

A325958


Sum of the corners of a 2n+1 X 2n+1 square spiral.


2



24, 76, 160, 276, 424, 604, 816, 1060, 1336, 1644, 1984, 2356, 2760, 3196, 3664, 4164, 4696, 5260, 5856, 6484, 7144, 7836, 8560, 9316, 10104, 10924, 11776, 12660, 13576, 14524, 15504, 16516, 17560, 18636, 19744, 20884, 22056, 23260, 24496, 25764, 27064, 28396
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OFFSET

1,1


COMMENTS

The 3 X 3 and 5 X 5 spirals are
.
789

6 12
 
543
.
with corners 7 + 9 + 5 + 3 = 24
and
.
2122232425

20 78910
  
19 6 12 11
   
18 543 12
 
1716151413
.
with corners 21 + 25 + 17 + 13 = 76.
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).
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.


LINKS



FORMULA

a(n) = 16*n^2 + 4*n + 4.
G.f.: 4*x*(6 + x + x^2) / (1  x)^3.
a(n) = 3*a(n1)  3*a(n2) + a(n3) for n>3.
(End)
E.g.f.: 4 + 4*exp(x)*(1 + 5*x + 4*x^2).  Stefano Spezia, Sep 11 2019


EXAMPLE

For n=1 (our first value) namely for a 3 X 3 spiral, we get a(1) = 24.
For n=2, for a 5 X 5 spiral, we get a(2) = 76.


MATHEMATICA



PROG

(PARI) a(n) = 16*n^2 + 4*n + 4;
(PARI) Vec(4*x*(6 + x + x^2) / (1  x)^3 + O(x^40)) \\ Colin Barker, Sep 10 2019


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



