 X 1 spiral. For ease, a 1 X 1 spiral happens to have no corners so the corresponding value might be considered as undefined (namely, undefined for n = 0).
%C However, from a theoretical perspective if n is allowed to be 0, meaning that a 1 X 1 spiral can have corners, the formulas below that include A114254 might need reconsideration. With the current formula a(n) = 16*n^2 + 4*n + 4, a(0) = 4, meaning that a 1 X 1 spiral (with value 1) has 4 corners with value 1, giving sum 4. This might pave the way to a discussion, considered parallel with A114254. With the given equations, a 1 X 1 spiral happens to have a corner sum of 4. However, a 1 X 1 spiral has a diagonal sum of 1, from A114254. This seems as to be a contradiction; namely, the first term of A114254 should at least be 4 in this case, as corners constitute a subset of diagonal elements.
%H Colin Barker, <a href="/A325958/b325958.txt">Table of n, a(n) for n = 1..1000</a>
%H Minh Nguyen, <a href="https://aquila.usm.edu/honors_theses/777/">2-adic Valuations of Square Spiral Sequences</a>, Honors Thesis, Univ. of Southern Mississippi (2021).
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = A114254(n) - A114254(n-1).
%F a(n) = 16*n^2 + 4*n + 4.
%F From _Colin Barker_, Sep 10 2019: (Start)
%F G.f.: 4*x*(6 + x + x^2) / (1 - x)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.
%F (End)
%F E.g.f.: -4 + 4*exp(x)*(1 + 5*x + 4*x^2). - _Stefano Spezia_, Sep 11 2019
%e For n=1 (our first value) namely for a 3 X 3 spiral, we get a(1) = 24.
%e For n=2, for a 5 X 5 spiral, we get a(2) = 76.
%t Table[ 16n^2+4n+4, {n, 42}] (* _Metin Sariyar_, Sep 14 2019 *)
%o (PARI) a(n) = 16*n^2 + 4*n + 4;
%o (PARI) Vec(4*x*(6 + x + x^2) / (1 - x)^3 + O(x^40)) \\ _Colin Barker_, Sep 10 2019
%Y Cf. A114254.
%K nonn,easy
%O 1,1
%A _Yigit Oktar_, Sep 10 2019