A069813 -id:A069813 - cp's OEIS frontend

A257594 Consider the hexagonal lattice packing of circles; a(n) is the maximal number of circles that can be enclosed by a closed chain of n circles.

0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 4, 5, 7, 8, 10

Offset: 0

N. J. A. Sloane, May 18 2015

			In the hexagonal lattice packing of pennies, one penny can be enclosed by 6 pennies, 2 pennies by eight pennies, 3 pennies by 9 pennies, 4 pennies by 10 pennies, 5 pennies by 11 pennies, and 7 pennies by 12 pennies.

R. L. Graham and N. J. A. Sloane, Penny-Packing and Two-Dimensional Codes, Discrete and Comput. Geom. 5 (1990), 1-11.
Craig Knecht, Classification of spaces between the pennies
Craig Knecht, Trying to relate Sloane's 1990 findings to intercircle volumes
R. J. Mathar, Illustration of conjectured a(9) to a(24)
Kival Ngaokrajang, Illustration of initial terms

Cf. A257481.

Conjecture (derived from Euler's F+V=E+1 formula): a(n) = 1+(A069813(n)-n)/2 = A001399(n-6), which means g.f. is x^6 / ( (1+x)*(1+x+x^2)*(1-x)^3 ). - R. J. Mathar, Jul 14 2015

a(13) and a(14) from R. J. Mathar, Jul 10 2015

A257481 Consider a hole-less cluster of n circles in the hexagonal lattice packing of circles; a(n) is the maximal number of circles that touch 6 circles.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 3, 3, 4

Offset: 1

Peter Woodward, Apr 26 2015

			For a(7), one circle can be completely enclosed by six surrounding circles, so a(7)=1, a(n)=0 for n<7.
For a(10), two circles can be completely enclosed by eight surrounding circles, so a(10)=2.

R. L. Graham and N. J. A. Sloane, Penny-Packing and Two-Dimensional Codes, Discrete and Comput. Geom. 5 (1990), 1-11.
Kival Ngaokrajang, Illustration of initial terms
Eric Weisstein's World of Mathematics, Hexagonal grid
Wikipedia, Circle packing

Cf. A182619, A257594, A069813.

Edited by N. J. A. Sloane, May 18 2015

A353091 Irregular triangle read by rows: T(n, k) is the number of n-step closed walks on the hexagonal lattice having algebraic area k.

Original entry on oeis.org

1, 6, 0, 6, 66, 0, 12, 0, 150, 0, 30, 1020, 0, 420, 0, 84, 0, 6, 0, 3444, 0, 1302, 0, 252, 0, 42, 19890, 0, 11952, 0, 4284, 0, 984, 0, 216, 0, 24, 0, 82062, 0, 42972, 0, 14814, 0, 4248, 0, 990, 0, 216, 0, 18, 449976, 0, 327420, 0, 158970, 0, 57180, 0, 18780, 0, 5190, 0, 1350, 0, 270, 0, 30

Offset: 0

Andrey Zabolotskiy, Apr 22 2022

Rows 0 and 2 have 1 element each; row 1 is empty; for n > 2, we have 0 <= k <= A069813(n).

Rows can be extended to negative k with T(n, -k) = T(n, k). Sums of such extended rows give A002898.

			The triangle begins:
[1]
[]
[6]
[0, 6]
[66, 0, 12]
[0, 150, 0, 30]
[1020, 0, 420, 0, 84, 0, 6]
[0, 3444, 0, 1302, 0, 252, 0, 42]
[19890, 0, 11952, 0, 4284, 0, 984, 0, 216, 0, 24]
[0, 82062, 0, 42972, 0, 14814, 0, 4248, 0, 990, 0, 216, 0, 18]
[449976, 0, 327420, 0, 158970, 0, 57180, 0, 18780, 0, 5190, 0, 1350, 0, 270, 0, 30]
...

Cf. A069813 (greatest area), A002898 (all closed walks), A352838 (square lattice).

For n > 1, row n seems to end with A109047(n).

A327896 a(n) is the minimum number of tiles needed for constructing a polyiamond with n holes.

Original entry on oeis.org

9, 14, 19, 23, 27, 31, 35, 39, 43, 47, 51, 54, 58, 62, 65, 69, 73, 76, 80, 83, 87, 90, 94, 97, 101, 104, 108, 111, 115, 118, 122, 125, 129, 132, 135, 139, 142, 146, 149, 152, 156, 159, 163, 166, 169, 173, 176, 179, 183, 186, 189, 193, 196, 199, 203, 206, 209, 213

Offset: 1

Stefano Spezia, Sep 29 2019

nonn

For n > 0, it is easy to prove that k(n) = floor((3 + sqrt(3*(3+8*n)))/6) is the unique integer that satisfies the inequalities 3*binomial(k,2) <= n <= 3*binomial(k+1,2) of Theorem 1.1 in Malen and Roldán.

Proof: solving in k the above inequalities for n > 0, one gets that x - 1 <= k <= x, where x = (3 + sqrt(3*(3+8*n)))/6. Since 3*(3+8*n) is never a perfect square, it follows that x is not an integer and k = floor(x). QED.

Greg Malen and Érika Roldán, Polyiamonds Attaining Extremal Topological Properties, arXiv:1906.08447 [math.CO], 2019.

Cf. A000217, A000577, A067628, A069813, A070764, A161680.

k:=n->floor((3+sqrt(3*(3+8*n)))/6): a:=n->3*(n+k(n))+1+ceil(2*n/k(n)): seq(a(n), n = 1 .. 58)

k[n_]:=Floor[(3+Sqrt[3*(3+8n)])/6]; a[n_]:=3(n+k[n])+1+Ceiling[2n/k[n]]; Array[a,58]

a(n) = 3*(n + k(n)) + 1 + ceiling(2*n/k(n)), where k(n) = floor((3 + sqrt(3*(3+8*n)))/6).

cp's OEIS Frontend

A257594 Consider the hexagonal lattice packing of circles; a(n) is the maximal number of circles that can be enclosed by a closed chain of n circles.

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A257481 Consider a hole-less cluster of n circles in the hexagonal lattice packing of circles; a(n) is the maximal number of circles that touch 6 circles.

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A353091 Irregular triangle read by rows: T(n, k) is the number of n-step closed walks on the hexagonal lattice having algebraic area k.

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A327896 a(n) is the minimum number of tiles needed for constructing a polyiamond with n holes.

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