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 nonzero digits of k in base p.
a(n)=(n mod p)!*product{00}(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
a(n)=n!, for 0<=n
a(k*p^m)=k*(p-1)!^(k*m*p^(m-1))*(k-1)!^(p^m) for 0<=k
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.
A131385
Product ceiling(n/1)*ceiling(n/2)*ceiling(n/3)*...*ceiling(n/n) (the 'ceiling factorial').
Original entry on oeis.org
1, 1, 2, 6, 16, 60, 144, 672, 1536, 6480, 19200, 76032, 165888, 1048320, 2257920, 8294400, 28311552, 126904320, 268738560, 1470873600, 3096576000, 16094453760, 51385466880, 175814737920, 366917713920, 2717245440000, 6782244618240, 22754631352320, 69918208819200
Offset: 0
From _Paul D. Hanna_, Nov 26 2012: (Start)
Illustrate initial terms using formula involving the floor function []:
a(1) = 1;
a(2) = [2/1] = 2;
a(3) = [3/1]*[4/2] = 6;
a(4) = [4/1]*[5/2]*[6/3] = 16;
a(5) = [5/1]*[5/2]*[7/3]*[8/4] = 60;
a(6) = [6/1]*[7/2]*[8/3]*[9/4]*[10/5] = 144.
Illustrate another alternative generating method:
a(1) = 1;
a(2) = (2/1)^[1/1] = 2;
a(3) = (2/1)^[2/1] * (3/2)^[2/2] = 6;
a(4) = (2/1)^[3/1] * (3/2)^[3/2] * (4/3)^[3/3] = 16;
a(5) = (2/1)^[4/1] * (3/2)^[4/2] * (4/3)^[4/3] * (5/4)^[4/4] = 60.
(End)
For n=3 the a(3)=6 functions f from subsets of {1,2} into {1,2} with f(x) == 0 (mod x) are the following: f=empty set (since null function vacuously holds), f={(1,1)}, f={(1,2)}, f={(2,2)}, f={(1,1),(2,2)}, and f={(1,2),(2,2)}. - _Dennis P. Walsh_, Nov 13 2015
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a:= n-> mul(ceil(n/k), k=1..n):
seq(a(n), n=0..40); # Dennis P. Walsh, Nov 13 2015
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Table[Product[Ceiling[n/k],{k,n}],{n,25}] (* Harvey P. Dale, Sep 18 2011 *)
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a(n)=prod(k=1,n-1,floor((n+k-1)/k)) \\ Paul D. Hanna, Feb 01 2013
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a(n)=prod(k=1,n-1,((k+1)/k)^floor((n-1)/k))
for(n=1,30,print1(a(n),", ")) \\ Paul D. Hanna, Feb 01 2013
A127482
Product of the nonzero digital products of all the prime numbers prime(1) to prime(n).
Original entry on oeis.org
2, 6, 30, 210, 210, 630, 4410, 39690, 238140, 4286520, 12859560, 270050760, 1080203040, 12962436480, 362948221440, 5444223321600, 244990049472000, 1469940296832000, 61737492466944000, 432162447268608000, 9075411392640768000, 571750917736368384000
Offset: 1
Alain Van Kerckhoven (alain(AT)avk.org), Sep 12 2007
a(7) = dp_10(2)*dp_10(3)*dp_10(5)*dp_10(7)*dp_10(11)*dp_10(13)*dp_10(17) = 2*3*5*7*(1*1)*(1*3)*(1*7) = 4410.
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a:= proc(n) option remember; `if`(n<1, 1, a(n-1)*mul(
`if`(i=0, 1, i), i=convert(ithprime(n), base, 10)))
end:
seq(a(n), n=1..30); # Alois P. Heinz, Mar 11 2022
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Rest[FoldList[Times,1,Times@@Cases[IntegerDigits[#],Except[0]]&/@ Prime[ Range[ 20]]]] (* Harvey P. Dale, Mar 19 2013 *)
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f(n) = vecprod(select(x->(x>1), digits(prime(n)))); \\ A101987
a(n) = prod(k=1, n, f(k)); \\ Michel Marcus, Mar 11 2022
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from math import prod
from sympy import sieve
def pod(s): return prod(int(d) for d in s if d != '0')
def a(n): return pod("".join(str(sieve[i+1]) for i in range(n)))
print([a(n) for n in range(1, 23)]) # Michael S. Branicky, Mar 11 2022
A145119
a(n) = Product_{k=1..n-1} (ceiling(n/k) - ceiling(n/k) mod 2).
Original entry on oeis.org
1, 2, 4, 16, 32, 96, 384, 1024, 2048, 10240, 30720, 73728, 294912, 688128, 1835008, 12582912, 25165824, 56623104, 283115520, 629145600, 1887436800, 11072962560, 26575110144, 57982058496, 231928233984, 753766760448, 1758789107712
Offset: 1
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