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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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29
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,11
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COMMENTS
| Also the highest power of 10 dividing n! (different from A054899). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 18 2007
a(n) = (n - A053824(n))/4. [From Lekraj Beedassy (blekraj(AT)yahoo.com), 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 (blekraj(AT)yahoo.com), Nov 03 2010]
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REFERENCES
| 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 J.M. Grau Ribas (grau(AT)uniovi.es), Feb 14 2010]
A. M. Oller-MarcAeeeen. A new lookat the trailing zeroes of n!. arXiv:0906.4868v1 [math.NT]. [From J.M. Grau Ribas (grau(AT)uniovi.es), Feb 14 2010]
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LINKS
| T. D. Noe, Table of n, a(n) for n=0..1000
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
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FORMULA
| 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
G.f.: g(x)=sum{k>0, x^(5^k)/(1-x^(5^k))}/(1-x). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 18 2007
a(n)=sum{5<=k<=n, sum{j|k,j>=5, floor(log_5(j))-floor(log_5(j-1))}}. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 25 2007
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. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 25 2007
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))}. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 25 2007
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). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
a(k*5^m)=k*(5^m-1)/4, for 0<=k<5, m>=0. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
Asymtotic behavior: a(n)=n/4+O(log(n)), a(n+1)-a(n)=O(log(n)), which follows from the inequalities below. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
a(n)<=(n-1)/4; equality holds for powers of 5. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
a(n)>=n/4-1-floor(log_5(n)); equality holds for n=5^m-1, m>0. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
lim inf (n/4-a(n))=1/4, for n-->oo. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
lim sup (n/4-log_5(n)-a(n))=0, for n-->oo. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
lim sup (a(n+1)-a(n)-log_5(n))=0, for n-->oo. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
a(n) <= A027869(n). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jan 27 2008
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MATHEMATICA
| 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}] (* RGWv *)
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CROSSREFS
| 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.
Cf. A067080, A098844, A132029, A054999.
Sequence in context: A154099 A105511 A187183 * A060384 A105564 A025811
Adjacent sequences: A027865 A027866 A027867 * A027869 A027870 A027871
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KEYWORD
| nonn,base,nice,easy
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AUTHOR
| Warut Roonguthai (warut822(AT)yahoo.com)
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