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A218976 a(n) is the smallest positive integer such that 10^(2 + floor(k/a(1)) + floor(k/a(2)) + ... + floor(k/a(n))) divides (k+9)! for all k > 0. 1
6, 16, 116, 241, 242, 491, 991, 2491, 3331, 14966, 15556, 62491, 78116, 83331, 249991, 264866, 546841, 1109366, 2265491, 4999861, 4999991, 5837041, 12499996, 25249861, 26011861, 36249091, 80070866, 190999991, 242090611, 365038621, 976562241, 1210466866, 1830622801 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Every factorial of the form (k+9)! for every integer k > 0 ends at least in two zeros. This sequence gives a lower bound on the number of zeros.
This sequence is infinite and increasing, with 1/a(1) + 1/a(2) + ... = 1/4.
Conjecture: All terms except a(5) are 1 mod 5. - R. J. Cano, Nov 11 2012
LINKS
Dickson, Leonard Eugene (2005) [1919], History of the theory of numbers. Vol. I: Divisibility and primality., New York: Dover Publications, ISBN 978-0-486-44232-7, [MR], page 263. De Polignac's formula
Wikipedia, Trailing zero
FORMULA
Let Psi(k) = 2 + sum_{n >= 1} floor(k/a(n)). Then 10^Psi(k) divides (k+9)!.
EXAMPLE
10^(2 + floor(5/5)) does not divide 14!, so a(1) > 5. But 10^(2 + floor(k/6)) divides (k+9)! for all k > 0, so a(1) = 6.
PROG
(PARI) searchLimit(s1)={
my(e, s2, f=(e, s)->(e+2-9*s)/(s-s1));
while(s2<=s1, s2 += 1/5^e++);
min(f(e, s2), f(e++, s2+=1/5^e))\1
};
v5(n)=my(s); while(n\=5, s+=n); s;
nxt(v=[6])={
my(S=sum(i=1, #v, 1/v[i]), candidate=max(v[#v], 1\(1/4-S))+1, k=candidate, lm=searchLimit(S+1/candidate));
while(k<=lm,
if(v5(k+9)<2+sum(i=1, #v, k\v[i])+k\candidate,
candidate++;
lm=searchLimit(S+1/candidate)
,
k++
)
);
candidate
};
steps(n)={
my(v=[6], t);
print1(6);
for(i=2, n,
t=nxt(v);
print1(", "t);
v=concat(v, t)
);
v
};
steps(20)
CROSSREFS
Sequence in context: A368875 A115331 A239027 * A173737 A152663 A229560
KEYWORD
nonn
AUTHOR
EXTENSIONS
a(32)-a(33) from Charles R Greathouse IV, Nov 19 2012
STATUS
approved

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Last modified April 18 11:52 EDT 2024. Contains 371779 sequences. (Running on oeis4.)