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Theorem.For level $N=6$ and all $m \ge 1$ such that $T(m)+3$ is even (equivalently, $m \equiv 1$ or $2 \pmod{4}$),$$\dim S_{T(m)+3}!\bigl(\Gamma_0(6)\bigr) = T(m).$$
In particular, every odd triangular number $T(m)$ is realised as a cusp-form dimension at weight $k=T(m)+3$ on $\Gamma_0(6)$.
Proof. From §3.3 and Stein, Proposition 6.1, $\dim S_k\!\bigl(\Gamma_0(6)\bigr) = k-3$ for even$k \ge 4$. The weight $k = T(m)+3$ is even precisely when $T(m)$ is odd, which occurs when $m \equiv 1$ or $2 \pmod{4}$. Substituting $k = T(m)+3$ gives $T(m)+3-3 = T(m)$. $\blacksquare$
The corollary gives a catalogue restricted to$m \equiv 1, 2 \pmod{4}$:
$m$
1
2
5
6
9
10
13
14
$T(m)$
1
3
15
21
45
55
91
105
$k=T(m)+3$
4
6
18
24
48
58
94
108
$m \bmod 4$
1
2
1
2
1
2
1
2
The differences between successive valid weights are $2, 12, 6, 24, 10, 36, 14, \ldots$, which do not form a simple arithmetic progression.
5.2. Eisenstein Twin Theorem
Theorem.For level $N=6$ and all even $k\ge 4$, the Eisenstein subspace dimension is constant:$$\dim E_k!\bigl(\Gamma_0(6)\bigr) = 2.$$
Proof. The total dimension formula for even$k$ gives
$$\dim M_k!\bigl(\Gamma_0(6)\bigr) = (k-1)(0-1) + 0 + 0 + \tfrac{k-1}{2}\cdot 4 = (k-1)(-1) + 2(k-1) = k - 1.$$
The full modular space at level $N$ satisfies $\dim M_k = \dim S_k + \dim E_k$, and therefore
$$\dim E_k!\bigl(\Gamma_0(6)\bigr) = \dim M_k - \dim S_k = (k-1) - (k-3) = 2.$$
Cusp analysis: $X_0(6)$ has four cusps at ${0,,1/2,,1/3,,1/6}$ with widths $1,,2,,3,,6$. The Eisenstein series at level $6$ consists of those of width $1$ (one series) plus one of each pair of wider cusps that couple under the Atkin-Lehner involution, giving exactly two linearly independent Eisenstein series for even $k\ge 4$. $\blacksquare$
Remark. For odd$k \ge 3$, every Eisenstein series at level $\ge 2$ vanishes by invariance under $-I$, so $\dim E_k=0$. The theorem therefore applies only to even weights.
CF period of $\sqrt{21}=[4;\overline{1,1,2,1,1,8}]$
$6$
6
$\dim S_{24}\!\bigl(\Gamma_0(6)\bigr)$
weight-$24$ cusp dimension on $\Gamma_0(6)$
$21$
Items $1$ through $5$ equal $6$ (the index). Item $6$ equals $21$ (the value).
5.4. Connection to $\dim S_k\!\bigl(\Gamma_0(21)\bigr)$
For comparison, level $N=21$ has modular signature $(g,\nu_2,\nu_3,\nu_\infty)=(1,0,2,4)$, using the counts in Stein, definitions preceding Proposition 6.1. Direct computation:
The level $21$ does not return to $21$ at any weight in the table above, in contrast to level $6$. The cusp dimension at level $6$, weight $24$ equals $21=T(6)$ (see §3.3), while at level $21$ the cusp dimension at weight $4$ equals $6=R(3,3)$ (table).
5.5. Polygonal Richness and the Fibonacci Intersection
Define the polygonal richness$\rho(N)$ as the number of distinct pairs $(s,n)$ with $s\ge 3$, $n\ge 2$ such that $P(s,n)=N$, where $P(s,n)=\frac{(s-2)n^2-(s-4)n}{2}$ is the $n$-th $s$-gonal number (polygonal number).
Computation. For $N=21$, solving $P(s,n)=21$ yields exactly three solutions:
Observation. Among Fibonacci numbers below $1000$, namely $F_n$ for $n\le 16$, the values $21$, $55$, and $144$ each have polygonal richness $\rho=3$. All other Fibonacci numbers below $1000$ have $\rho\le 2$.
The general formula for polygonal richness: $s(n)=\frac{2n^2-4n+2N}{n(n-1)}$ must be a positive integer $\ge 3$. For $N=21$, this holds at $n=2,3,6$. For $N=55$, it holds at $n=2,5,10$. For $N=144$, it holds at $n=2,3,12$.
Proof. The equation $\lambda(n)=6$ has exactly $14$ solutions below $200$:
$$n\in{7,9,14,18,21,28,36,42,56,63,72,84,126,168}.$$
Among these, the triangular numbers are ${21,28,36}$ because $T(6)=21$, $T(7)=28$, and $T(8)=36$. Of these, only $21$ is a Fibonacci number, namely $F_8=21$. $\blacksquare$
This establishes the Carmichael function constraint as the binding condition in the quintuple identity. It eliminates $55$, which has $\lambda(55)=20\neq 6$, and all other triangular-Fibonacci intersections.
5.7. The Genus Transition at Triangular Levels
For the modular curve $X_0(N)$ with $N=T(n)$ (modular curve), the genus $g$ is:
$$g=1+\frac{\mu}{12}-\frac{\nu_2}{4}-\frac{\nu_3}{3}-\frac{\nu_\infty}{2},$$
where $\mu=N\prod_{p|N}(1+1/p)$ is the index, $\nu_2, \nu_3$ are elliptic point counts, and $\nu_\infty$ is the cusp count.
Computation. For triangular levels $N=T(n)$ with $1\le n\le 7$:
$n$
$T(n)$
$\mu$
$\nu_\infty$
$g$
$1$
$1$
$1$
$1$
$0$
$2$
$3$
$4$
$2$
$0$
$3$
$6$
$12$
$4$
$0$
$4$
$10$
$18$
$4$
$0$
$5$
$15$
$24$
$4$
$1$
$6$
$21$
$32$
$4$
$1$
$7$
$28$
$48$
$6$
$2$
Theorem.The genus of $X_0(T(n))$ transitions from $g=0$ to $g=1$ at $n=5$ ($T(5)=15$). The levels $T(6)=21$ and $T(7)=28$ both have positive genus, with $g=1$ and $g=2$ respectively.
This genus transition marks a structural change in the geometric complexity of the modular curve: for $n\le 4$, $X_0(T(n))$ is a rational curve (genus $0$), while for $n\ge 5$, it has non-trivial topology.
5.8. The Quadratic Field Connection
The fields $\mathbb{Q}(\sqrt{5})$ and $\mathbb{Q}(\sqrt{21})$ are both real quadratic fields with class number $h=1$ (class-number-one fields).
For $\mathbb{Q}(\sqrt{5})$:
Ring of integers: $\mathbb{Z}[\varphi]$ where $\varphi=(1+\sqrt{5})/2$
Fundamental unit: $\varphi$
Discriminant: $\Delta=5$
For $\mathbb{Q}(\sqrt{21})$:
Ring of integers: $\mathbb{Z}[(1+\sqrt{21})/2]$ (since $21\equiv 1\pmod{4}$)
Fundamental unit: $\varepsilon=(5+\sqrt{21})/2$
Discriminant: $\Delta=21$
Theorem.Both $\mathbb{Q}(\sqrt{5})$ and $\mathbb{Q}(\sqrt{21})$ have class number $1$, establishing unique factorization in their rings of integers.
This class-number-one property, tabulated for both fields in the class-number-one list, connects Fibonacci numbers, which are governed by $\mathbb{Q}(\sqrt{5})$, with the invariant $\Lambda(6)=1/\sqrt{21}$, which is governed by $\mathbb{Q}(\sqrt{21})$.
Ramification. The discriminant $\Delta=21=3\times 7$ determines that exactly the primes $3$ and $7$ ramify in $\mathbb{Q}(\sqrt{21})$ (fundamental discriminant). This is the arithmetic reason why $21=3\times 7$ appears in the class number formula and L-function identities.
6. Padovan, Narayana, Catalan, and Aliquot Structure
6.1. Padovan Sequence
The Padovan sequence is defined by $P(0)=P(1)=P(2)=1$ and $P(n)=P(n-2)+P(n-3)$ for $n\ge 3$ (Padovan sequence). The first thirteen terms are
$$1,\ 1,\ 1,\ 2,\ 2,\ 3,\ 4,\ 5,\ 7,\ 9,\ 12,\ 16,\ \mathbf{21},$$
with $P(12)=21$.
Closed form. The characteristic polynomial is $x^3 - x - 1 = 0$, whose real root is the plastic constant $\rho \approx 1.3247$. The two complex roots are conjugate with modulus $|\rho'| \approx 0.8688 < 1$. The closed form is
$$P(n) = \frac{\rho^{n}}{(\rho-1)(\rho+1)} + \mathrm{(conjugate\ terms)},$$
and $P(n) \sim \rho^n/((\rho-1)(\rho+1))$ for large $n$.
Membership. The integer $21$ appears in each of the triangular, Fibonacci, Jacobsthal, and Padovan sequences at distinct indices. The Padovan index is $12$ above, and the Jacobsthal partial sum is at index $5$ in §12.3.
6.2. Narayana Numbers
The Narayana numbers refine the Catalan numbers (Narayana number) by
$$N(n,k) = \frac{1}{n}\binom{n}{k}\binom{n}{k-1}, \qquad C_n = \sum_{k=1}^{n} N(n,k).$$
The seventh row is
$$N(7,\ast) = \big[,1,\ 21,\ 105,\ 175,\ 105,\ 21,\ 1,\big].$$
Direct computation for $N(7,2)$:
$$N(7,2) = \frac{1}{7}\binom{7}{2}\binom{7}{1} = \frac{1}{7}\cdot 21 \cdot 7 = 21.$$
The Narayana numbers count triangulations of a polygon by an orientation-compatible class. $N(7,2)=21$ counts the triangulations of a heptagon by $2$-ear removal, and $N(7,6)=21$ counts those by the dual $(7-2)$-ear removal. The symmetry $N(n,k)=N(n,n+1-k)$ is reflected in the palindrome of the row.
6.3. One-Half of the Fifth Catalan Number
By the standard formula $C_n = \frac{1}{n+1}\binom{2n}{n}$ (Catalan number), the fifth Catalan number is
$$C_5 = \frac{1}{6}\binom{10}{5} = \frac{1}{6}\cdot 252 = 42 = 2\cdot 21.$$
Therefore $21 = C_5/2$. The half-Catalan sequence $C_n/2$ for small $n$ gives
$$\tfrac{1}{2},\ 1,\ 2,\ 5,\ \tfrac{21}{2},\ 33,\ \tfrac{429}{2},\ 715,\ \ldots$$
Few of these are integers (only when $C_n \equiv 0 \pmod 2$). The first such at $n=15$ gives $C_{15}/2 = 9694845$. This note does not bear on the central identity $21 = C_5/2$.
6.4. Aliquot Sequence
The aliquot sequence of $n$ is defined by $a_0 = n$ and $a_{k+1} = \sigma(a_k) - a_k$ (aliquot sequence). For $n=21$,
$$21 \to (\sigma(21) - 21) = (1+3+7+21-21) = 11 \to \sigma(11)-11 = (1+11-11) = 1 \to \sigma(1)-1 = 0 \to 0 \to \cdots$$
The sequence reaches the absorbing state $0$ in three steps from $21$. The sequence is deficient-terminating since $21$ is deficient ($\sigma(21) = 32 < 42 = 2\cdot 21$) and $11$ is prime (deficient).
6.5. Sum of Three Squares
By Lagrange's four-square theorem every positive integer is a sum of four squares, and by Legendre's three-square theorem an integer $n$ is a sum of three squares if and only if $n$ is not of the form $4^a(8b+7)$. Since $21 \equiv 5 \pmod 8$, it is a sum of three squares. The unique representation with $a \le b \le c$ is
$$21 = 1^2 + 2^2 + 4^2.$$
Verification: $1 + 4 + 16 = 21$. No representation as a sum of two positive squares exists because $21 \equiv 1 \pmod 4$, but the prime factor $3 \equiv 3 \pmod 4$ appears to an odd power (sum of two squares theorem).