A218146 The number of centered circles that can form hexagonal symmetry. Contains all hexagonal numbers.

1, 7, 13, 19, 31, 37, 43, 55, 61, 91, 97, 109, 121, 127, 139, 151, 163, 169, 217, 229, 241, 253, 265, 271, 283, 295, 307, 319, 331, 343, 355, 367, 379, 391, 397

Offset: 1

Author

Jason Betts, Oct 21 2012

Keywords

• nonn

Comments

B(n) contain all of the hexagonal numbers such that: 1. Bn(m) = H(n), each subsequence always starts with a hexagonal number; 2. Bn(m) < H(n+1), all values of Bn(m) are less than the next hexagonal number; 3. If n=odd, the number of terms of Bn(m), m = { (n-1)/2 + 1}; 4. If n=even, the number of terms of Bn(m), m = { (n/2) + 1}.

1. All items describe a radial symmetrical geometric shape around a central circle.

2. All items are symmetrically reflective and are never non-trigional in geometry.

3. All hexagonal numbers are contained within this set of Radial Numbers, B(n).

4. The first term is 1, which is also the first Pythagorean Square and Triangular number. However, as the set B(n) must contain a central circle, this precludes most Pythagorean Triangular Numbers as they do not have a central circle, with 91 being the next one after 1.

5. Start with n=1, where n is the number of circles from the radius to the circumference of the figure and is also the number of circles along the side of the complete hexagonal figure, having the total hexagonal number of circles.

6. If n=odd, then the number of terms from the last hexagonal number to the next hexagonal number are {((n-1)/2) +1}, by adding 2 circles to the center of each side, symmetrically from middle outwards, for each term (12 circles), except for the last term which will always be 1 circle that fills in the corners to make the complete hexagonal geometry (6 circles). Thus { (m-1)x12 + 6 }

7. If n=even, then the number of terms from the last hexagonal number to the next hexagonal number are {(n/2) +1}, by adding 1 circle to the middle of each side as the first term (6 circles), then add 2 circles to each side for each consequent term (12 circles), except for the last term which will always be 1 circle that fills in the corners to make the complete hexagonal geometry (6 circles). Thus {6 + (m-2)x12 + 6} = { (m-1) x 12 }, for all m>2.

The definition is not clear to me, but it sounds like this should be the same as A038590 (the partial sums of A035019). See also A004016. - N. J. A. Sloane, Dec 08 2012

Examples

			First terms for Bn, where n denotes the number of circles of the radius, including the center:
B1(1) = H(1) = 1.
B1(2) = B(1) + 6 = 7, as n=odd, m=1 term, the only term is 6; H(2)=7.
B2(3) = B(2) + 6 = 13, as n=even, number of terms m=2, the first term is 6.
B2(4) = B(3) + 6 = 19, as n=even, number of terms m=2, last term is 6; H(3)=19.
B3(5) = B(4) + 12 = 31, as n=odd, number of terms m=2, first term is 12.
B3(6) = B(5) + 6 = 37, as n=odd, number of terms m=2, last term is 6; H(4)=37.
B4(7) = B(6) + 6 = 43, as n=even, number of terms m=3, first term is 6.
B4(8) = B(7) + 12 = 55, as n=even, number of terms m=3, second term is 12.
B4(9) = B(8) + 6 = 61, as n=even, number of terms m=3, third term is 6; H(5)=61.
B5(10) = B(9) + 12 = 73, as n=odd, number of terms m=3, first term is 12.
B5(11) = B(10) + 12 = 85, as n=odd, number of terms m=3, second term is 12.
B5(12) = B(11) + 6 = 91, as n=odd, number of terms m=3, third term is 6; H(6)=91.
		

References

• Jason Betts, Maths Experiments, Software Publications, 2005, 36-40.

Links

• Jason Betts, Illustration of initial terms

Crossrefs

A003215 is a subsequence.

Cf. A004016, A035019, A038589, A038590. - N. J. A. Sloane, Dec 08 2012

Formula

For all hexagonal numbers H(n), there exists a sequence B(n) such that H(n) < B(n) < .. < B(n+m) < H(n+1), where m = {(n/2)+1} if n=even and m={(n-1)/2 +1} if n=odd.

1. Bn(m) = H(n), each subsequence always starts with a hexagonal number.

2. Bn(m) < H(n+1), all values of Bn(m) are less than the next hexagonal number.

3. If n=odd, the number of terms of Bn(m), m = { (n-1)/2 + 1}

4. If n=even, the number of terms of Bn(m), m = { (n/2) + 1}