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A027868
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Number of trailing zeros in n!; highest power of 5 dividing n!.
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53
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0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 19
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OFFSET
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0,11
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COMMENTS
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Also the highest power of 10 dividing n! (different from A054899). - Hieronymus Fischer, Jun 18 2007
a(n) = (n - A053824(n))/4. [From Lekraj Beedassy, Nov 01 2010]
Alternatively, a(n) equals the expansion of the base-5 representation A007091(n) of n (i.e., where successive positions from right to left stand for 5^n or A000351(n)) under a scale of notation whose successive positions from right to left stand for (5^n - 1)/4 or A003463(n) ; For instance, n = 7392 has base-5 expression 2*5^5 + 1*5^4 + 4*5^3 + 0*5^2 + 3*5^1 + 2*5^0, so that a(7392) = 2*781 + 1*156 + 4*31 + 0*6 + 3*1 + 2*0 = 1845. [From Lekraj Beedassy, Nov 03 2010]
Partial sums of A112765. - Hieronymus Fischer, Jun 06 2012
10^a(n) = A000142(n) / A004154(n). - Reinhard Zumkeller, Nov 24 2012
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REFERENCES
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David S. Hart, James E. Marengo, Darren A. Narayan and David S. Ross. On the number of trailing zeros in n! College Math. J., 39(2):139-145, 2008. [From José María Grau Ribas, Feb 14 2010]
A. M. Oller-MarcAeeeen. A new lookat the trailing zeroes of n!. arXiv:0906.4868v1 [math.NT]. [From José María Grau Ribas, Feb 14 2010]
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LINKS
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T. D. Noe and Hieronymus Fischer, Table of n, a(n) for n = 0..10000 (first 1000 terms from T. D. Noe)
Eric Weisstein's World of Mathematics, Factorial
Index entries for sequences related to factorial numbers
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FORMULA
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Floor[n/5] + floor[n/25] + floor[n/125] + floor[n/625] + ....
Sum [ n/5^i ] from i=1 to infinity.
a(n)=(n-A053824(n))/4
Contribution from Hieronymus Fischer, Jun 25 2007 and Aug 13 2007 (Start):
G.f.: g(x) = sum{k>0, x^(5^k)/(1-x^(5^k))}/(1-x).
a(n) = sum{5<=k<=n, sum{j|k, j>=5, floor(log_5(j)) - floor(log_5(j-1))}}.
G.f.: g(x) = L[b(k)](x)/(1-x)
where L[b(k)](x) = sum{k>=0, b(k)*x^k/(1-x^k)} is a Lambert series with b(k) = 1, if k>1 is a power of 5, else b(k) = 0.
G.f.: g(x) = sum{k>0, c(k)*x^k}/(1-x),
where c(k) = sum{j>1, j|k, floor(log_5(j)) - floor(log_5(j - 1))}.
Recurrence:
a(n) = floor(n/5) + a(floor(n/5));
a(5*n) = n + a(n);
a(n*5^m) = n*(5^m-1)/4 + a(n).
a(k*5^m) = k*(5^m-1)/4, for 0<=k<5, m>=0.
Asymptotic behavior:
a(n) = n/4 + O(log(n)),
a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
a(n) <= (n-1)/4; equality holds for powers of 5.
a(n) >= n/4 - 1 - floor(log_5(n)); equality holds for n=5^m-1, m>0.
lim inf (n/4 - a(n)) = 1/4, for n-->oo.
lim sup (n/4 - log_5(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_5(n)) = 0, for n-->oo. (End)
a(n) <= A027869(n). - Reinhard Zumkeller, Jan 27 2008
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EXAMPLE
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a(100) = 24.
a(10^3) = 249.
a(10^4) = 2499.
a(10^5) = 24999.
a(10^6) = 249998.
a(10^7) = 2499999.
a(10^8) = 24999999.
a(10^9) = 249999998.
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MATHEMATICA
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Table[t = 0; p = 5; While[s = Floor[n/p]; t = t + s; s > 0, p *= 5]; t, {n, 0, 100} ]
Table[ IntegerExponent[n!], {n, 0, 80}] (* Robert G. Wilson v *)
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PROG
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(Haskell)
a027868 n = sum $ takeWhile (> 0) $ map (n `div`) $ tail a000351_list
-- Reinhard Zumkeller, Oct 31 2012
(PARI) a(n)=my(s); while(n, s+=n\=5); s \\ Charles R Greathouse IV, Nov 08 2012
(PARI) a(n)=valuation(n!, 5) \\ Charles R Greathouse IV, Nov 08 2012
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CROSSREFS
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See A000966 for the missing numbers. Cf. A011371 and A054861 for analogues involving powers of 2 and 3.
Cf. A054899, A007953, A112765, A067080, A098844, A132027, A067080, A098844, A132029, A054999, A112765, A191610, A000351.
Sequence in context: A154099 A105511 A187183 * A060384 A105564 A025811
Adjacent sequences: A027865 A027866 A027867 * A027869 A027870 A027871
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KEYWORD
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nonn,base,nice,easy
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AUTHOR
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Warut Roonguthai
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EXTENSIONS
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Examples added by Hieronymus Fischer, Jun 06 2012
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STATUS
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approved
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