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%I #79 Jan 01 2020 22:00:59
%S 2,6,4,6,12,15,8,10,21,12,14,50,27,30,16,18,36,20,22,85,45,24,26,100,
%T 28,30,57,60,182,63,32,34,135,36,38,78,150,40,42,81,44,46,175,90,93,
%U 48,50,99,52,54,210,215,56,58,114,60,62,120,123,364,126,129,64
%N a(n) is the smallest k such that in the prime power factorization of k! there exists at least one exponent n.
%C For n = 1, 2, 3, etc., a(n)! contains 2^1, 3^2, 2^3, 2^4, 3^5, 3^6, 2^7, etc.
%C 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.
%C Sum_{i>=1} n*(p-1)/p^i} = n and Sum_{i=1..m} floor(n*(p-1)/p^i)) < n where m = floor(log(n*(p-1)/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*(p-1) and p|k. - _David A. Corneth_, Mar 21 2017
%C 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
%H David A. Corneth, <a href="/A240751/b240751.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Peter J. C. Moses)
%H Vladimir Shevelev, Charles R. Greathouse IV, and Peter J. C. Moses, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Moses/moses1.html">On intervals (kn, (k+1)n) containing a prime for all n>1</a>, Journal of Integer Sequences, Vol. 16 (2013), Article 13.7.3. <a href="http://arxiv.org/abs/1212.2785">arXiv:1212.2785</a>
%H Robert G. Wilson v, <a href="/A240751/a240751.pdf">Graph of the first 10000 terms</a>
%H Robert G. Wilson v, <a href="/A240751/a240751_1.pdf">Graph of the first 2500000 terms.</a>
%F a(n) <= n*t, where t is such that t*(1-1300/log^4(t))/log(t) >= n+1. Cf. Shevelev, Greathouse IV, and Moses link, Proposition 6.
%F a(n) = A284050(n)*n + A284051(n). - _Robert G. Wilson v_, Apr 15 2017
%e 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.
%e From _David A. Corneth_, Mar 21 2017: (Start)
%e To compute a(5) we first see if there is a factorial k! such that 2^5||k!. I.e., p = 2. The next multiple of p = 2 and larger than n * (p-1) = 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.
%e 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)
%t 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 *)
%t Table[k = 2; While[! MemberQ[FactorInteger[k!][[All, -1]], n], k++]; k, {n, 63}] (* _Michael De Vlieger_, Mar 24 2017 *)
%t 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 *)
%o (PARI) hasexp(k, n)=f = factor(k!); for (i=1, #f~, if (f[i, 2] == n, return (1));); return (0);
%o a(n) = {k = 2; while (!hasexp(k, n), k++); k;} \\ _Michel Marcus_, Apr 12 2014
%o (PARI) a(n)=my(r = 0, m, p = 2, cn, cm); while(1,cn = n * (p-1); 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
%o (PARI) valp(n,p)=my(s=n); while(n\=p, s+=n); s
%o 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(upper-lower>1, m=(lower+upper)\2; mV=f(m); if(mV<n, lower=m; lV=mV, upper=m; uV=mV)); if(uV==n, upper, oo)
%o 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.");
%o a(n)=my(t); forprime(p=2,, t=(n+1)*(p-1)\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
%Y Cf. A240537, A240606, A240619, A240620, A240668, A240669, A240670, A240672, A240695, A284050, A284051.
%K nonn,easy
%O 1,1
%A _Vladimir Shevelev_, Apr 12 2014
%E More terms from _Michel Marcus_, Apr 12 2014
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