A240751 a(n) is the smallest k such that in the prime power factorization of k! there exists at least one exponent n.
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
Offset: 1
Examples
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 2^5||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. 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)
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. Also arXiv preprint, arXiv:1212.2785 [math.NT], 2012.
- Robert G. Wilson v, Graph of the first 10000 terms
- Robert G. Wilson v, Graph of the first 2500000 terms.
Crossrefs
Programs
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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 *)
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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 * (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
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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
Formula
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.
Extensions
More terms from Michel Marcus, Apr 12 2014
Comments