A325958 Sum of the corners of a 2n+1 X 2n+1 square spiral.

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

Offset: 1

Author

Yigit Oktar, Sep 10 2019

Keywords

• nonn
• easy

Comments

The 3 X 3 and 5 X 5 spirals are

.

7---8---9

|

6 1---2

| |

5---4---3

.

with corners 7 + 9 + 5 + 3 = 24

and

.

21--22--23--24--25

|

20 7---8---9--10

| | |

19 6 1---2 11

| | | |

18 5---4---3 12

| |

17--16--15--14--13

.

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.

Examples

			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.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Minh Nguyen, 2-adic Valuations of Square Spiral Sequences, Honors Thesis, Univ. of Southern Mississippi (2021).
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A114254.

Programs

Mathematica

Table[ 16n^2+4n+4, {n, 42}] (* Metin Sariyar, Sep 14 2019 *)

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

Formula

a(n) = A114254(n) - A114254(n-1).

a(n) = 16*n^2 + 4*n + 4.

From Colin Barker, Sep 10 2019: (Start)

G.f.: 4*x*(6 + x + x^2) / (1 - x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.

(End)

E.g.f.: -4 + 4*exp(x)*(1 + 5*x + 4*x^2). - Stefano Spezia, Sep 11 2019