Complexity of computing expansion of a newform level 18 weight 3 and character [3] - OEIS A116418 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:09:13Z http://mathoverflow.net/feeds/question/84443 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84443/complexity-of-computing-expansion-of-a-newform-level-18-weight-3-and-character-3 Complexity of computing expansion of a newform level 18 weight 3 and character [3] - OEIS A116418 joro 2011-12-28T13:07:49Z 2011-12-31T04:06:44Z <p>I am not familiar with newforms, so this may not make any sense.</p> <p>OEIS sequence <a href="https://oeis.org/A116418" rel="nofollow">A116418</a> is Expansion of a newform level 18 weight 3 and character [3]</p> <p>Numerical evidence suggest that up to $10^5$ $$\text{A116418}[n] \equiv \sigma(3n+1) \pmod 3$$</p> <blockquote> <p>What is the complexity of computing A116418[n], possibly assuming $n$ is factored (for modular form coefficients after $n$ is factored the coefficient is efficiently computable).</p> </blockquote> <p>Gjergji Zaimi <a href="http://mathoverflow.net/questions/59178/up-to-106-sigma8n1-mod-4-oeis-a001935n-mod-4-number-of-partition" rel="nofollow">proved</a> a similar congruence involving eta and A116418 is expansion of an eta formula.</p> <p><em>Added</em> My main interest is computing $\sigma(3n+1) \mod 3$ and a comment by Dror Speiser suggests the coefficient of the newform is computable in polynomial time assuming $n$ is factored.</p> <p>The factorization of $n$ is not related related to the factorization of $3n+1$ and for numbers of form $3 \cdot 2^n + 1$ the factorization is trivial.</p> <blockquote> <p>Is A116418 really the expansion of the newform or is it a typo in OEIS?</p> <p>Is the congruence $\text{A116418}[n] \equiv \sigma(3n+1) \pmod 3$ identity or just coincidence for the the first $10^5$ terms?</p> </blockquote> http://mathoverflow.net/questions/84443/complexity-of-computing-expansion-of-a-newform-level-18-weight-3-and-character-3/84489#84489 Answer by Noam D. Elkies for Complexity of computing expansion of a newform level 18 weight 3 and character [3] - OEIS A116418 Noam D. Elkies 2011-12-29T03:33:55Z 2011-12-31T04:06:44Z <p>[More a comment than an answer, but too long for the comment space]</p> <p>Call this form $$\varphi := \frac{\eta(q^3)^2 \eta(q^6)^3 \eta(q^9)^2}{\eta(q^{18})} = q - 2q^4 - 4q^7 + 6q^{10} + 8q^{13} \cdots.$$ The listing of coefficients in the OEIS is correct as far as it goes (checked with copy-and-paste to <strong>gp</strong>). The form is not CM: the coefficients are supported on $q^n$ with $n \equiv 1 \bmod 3$ but do not vanish even for $n$ such as $10$ and $22$ that are $1 \bmod 3$ but not norms from ${\bf Q}(\sqrt{-3})$. In particular the coefficients aren't multiplicative, so $\varphi$ isn't quite an eigenform. It seems that the relevant eigenforms are obtained as follows. Apply $w_{18}$ to get (within a multiplicative factor) $$\phi := \frac{\eta(q^6)^2 \eta(q^3)^3 \eta(q^2)^2}{\eta(q)} = q + q^2 - 2q^4 - 3q^5 - 4q^7 - 2q^8 + 6q^{10} + 12q^{11} + 8q^{13} - 4q^{14} \cdots,$$ whose $q^n$ coefficient is 0 if $n \equiv 0 \bmod 3$, and coincides with the $q^n$ coefficient of $\varphi$ also when $n \equiv 1 \bmod 3$, but need not vanish for $n \equiv 2 \bmod 3$. Then "experimentally" if $m,n$ are relatively prime then the $q^{mn}$ coefficient of $\phi$ equals the product of the $q^m$ and $q^n$ coefficients, <em>unless</em> both $m$ and $n$ are $2 \bmod 3$, when the $q^{mn}$ coefficient is $-2$ times that product. Hence we obtain an eigenform by choosing a square root of $-2$ and multiplying the $q^n$ coefficient of $\phi$ by that square root for each $n \equiv 2 \bmod 3$.</p> <p>As Dror Speiser notes, the Edixhoven program promises to compute the $q^n$ coefficient of such a form in time $\log^{O(1)}n$ for $n$ prime, and thus for all $n$ given the factorization of $n$; but I don't think this has been implemented yet to the point that one could actually carry out the computation this way. For specific forms there can be shortcuts that make a $\log^{O(1)}n$ computation practical (still assuming $n$ is factored), but here I've tried a few things and not yet(?) found such a shortcut.</p> <p><em>[added later]</em> Curiously the images of $\phi$ under the other two $w$ operators are in the linear span of $\varphi$ and $\phi$: if we write $$\psi = \frac{\eta(q^3)^3 \eta(q^6)^2 \eta(q^{18})^2}{\eta(q^9)} = q^2 - 3q^5 - 2q^8 + 12q^{11} - 4q^{14} \cdots$$ for (a multiple of) the $w_2$ image, then $\phi = \varphi + \psi$, while $\varphi - 2 \psi$ is the multiple $$\frac{\eta(q)^2 \eta(q^3)^2 \eta(q^6)^3}{\eta(q^2)} = q - 2q^2 - 2q^4 + 6q^5 - 4q^7 + 4q^8 + 6q^{10} - 24q^{11} + 8q^{13} + 8q^{14} \ldots$$ of the $w_9$ image.</p>