Seqsplit fail in mathmode

Question

I see from the documentation that seqsplit is supposed to work in mathmode.

However, I have the following:

Since we know $1000! = \seqsplit{4.0238726007709377354370243392300398571937486421...105933983835777939410970027753472e^{2567}}$, we can represen...

And get this error:

! Missing { inserted.
<to be read again> 
                   \futurelet 
l.487 ...15933983835777939410970027753472e^{2567}}
                                                  $, we can represen...
? [ENTER]
! Missing } inserted.
<inserted text> 
                }
l.487 ...15933983835777939410970027753472e^{2567}}
                                                  , we can represent...
? [ENTER]

It's not clear to me which part of this is underspecified in { or }. Is it fighting with the e^{}, thinking these are closing seqsplit?

What's strange is the output is correct, but I still have to answer the prompts.

Answer 1

I propose a different macro:

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\longnumber}{m}
 {
  \group_begin:
  \int_step_inline:nnn { `0 } { `9 }
   {
    \char_set_active_eq:nc { ##1 } { __mittenchops_activenumber_##1: }
    \mathcode##1="8000\scan_stop:
   }
  \char_set_active_eq:nN { `. } \__mittenchops_activeperiod:
  \mathcode`.="8000\scan_stop:
  #1
  \group_end:
 }
\int_step_inline:nnn { `0 } { `9 }
 {
  \cs_new:cpn { __mittenchops_activenumber_#1: }
   {
    \mathchar#1\scan_stop:
    \mspace{0mu plus 2mu}
    \penalty0\scan_stop:
   }
 }
\cs_new:Npn \__mittenchops_activeperiod:
 {
  \unpenalty
  \mathchar`.\scan_stop:
  \mspace{0mu plus 2mu}
 }
\ExplSyntaxOff

\begin{document}

Since we know $1000! = \longnumber{4.023872600770937735437024339230
0398571937486421\dots105933983835777939410970027753472e^{2567}}$, 
we can represent...

\end{document}

Each digit is turned into a macro that typesets the digit followed by a glob of zero glue, but stretchable up to 2mu, and by a penalty that allows a line break. The period, however, removes the penalty, so there is no allowed break before and after it. There is no problem with the exponent, because penalties there do nothing.

Answer 2

It would be a pity to hardcode your 1000! in the LaTeX source.

\documentclass{article}
\usepackage{amsmath}% provides \mspace

\usepackage{xintfrac}% provides \xintFac, \xintREZ

\makeatletter
\newcommand\IntToSplitableFloat[1]{%
% We assume input will expand to some (long, *positive*) integer
   % first we handle trailing zeros
   % (unfortunately xint does not provide auxiliaries to get the A, B, C
   %  from the \xintREZ output A/B[C])
   \expandafter\ITSF@parseinput
               \romannumeral0\xintrez{#1}%
   % attention that \xintLen expands its argument but treats it as a
   % number to normalize, so it removes leading zeros, and they may
   % occur as we have trimmed the leading digit. So we use \xintLength
   % but some \expandafter are needed.
   \edef\ITSF@nbofnextdigits
       {\expandafter\xintLength\expandafter{\ITSF@nextdigits}}%
   % compute the final exponent
   \edef\ITSF@exponent{\the\numexpr\ITSF@nbofnextdigits+\ITSF@exp}%
   \ITSF@lead.\mspace{0mu plus 2mu}%
   \expandafter\ITSF@loop\ITSF@nextdigits\relax
   \unpenalty\unskip\nobreak\;10^{\ITSF@exponent}%
}

\def\ITSF@parseinput #1#2/#3[#4]{%
% make global definitions to keep possibility to reuse the
% parsed integer afterwards without recomputing it
  \gdef\ITSF@lead{#1}\gdef\ITSF@nextdigits{#2}\gdef\ITSF@exp{#4}%
  % we assume here #3 is 1 with no error checking...
}

\def\ITSF@loop #1{%
  \if#1\relax\else#1\mspace{0mu plus 2mu}\penalty0 \expandafter\ITSF@loop\fi
}
\makeatother

\begin{document}
Since we know that $1000! = \IntToSplitableFloat{\xintFac{1000}}$, we
feel that much happier.
\end{document}

I deliberately do not use e^{..} notation, see my comment