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User: Jason Betts

Jason Betts has authored 3 sequences.

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

Jason Betts, Oct 21 2012

nonn

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}.
All items describe a radial symmetrical geometric shape around a central circle.
All items are symmetrically reflective and are never non-trigional in geometry.
All hexagonal numbers are contained within this set of Radial Numbers, B(n).
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.
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.
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 }
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

			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.

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

Jason Betts, Illustration of initial terms

A003215 is a subsequence.

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

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

A126359 A sequence of prime numbers that can be expressed using only digits 0 and 1 in minimum ascending bases.

Original entry on oeis.org

3, 5, 7, 37, 43, 223, 1297, 1303, 1549, 2801, 4673, 6571, 10111, 101111

Offset: 3

Jason Betts, Aug 14 2012

= 10 base 3 = 3 + 0
= 10 base 5 = 5 + 0
= 11 base 6 = 6 + 1
= 101 base 6 = 36 + 0 + 1
= 111 base 6 = 36 + 6 + 1
= 1011 base 6 = 216 + 0 + 6 + 1
= 10001 base 6 = 1296 + 0 + 0 + 0 + 1
= 10011 base 6 = 1296 + 0 + 0 + 6 + 1
= 11101 base 6 = 1296 + 216 + 36 + 0 + 1
= 11111 base 7 = 2401 + 343 + 49 + 7 + 1
= 11101 base 8 = 4096 + 512 + 64 + 0 + 1
= 10011 base 9 = 6561 + 0 + 0 + 9 + 1
= 10111 base 10 = 10000 + 0 + 100 + 10 + 1
101111 = 101111 base 10 = 100000 + 0 + 1000 + 100 + 10 + 1

			2801 is a prime in this sequence in base 7. The next prime in base 7 is 17207 but this exceeds the value of a prime in base 8, such that A(n) (base y) < A(n+1) (base y+1) < A(n+1) (base y), so the next number in this sequence must go to the next base 8, which is 4673, because the number after 2081 in base 7 is 17207, and 4673 < 17207.

A215511

Step 1: Starting at the first prime number (3), convert to the minimum base (3, as all primes may be expressed in binary).
Step 2: If the next prime number can be converted into the same base using only 0 and 1 without exceeding the value of the next prime number in the next base, this is the next item in the sequence.
Step 3: If the next prime number cannot be expressed in this base before exceeding the value of the next prime number in the next base, skip this prime number and move on to the next prime number and repeat Step 2.
Step 4: If the next prime number cannot be expressed in this base before exceeding the value of the next prime number in the next base, but can be expressed in the next base, this is the next item in the sequence.

A215511 A sequence of prime numbers expressed as minimum bases using only digits 0 and 1.

Original entry on oeis.org

10, 10, 11, 101, 111, 1011, 10001, 10011, 11101, 11111, 11101, 10011, 10111, 101111

Offset: 3

Jason Betts, Aug 14 2012

= 10 base 3 = 3 + 0
= 10 base 5 = 5 + 0
= 11 base 6 = 6 + 1
= 101 base 6 = 36 + 0 + 1
= 111 base 6 = 36 + 6 + 1
= 1011 base 6 = 216 + 0 + 6 + 1
= 10001 base 6 = 1296 + 0 + 0 + 0 + 1
= 10011 base 6 = 1296 + 0 + 0 + 6 + 1
= 11101 base 6 = 1296 + 216 + 36 + 0 + 1
= 11111 base 7 = 2401 + 343 + 49 + 7 + 1
= 11101 base 8 = 4096 + 512 + 64 + 0 + 1
= 10011 base 9 = 6561 + 0 + 0 + 9 + 1
= 10111 base 10 = 10000 + 0 + 100 + 10 + 1
101111 = 101111 base 10 = 100000 + 0 + 1000 + 100 + 10 + 1

			The first term is 3 in base 3. The next prime in that base is 13, which is greater than the value of the prime in the next base, which is 5 in base 4, so the second term is 5 in base 4.

A126359

Step 1: Starting at the first prime number (3), convert to the minimum base (3, as all primes may be expressed in binary).
Step 2: If the next prime number can be converted into the same base using only 0 and 1 without exceeding the value of the next prime number in the next base, this is the next item in the sequence.
Step 3: If the next prime number cannot be expressed in this base before exceeding the value of the next prime number in the next base, skip this prime number and move on to the next prime number and repeat Step 2.
Step 4: If the next prime number cannot be expressed in this base before exceeding the value of the next prime number in the next base, but can be expressed in the next base, this is the next item in the sequence.

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User: Jason Betts

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

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A126359 A sequence of prime numbers that can be expressed using only digits 0 and 1 in minimum ascending bases.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A215511 A sequence of prime numbers expressed as minimum bases using only digits 0 and 1.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula