1,2

Hieronymus Fischer, Table of n, a(n) for n = 1..102

The following formulas are given for general bases p>1:

a(n)=product{1<=k<=n, dp_p(k)} where dp_p(k) = product of the non-zero digits of k in base p.

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).

Recurrence: a(n+k*p^m)=a(n)*k^n*a(k*p^m) for 0<=k<p, 0<=n<p^m.

a(n)=n!, for 0<=n<p.

a(k*p^m)=k*(p-1)!^(k*m*p^(m-1))*(k-1)!^(p^m) for 0<=k<p.

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.

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).

For p=10: a(10^n)=9!^(n*10^(n-1)).

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.

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).

a(345)=3*4*5*3^45*4^5*(3-1)!^100*(4-1)!^10*(5-1)!^1*9!^64.

a(1000)=9!^300. a(1111)=9!^321.

Cf. A131385, A010786, A131387, A007953, A007954, A037131.

Sequence in context: A061602 A033647 A109834 this_sequence A084012 A062919 A154657

Adjacent sequences: A131448 A131449 A131450 this_sequence A131452 A131453 A131454

nonn

Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jul 11 2007

New b-file from Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Sep 10 2007