Figurate Numbers

CS(n) = 2*f(n) + 3

Centered square numbers CS(n) = 2n^2 - 2n + 1 equal twice the quadratic f(n) = n^2 - n - 1, plus 3. The same quadratic whose roots are the golden ratio appears again, now dressed as figurate geometry. At positions 1 through 6 the values are {1, 5, 13, 25, 41, 61} -- four primes, one perfect square (25 = 5^2), and the identity.

The Centered Square Theorem

Centered Square Identity (PROVED)
CS(n) = 2n^2 - 2n + 1 = 2*f(n) + 3, where f(n) = n^2 - n - 1 is the golden-ratio quadratic. Proof: 2*f(n) + 3 = 2(n^2 - n - 1) + 3 = 2n^2 - 2n - 2 + 3 = CS(n). QED.
nf(n)NoteCS(n) = 2f+3
1-1f(1) = -11
21f(2) = 15
35f(3) = 5 (prime)13
411f(4) = 11 (prime)25 = 5^2
519f(5) = 19 (prime)41
629f(6) = 29 (prime)61
741f(7) = 41 (prime)85 = 5*17
11109f(11) = 109 (prime)221 = 13*17

At chain primes, f(n) often produces the next chain prime: f(2) = 1, f(3) = 5, f(5) = 19, f(7) = 41. The centered square maps each n to 2*f(n) + 3: so 2 maps to 5, 3 maps to 13, 5 maps to 41.

The Difference Theorem

Centered Square Difference (PROVED)
CS(n+1) - CS(n) = 4n. Proof: CS(n+1) - CS(n) = 2(n+1)n + 1 - 2n(n-1) - 1 = 2n[(n+1) - (n-1)] = 4n. QED.
Gap at n = 1
4
CS(2) - CS(1) = 5 - 1.
Gap at n = 2
8
CS(3) - CS(2) = 13 - 5.
Gap at n = 3
12
CS(4) - CS(3) = 25 - 13.
Gap at n = 7
28
CS(8) - CS(7) = 113 - 85.
Gap at n = 11
44
CS(12) - CS(11) = 265 - 221.

Every gap is 4n. At n = 1, 2, 3, 5, 7, 11: gaps are 4, 8, 12, 20, 28, 44.

The Star-Hex Bridge

Star-Hexagonal Duality (PROVED)
S(n) = 2*CH(n) - 1, where S(n) = 6n(n-1)+1 (star) and CH(n) = 3n(n-1)+1 (centered hexagonal). Proof: 2*CH(n) - 1 = 2(3n(n-1)+1) - 1 = 6n(n-1)+1 = S(n). QED.
nCH(n)S(n) = 2*CH-1Note
111Identity
2713Both prime
319 = f(5)3719 = f(5); 37 prime
43773Both prime
561121 = 11^2Prime -> square
7127 = 2^7-1253 = 11*23Mersenne prime -> composite

At n = 2, the centered hexagonal is 7 and the star number is 13. At n = 3, they are 19 = f(5) and 37 (both prime). Doubling and subtracting 1 transforms centered hexagonals into star numbers.

210 as Figurate Number

T(20) = 210
20th triangular number
20 = 4*5. The ring Z/210 = 2*3*5*7 is the 20th triangular number.
P_5(12) = 210
12th pentagonal number
12 = lcm of exponents in Z/210. Pentagonal at the Carmichael lambda.
42 = P_15(3)
3rd 15-gonal number
42 = 2*3*7, a product of three chain primes.

210 sits at position 20 = 4*5 in the triangular family and position 12 in the pentagonal family. Both positions factor into small chain primes.

Explore: Figurate Calculator

Enter n. See centered square CS(n), the quadratic f(n), gap to next CS, centered hexagonal CH(n), and star number S(n).

Compute figurate numbers at position:

Try: 2 (CS=5), 3 (CS=13), 5 (CS=41), 7 (CS=85=5*17), 11 (CS=221=13*17).

Contrast

AspectStandard viewRing structure
CS values1, 5, 13, 25, 41, 61 -- pattern of 2n^2-2n+1Four of the first six are prime (5, 13, 41, 61); 25 = 5^2
IdentityCS(n) = 2n^2-2n+1, a formula= 2*f(n) + 3: twice the golden-ratio quadratic plus 3
Gaps4n = linear in nAt chain primes: 4, 8, 12, 20, 28, 44
Star-HexS = 2*CH - 1, algebraicAt n=2: CH=7, S=13. At n=3: CH=19=f(5), S=37 (both prime)

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