

A240751


a(n) is the smallest k such that in the prime power factorization of k! there exists at least one exponent n.


24



2, 6, 4, 6, 12, 15, 8, 10, 21, 12, 14, 50, 27, 30, 16, 18, 36, 20, 22, 85, 45, 24, 26, 100, 28, 30, 57, 60, 182, 63, 32, 34, 135, 36, 38, 78, 150, 40, 42, 81, 44, 46, 175, 90, 93, 48, 50, 99, 52, 54, 210, 215, 56, 58, 114, 60, 62, 120, 123, 364, 126, 129, 64
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OFFSET

1,1


COMMENTS

For n = 1, 2, 3, etc., a(n)! contains 2^1, 3^2, 2^3, 2^4, 3^5, 3^6, 2^7, etc.
Note that for number N and for sufficiently large k=k(N), in interval (k/(N+1), k/N] there exists a prime, and in case sqrt(k) < k/(N+1), p^N  k!. Therefore the sequence is infinite.
Sum_{i>=1} n*(p1)/p^i} = n and Sum_{i=1..m} floor(n*(p1)/p^i)) < n where m = floor(log(n*(p1)/log(p)). Therefore, we can test exponents of primes in k! to see if the exponent of p is n, where k is the least k > n*(p1) and pk.  David A. Corneth, Mar 21 2017
Record k's are 2, 6, 12, 15, 21, 50, 85, 100, 182, 210, 215, 364, 553, 560, 854, 931, 1120, etc., at indices 1, 2, 5, 6, 9, 12, 20, 24, 29, 51, 52, 60, 91, 92, 141, 154, 185, 186, 342, 403, 441, 447, 635, 765, 1035, 1092, 1378, 1435, 1540, 2015, 2553, 2740, 2808, 2865, 3265, 4922, 5322, 7209, etc.  Robert G. Wilson v, Apr 13 2017


LINKS

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from Peter J. C. Moses)
Vladimir Shevelev, Charles R. Greathouse IV, and Peter J. C. Moses, On intervals (kn, (k+1)n) containing a prime for all n>1, Journal of Integer Sequences, Vol. 16 (2013), Article 13.7.3. arXiv:1212.2785
Robert G. Wilson v, Graph of the first 10000 terms
Robert G. Wilson v, Graph of the first 2500000 terms.


FORMULA

a(n) <= n*t, where t is such that t*(11300/log^4(t))/log(t) >= n+1. Cf. Shevelev, Greathouse IV, and Moses link, Proposition 6.
a(n) = A284050(n)*n + A284051(n).  Robert G. Wilson v, Apr 15 2017


EXAMPLE

a(2)=6, since 6!=2^4*3^2*5, and there is no k<6 such that the factorization of k! contains a power p^2, where p is prime.
From David A. Corneth, Mar 21 2017: (Start)
To compute a(5) we first see if there is a factorial k! such that 2^5k!. I.e., p = 2. The next multiple of p = 2 and larger than n * (p1) = 5 is 6. The exponent of 2 in 6! Is 3 + 1 = 4 < 5. Therefore, we try the next multiple of p = 2 and larger than 6 which is 8. 8 has three factors 2. Therefore, 8! has 4 + 3 = 7 > 5 factors 2 and no factorial exists that properly divides 2^5.
So we try the next prime larger than 2, which is p = 3. We start with the next multiple of p and larger than n * (p  1) = 10, which is 12. The exponent of 3 in 12! is floor(12/3) + floor(4/3) = 5. Therefore, 12! is properly divisible by 3^5 and 12 is the least k such that k! has 5 as an exponent in the prime factorization. (End)


MATHEMATICA

fi[n_]:=fi[n]=FactorInteger[n!]; A240751={2}; Do[AppendTo[A240751, NestWhile[#+1 &, n+1, !MemberQ[Last[Transpose[fi[#]]], n]&]], {n, 2, 100}]; A240751 (* Peter J. C. Moses, Apr 12 2014 *)
Table[k = 2; While[! MemberQ[FactorInteger[k!][[All, 1]], n], k++]; k, {n, 63}] (* Michael De Vlieger, Mar 24 2017 *)
f[n_] := Block[{k = 0, p = 2, s}, While[True, While[s = Plus @@ Rest@ NestWhileList[ Floor[#/p] &, (p 1)n +k, # > 0 &]; s < n, k++]; If[s == n, Goto[fini]]; k = 0; p = NextPrime@ p]; Label[fini]; (p 1)n +k]; Array[f, 70] (* Robert G. Wilson v, Apr 15 2017, revised Apr 16 2017 and Apr 19 2017 *)


PROG

(PARI) hasexp(k, n)=f = factor(k!); for (i=1, #f~, if (f[i, 2] == n, return (1)); ); return (0);
a(n) = {k = 2; while (!hasexp(k, n), k++); k; } \\ Michel Marcus, Apr 12 2014
(PARI) a(n)=my(r = 0, m, p = 2, cn, cm); while(1, cn = n * (p1); m = p*(cn\p+1); r = 0; cm = m; while(cm, r+=cm\=p); while(r < n, m += p; r += valuation(m, p)); if(r==n, return(m)); p = nextprime(p + 1)) \\ David A. Corneth, Mar 20 2017
(PARI) valp(n, p)=my(s=n); while(n\=p, s+=n); s
findLower(f, n, lower, upper)=my(lV=f(lower), uV, m, mV); if(lV>=n, return(if(lV==n, lower, oo))); uV=f(upper); if(uV<n, return(oo)); while(upperlower>1, m=(lower+upper)\2; mV=f(m); if(mV<n, lower=m; lV=mV, upper=m; uV=mV)); if(uV==n, upper, oo)
addhelp(findLower, "findLower(f, n, lower, upper): Given a nondecreasing function f on [lower, upper], find the least integer m, lower <= m <= upper, such that f(m) = n, or an infinite value if no such m exists.");
a(n)=my(t); forprime(p=2, , t=(n+1)*(p1)\p; t=findLower(k>valp(k, p), n, t, logint(t, p)+t); if(t!=oo, return(t*p))) \\ Charles R Greathouse IV, Jul 27 2017


CROSSREFS

Cf. A240537, A240606, A240619, A240620, A240668, A240669, A240670, A240672, A240695, A284050, A284051.
Sequence in context: A119250 A059773 A127399 * A212283 A247566 A151689
Adjacent sequences: A240748 A240749 A240750 * A240752 A240753 A240754


KEYWORD

nonn,easy


AUTHOR

Vladimir Shevelev, Apr 12 2014


EXTENSIONS

More terms from Michel Marcus, Apr 12 2014


STATUS

approved



