Properties

Degree 4
Conductor $ 2^{2} \cdot 11^{2} \cdot 13^{2} $
Sign $-1$
Motivic weight 1
Primitive no
Self-dual yes
Analytic rank 1

Origins

Origins of factors

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Normalization:  

Dirichlet series

L(s)  = 1  − 6·3-s + 4-s − 2·5-s + 21·9-s − 2·11-s − 6·12-s + 12·15-s + 16-s − 2·20-s − 8·23-s − 7·25-s − 54·27-s + 8·31-s + 12·33-s + 21·36-s + 6·37-s − 2·44-s − 42·45-s + 26·47-s − 6·48-s − 13·49-s + 24·53-s + 4·55-s − 20·59-s + 12·60-s + 64-s − 4·67-s + ⋯
L(s)  = 1  − 3.46·3-s + 1/2·4-s − 0.894·5-s + 7·9-s − 0.603·11-s − 1.73·12-s + 3.09·15-s + 1/4·16-s − 0.447·20-s − 1.66·23-s − 7/5·25-s − 10.3·27-s + 1.43·31-s + 2.08·33-s + 7/2·36-s + 0.986·37-s − 0.301·44-s − 6.26·45-s + 3.79·47-s − 0.866·48-s − 1.85·49-s + 3.29·53-s + 0.539·55-s − 2.60·59-s + 1.54·60-s + 1/8·64-s − 0.488·67-s + ⋯

Functional equation

\[\begin{aligned} \Lambda(s)=\mathstrut & 81796 ^{s/2} \, \Gamma_{\C}(s)^{2} \, L(s)\cr =\mathstrut & -\, \Lambda(2-s) \end{aligned} \]
\[\begin{aligned} \Lambda(s)=\mathstrut & 81796 ^{s/2} \, \Gamma_{\C}(s+1/2)^{2} \, L(s)\cr =\mathstrut & -\, \Lambda(1-s) \end{aligned} \]

Invariants

\( d \)  =  \(4\)
\( N \)  =  \(81796\)    =    \(2^{2} \cdot 11^{2} \cdot 13^{2}\)
\( \varepsilon \)  =  $-1$
motivic weight  =  \(1\)
character  :  $\chi_{81796} (1, \cdot )$
Sato-Tate  :  $\mathrm{SU}(2)$
primitive  :  no
self-dual  :  yes
analytic rank  =  1
Selberg data  =  $(4,\ 81796,\ (\ :1/2, 1/2),\ -1)$
$L(1)$  $=$  $0$
$L(\frac12)$  $=$  $0$
$L(\frac{3}{2})$   not available
$L(1)$   not available

Euler product

\[L(s) = \prod_{p \text{ prime}} F_p(p^{-s})^{-1} \] where, for $p \notin \{2,\;11,\;13\}$, \[F_p(T) = 1 - a_p T + b_p T^2 - a_p p T^3 + p^2 T^4 \]with $b_p = a_p^2 - a_{p^2}$. If $p \in \{2,\;11,\;13\}$, then $F_p$ is a polynomial of degree at most 3.
$p$$\Gal(F_p)$$F_p$
bad2$C_1$$\times$$C_1$ \( ( 1 - T )( 1 + T ) \)
11$C_2$ \( 1 + 2 T + p T^{2} \)
13$C_1$$\times$$C_1$ \( ( 1 - T )( 1 + T ) \)
good3$C_2$ \( ( 1 + p T + p T^{2} )^{2} \)
5$C_2$ \( ( 1 + T + p T^{2} )^{2} \)
7$C_2$ \( ( 1 - T + p T^{2} )( 1 + T + p T^{2} ) \)
17$C_2$ \( ( 1 - 3 T + p T^{2} )( 1 + 3 T + p T^{2} ) \)
19$C_2$ \( ( 1 - 6 T + p T^{2} )( 1 + 6 T + p T^{2} ) \)
23$C_2$ \( ( 1 + 4 T + p T^{2} )^{2} \)
29$C_2$ \( ( 1 - 2 T + p T^{2} )( 1 + 2 T + p T^{2} ) \)
31$C_2$ \( ( 1 - 4 T + p T^{2} )^{2} \)
37$C_2$ \( ( 1 - 3 T + p T^{2} )^{2} \)
41$C_2$ \( ( 1 + p T^{2} )^{2} \)
43$C_2$ \( ( 1 - 5 T + p T^{2} )( 1 + 5 T + p T^{2} ) \)
47$C_2$ \( ( 1 - 13 T + p T^{2} )^{2} \)
53$C_2$ \( ( 1 - 12 T + p T^{2} )^{2} \)
59$C_2$ \( ( 1 + 10 T + p T^{2} )^{2} \)
61$C_2$ \( ( 1 - 8 T + p T^{2} )( 1 + 8 T + p T^{2} ) \)
67$C_2$ \( ( 1 + 2 T + p T^{2} )^{2} \)
71$C_2$ \( ( 1 + 5 T + p T^{2} )^{2} \)
73$C_2$ \( ( 1 - 10 T + p T^{2} )( 1 + 10 T + p T^{2} ) \)
79$C_2$ \( ( 1 - 4 T + p T^{2} )( 1 + 4 T + p T^{2} ) \)
83$C_2$ \( ( 1 + p T^{2} )^{2} \)
89$C_2$ \( ( 1 - 6 T + p T^{2} )^{2} \)
97$C_2$ \( ( 1 - 14 T + p T^{2} )^{2} \)
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\[\begin{aligned} L(s) = \prod_p \ \prod_{j=1}^{4} (1 - \alpha_{j,p}\, p^{-s})^{-1} \end{aligned}\]

Imaginary part of the first few zeros on the critical line

−9.982600302630202160398757175989, −9.182231187189023046155985298119, −8.174465989121661359437511477893, −7.52260855659566213355764232106, −7.45985560667588965978165906144, −6.69700065044511212992831284617, −6.06963680351175114271015390023, −5.88908450915949561478136129218, −5.58007316868712318029783390276, −4.61153453491182141362175201622, −4.43660190737283567298107688401, −3.76677853005733776527358303686, −2.25144343695204805750190070200, −0.973879971324260460585952773727, 0, 0.973879971324260460585952773727, 2.25144343695204805750190070200, 3.76677853005733776527358303686, 4.43660190737283567298107688401, 4.61153453491182141362175201622, 5.58007316868712318029783390276, 5.88908450915949561478136129218, 6.06963680351175114271015390023, 6.69700065044511212992831284617, 7.45985560667588965978165906144, 7.52260855659566213355764232106, 8.174465989121661359437511477893, 9.182231187189023046155985298119, 9.982600302630202160398757175989

Graph of the $Z$-function along the critical line