cp's OEIS Frontend

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

%I A218146 #29 Sep 01 2013 16:25:57
%S A218146 1,7,13,19,31,37,43,55,61,91,97,109,121,127,139,151,163,169,217,229,
%T A218146 241,253,265,271,283,295,307,319,331,343,355,367,379,391,397
%N A218146 The number of centered circles that can form hexagonal symmetry. Contains all hexagonal numbers.
%C A218146 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}.
%C A218146 1. All items describe a radial symmetrical geometric shape around a central circle.
%C A218146 2. All items are symmetrically reflective and are never non-trigional in geometry.
%C A218146 3. All hexagonal numbers are contained within this set of Radial Numbers, B(n).
%C A218146 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.
%C A218146 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.
%C A218146 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 }
%C A218146 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.
%C A218146 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
%D A218146 Jason Betts, Maths Experiments, Software Publications, 2005, 36-40.
%H A218146 Jason Betts, <a href="/A218146/a218146.jpg">Illustration of initial terms</a>
%F A218146 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.
%F A218146 1. Bn(m) = H(n), each subsequence always starts with a hexagonal number.
%F A218146 2. Bn(m) < H(n+1), all values of Bn(m) are less than the next hexagonal number.
%F A218146 3. If n=odd, the number of terms of Bn(m), m = { (n-1)/2 + 1}
%F A218146 4. If n=even, the number of terms of Bn(m), m = { (n/2) + 1}
%e A218146 First terms for Bn, where n denotes the number of circles of the radius, including the center:
%e A218146 B1(1) = H(1) = 1.
%e A218146 B1(2) = B(1) + 6 = 7, as n=odd, m=1 term, the only term is 6; H(2)=7.
%e A218146 B2(3) = B(2) + 6 = 13, as n=even, number of terms m=2, the first term is 6.
%e A218146 B2(4) = B(3) + 6 = 19, as n=even, number of terms m=2, last term is 6; H(3)=19.
%e A218146 B3(5) = B(4) + 12 = 31, as n=odd, number of terms m=2, first term is 12.
%e A218146 B3(6) = B(5) + 6 = 37, as n=odd, number of terms m=2, last term is 6; H(4)=37.
%e A218146 B4(7) = B(6) + 6 = 43, as n=even, number of terms m=3, first term is 6.
%e A218146 B4(8) = B(7) + 12 = 55, as n=even, number of terms m=3, second term is 12.
%e A218146 B4(9) = B(8) + 6 = 61, as n=even, number of terms m=3, third term is 6; H(5)=61.
%e A218146 B5(10) = B(9) + 12 = 73, as n=odd, number of terms m=3, first term is 12.
%e A218146 B5(11) = B(10) + 12 = 85, as n=odd, number of terms m=3, second term is 12.
%e A218146 B5(12) = B(11) + 6 = 91, as n=odd, number of terms m=3, third term is 6; H(6)=91.
%Y A218146 A003215 is a subsequence.
%Y A218146 Cf. A004016, A035019, A038589, A038590. - _N. J. A. Sloane_, Dec 08 2012
%K A218146 nonn
%O A218146 1,2
%A A218146 _Jason Betts_, Oct 21 2012