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%I #21 Aug 09 2015 23:37:57
%S 1,2,6,24,120,720,5040,40320,362880,362880,362880,725760,2177280,
%T 8709120,43545600,261273600,1828915200,14631321600,131681894400,
%U 263363788800,526727577600,2106910310400,12641461862400,101131694899200
%N Product of the nonzero digital products of all the numbers 1 to n (a 'total digital-product factorial' in base 10).
%H Hieronymus Fischer, <a href="/A131451/b131451.txt">Table of n, a(n) for n = 1..102</a>
%F The following formulas are given for general bases p>1:
%F a(n)=product{1<=k<=n, dp_p(k)} where dp_p(k) = product of the nonzero digits of k in base p.
%F a(n)=(n mod p)!*product{0<j<=log_p(n)}(p-1)!^(floor(n/p^j)*p^(j-1)) * product{0<j<=log_p(n),(floor(n/p^j)mod p)>0}(floor(n/p^j)mod p)^(1+(n mod p^j))*((floor(n/p^j)mod p)-1)!^(p^j).
%F Recurrence: a(n+k*p^m)=a(n)*k^n*a(k*p^m) for 0<=k<p, 0<=n<p^m.
%F a(n)=n!, for 0<=n<p.
%F a(k*p^m)=k*(p-1)!^(k*m*p^(m-1))*(k-1)!^(p^m) for 0<=k<p.
%F a(n)=(p-1)!^((m*p^(m+1)-(m+1)*p^m+1)/(p-1)^2)=(p-1)!^(1+2*p+3*p^2+...+m*p^(m-1)) for n=1+p+p^2+...+p^m.
%F a(n)=(p-1)!^(k*(m*p^(m+1)-(m+1)*p^m+1)/(p-1)^2)*(k-1)!^(p*(p^m-1)/(p-1))*k^(k*(p^(m+1)-(m+1)*p+m)/(p-1)^2)*k!*k^m, for n=k*(1+p+p^2+...+p^m).
%F For p=10: a(10^n)=9!^(n*10^(n-1)).
%F Asymptotic behavior: a(10^n)=10^(0.5559763...*n*10^n). Hence it grows slower than the factorial A000142(10^n) for which we have (10^n)!=10^((n-0.43429448...)*10^n+n/2+0.3990899...+o(1/n)). Example: a(1000) has 1668 digits, whereas 1000! has 2568 digits.
%e a(12)=dp_10(1)*dp_10(2)*dp_10(3)*...*dp_10(11)*dp_10(12)=1*2*3*4*5*6* 7*8*9*1*(1*1)*(1*2).
%e a(345)=3*4*5*3^45*4^5*(3-1)!^100*(4-1)!^10*(5-1)!^1*9!^64.
%e a(1000)=9!^300. a(1111)=9!^321.
%p with transforms;
%p f:=proc(n) option remember; if n = 0 then 1 else f(n-1)*digprod0(n); fi; end;[seq(f(n),n=0..40)]; # _N. J. A. Sloane_, Oct 12 2013
%Y Cf. A131385, A010786, A131387, A007953, A007954, A037131.
%K nonn,base
%O 1,2
%A _Hieronymus Fischer_, Jul 11 2007
%E New b-file from _Hieronymus Fischer_, Sep 10 2007
%E 2 typos in the formula section removed by _Hieronymus Fischer_, Dec 05 2011
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