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.
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
Offset: 1
Keywords
Examples
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.
Links
- Table of n, a(n) for n = 1..33
- 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, De Polignac's formula
- Wikipedia, Trailing zero
Programs
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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)
Formula
Let Psi(k) = 2 + sum_{n >= 1} floor(k/a(n)). Then 10^Psi(k) divides (k+9)!.
Extensions
a(32)-a(33) from Charles R Greathouse IV, Nov 19 2012
Comments