A240606 -id:A240606 - cp's OEIS frontend

A240619 Positions of records in A240606.

1, 3, 5, 10, 24, 27, 169, 408, 18757, 134418, 295680, 427684, 737564, 245340483, 1325085081, 8036409193, 152157373453, 191309317101, 2462443815565, 3205600889069, 9743167666007, 20076829452830

Offset: 1

Vladimir Shevelev, Apr 09 2014

The corresponding records are 0,2,3,4,5,6,7,12,16,...

Cf. A240537, A240606.

A240606[n_] := Count[First[Split[Mod[Last[Transpose[FactorInteger[(2*n)!]]], 2]]], 0]; (* From Peter J. C. Moses *)
k = -1; A240619 = {};
For[i = 1, i <= 2000, i++,
If[A240606[i] > k, k = A240606[i]; AppendTo[A240619, i]]]; A240619 (* Robert Price, Mar 26 2020 *)

a(10)-a(16) from Hiroaki Yamanouchi, Sep 30 2014

a(17)-a(22) from Giovanni Resta, Apr 07 2020

A240751 a(n) is the smallest k such that in the prime power factorization of k! there exists at least one exponent n.

Original entry on oeis.org

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

Vladimir Shevelev, Apr 12 2014

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*(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
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

			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)

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.

Cf. A240537, A240606, A240619, A240620, A240668, A240669, A240670, A240672, A240695, A284050, A284051.

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 *)

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

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

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(uV1, m=(lower+upper)\2; mV=f(m); if(mVvalp(k,p), n, t, logint(t,p)+t); if(t!=oo, return(t*p))) \\ Charles R Greathouse IV, Jul 27 2017

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.

a(n) = A284050(n)*n + A284051(n). - Robert G. Wilson v, Apr 15 2017

More terms from Michel Marcus, Apr 12 2014

A240620 a(n) is the smallest k such that in the prime power factorization of k! at least the first n positive exponents are even.

Original entry on oeis.org

6, 6, 10, 20, 48, 54, 338, 816, 816, 816, 816, 816, 37514, 37514, 37514, 37514, 268836, 268836, 591360, 855368, 1475128, 1475128, 1475128, 1475128, 1475128, 1475128, 127632241, 472077979, 472077979, 472077979, 472077979, 472077979, 472077979, 16072818386

Offset: 1

Vladimir Shevelev, Apr 09 2014

nonn

The sequence is nondecreasing and, by Berend's theorem, a(n) --> infinity as n goes to infinity.

The distinct terms 6, 10, 20, 48, 54, 338, 816, 37514, 268836, ... repeat 2, 1, 1, 1, 1, 1, 5, 4, 2, ... times.

			Prime power factorizations of k! for k = 2, 3, 4, 5, 6 are 2, 2*3, 2^3*3, 2^3*3*5, 2^4*3^2*5. Thus the least k having at least 1 first even exponent is 6, and 6 is also the least k having at least 2 first even exponents. So a(1) = a(2) = 6.

P. Erdős, P. L. Graham, Old and new problems and results in combinatorial number theory, L'Enseignement Mathématique, Imprimerie Kunding, Geneva, 1980.

Giovanni Resta, Table of n, a(n) for n = 1..46 (first 36 terms from Hiroaki Yamanouchi)
D. Berend, Parity of exponents in the factorization of n!, J. Number Theory, 64 (1997), 13-19.
Y.-G. Chen, On the parity of exponents in the standard factorization of n!, J. Number Theory, 100 (2003), 326-331.

Cf. A240537, A240606, A240619.

isokp(n,k) = {my(fk = k!, f = factor(fk)); if (#f~ < n, return (0)); if (f[n,1] != prime(n), return (0)); for (j=1, n, if (f[j,2] % 2, return(0));); return(1);}
a(n) = {my(k=1); while (! isokp(n,k), k++); k;} \\ Michel Marcus, Feb 04 2016

a(17)-a(18) corrected and a(19)-a(34) added by Hiroaki Yamanouchi, Sep 05 2014

A240668 Number of the first odd exponents in the prime power factorization of (2*n)!.

Original entry on oeis.org

1, 2, 0, 1, 0, 0, 2, 1, 0, 0, 2, 0, 1, 2, 0, 1, 0, 0, 2, 0, 3, 3, 0, 0, 1, 2, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 2, 0, 0, 1, 5, 0, 1, 0, 0, 3, 0, 1, 1, 0, 2, 0, 0, 2, 1, 0, 0, 3, 0, 1, 2, 0, 3, 0, 0, 2, 0, 5, 2, 0, 0, 1, 3, 0, 1, 0, 0, 2, 0, 1, 1, 0, 1, 0, 0, 4

Offset: 1

Vladimir Shevelev, Apr 10 2014

nonn

According to Chen's theorem, the sequence is unbounded.

			32! = 2^31*3^14*5^7*7^4*11^2*13^2*17*19*23*29*31, and only the first 1 exponent is odd, so a(16) = 1.

Peter J. C. Moses, Table of n, a(n) for n = 1..10000
D. Berend, Parity of exponents in the factorization of n!, J. Number Theory, 64 (1997), 13-19.
Y.-G. Chen, On the parity of exponents in the standard factorization of n!, J. Number Theory, 100 (2003), 326-331.

Cf. A240537, A240606, A240619, A240620.

Map[Count[First[Split[Mod[Last[Transpose[FactorInteger[(2*#)!]]],2]]],1]&,Range[75]] (* Peter J. C. Moses, Apr 10 2014 *)

a(n) = {my(f = factor((2*n)!)); my(nb = 0); my(i = 1); while((i <= #f~) && (f[i, 2] % 2), nb++; i++;); nb;} \\ Michel Marcus, Apr 10 2014

a(n)*A240606(n) = 0.

More terms from Michel Marcus, Apr 10 2014

A240669 Number of the first odious exponents (A000069) in the prime power factorization of (2n)!.

Original entry on oeis.org

1, 0, 3, 4, 4, 0, 1, 0, 2, 0, 1, 1, 0, 2, 10, 11, 1, 0, 1, 1, 0, 2, 2, 0, 2, 1, 2, 0, 0, 3, 0, 0, 2, 0, 4, 1, 0, 2, 1, 0, 1, 5, 2, 0, 0, 6, 0, 1, 0, 1, 2, 0, 0, 1, 0, 1, 3, 2, 0, 0, 1, 0, 0, 3, 3, 0, 1, 1, 0, 2, 1, 0, 8, 1, 1, 0, 0, 1, 0, 2, 0, 1, 2, 0, 0, 3

Offset: 1

Vladimir Shevelev, Apr 10 2014

nonn

Conjecture: The sequence is unbounded. (This conjecture does not follow from Chen's theorem.)

			28! = 2^25*3^13*5^6*7^4*11^2*13^2*17*19*23, and only the first 2 exponents are odious, so a(14) = 2.

Peter J. C. Moses, Table of n, a(n) for n = 1..2000
Y.-G. Chen, On the parity of exponents in the standard factorization of n!, J. Number Theory, 100 (2003), 326-331.

Cf. A000069, A240537, A240606, A240619, A240620, A240668.

Map[Count[First[Split[Map[OddQ[DigitCount[#,2][[1]]]&,Last[Transpose[FactorInteger[(2*#)!]]&[#]]]]],True]&,Range[75]] (* Peter J. C. Moses, Apr 10 2014 *)

A240670 Numbers n for which all exponents in the prime power factorization of (2*n)! are odious (A000069).

Original entry on oeis.org

1, 3, 4, 5, 15, 16

Offset: 1

Vladimir Shevelev, Apr 10 2014

The next term, if it exists, must be more than 45000. - Peter J. C. Moses, Apr 11 2014
The sequence is finite.
Proof. For sufficiently large n, we always have a prime in (n/4, n/3]. Such primes p divide n! and, at the same time, for them we have 3<=n/p<4. Thus floor(n/p)=3, and in case sqrt(n)=93 in order for the above arguments to be true. So 16 is the last term of the sequence. - Vladimir Shevelev, Apr 11 2014

			32! = 2^31*3^14*5^7*7^4*11^2*13^2*17*19*23*29*31, and all exponents: 31,14,7,4,2,2,1,1,1,1,1 are odious, so 16 is in the sequence.

D. Berend, G. Kolesnik, Regularity of patterns in the factorization of n!, J. Number Theory, 124 (2007), no. 1, 181-192.
Vladimir Shevelev, Charles R. Greathouse IV, 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

Cf. A000069, A240537, A240606, A240619, A240620, A240668, A240669.

odiousQ[n_] := OddQ[DigitCount[n, 2][[1]]];
For[n = 1, True, n++, If[AllTrue[FactorInteger[(2 n)!][[All, 2]], odiousQ], Print[n]]] (* Jean-François Alcover, Sep 20 2018 *)

isok(n) = {f = factor((2*n)!); sum(i=1, #f~, hammingweight(f[i, 2]) % 2) == #f;} \\ Michel Marcus, Apr 11 2014

A240672 Number of the first evil exponents (A001969) in the prime power factorization of (2n)!.

Original entry on oeis.org

0, 1, 0, 0, 0, 2, 0, 3, 0, 1, 0, 0, 4, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 2, 0, 1, 2, 0, 1, 0, 0, 2, 0, 0, 2, 0, 0, 0, 1, 1, 0, 2, 0, 2, 0, 0, 1, 1, 0, 2, 0, 0, 0, 9, 2, 0, 1, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 0, 2, 1, 0, 2, 0, 3, 0, 0, 1, 1, 0, 2

Offset: 1

Vladimir Shevelev, Apr 10 2014

nonn

Conjecture: The sequence is unbounded. (This conjecture does not follow from Chen's theorem.)

			26! = 2^23*3^10*5^6*7^3*11^2*13^2*17*19*23, and the first 4 exponents (23,10,6,3) are evil, so a(13) = 4.

Peter J. C. Moses, Table of n, a(n) for n = 1..2000
Y.-G. Chen, On the parity of exponents in the standard factorization of n!, J. Number Theory, 100 (2003), 326-331.

Cf. A001969, A240537, A240606, A240619, A240620, A240668, A240669, A240670.

Map[Count[First[Split[Map[EvenQ[DigitCount[#,2][[1]]]&,Last[Transpose[FactorInteger[(2*#)!]]&[#]]]]],True]&,Range[75]] (* Peter J. C. Moses, Apr 10 2014 *)

a(n)*A240669(n) = 0.

A240695 a(n) is the smallest k such that a unique product of distinct terms of A050376 which is equal to k! contains at least the first n terms of A050376.

Original entry on oeis.org

2, 3, 4, 5, 125, 125, 138, 220, 220, 1766, 5526, 10351, 12365, 65653, 65653, 202738, 490333, 808762, 1478432, 1971352, 1971352, 1971352, 14798206, 14798206, 14798206, 14798206, 161974053, 547880880, 1619543840, 1619543840, 1619543840, 2103844465, 6435961044

Offset: 1

Vladimir Shevelev, Apr 10 2014

nonn

By the definition, the representation of a(n)! as a product of distinct terms of A050376 should contain the first n terms of A050376 and there is no restriction on the distribution of other factors of this product.

a(38) > 2 * 10^11. - Hiroaki Yamanouchi, Oct 01 2014

			5! = 2*3*4*5. We have the first 4 terms of A050376, so a(4) = 5.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..37

Cf. A240537, A240606, A240619, A240620, A240668, A240669, A240670, A240672.

bad[n_, pp_, mo_] := Catch[Do[If[ Mod[(n - Total@ IntegerDigits[n, pp[[i]]]) /(pp[[i]] - 1), mo[[i]] + 1] != mo[[i]], Throw@ True], {i, Length@ pp}]; False]; a[n_]:= Block[{fa, mo, pp, k},fa = FactorInteger[ Times @@ Select[Range[2, Prime[n]], (f = FactorInteger@# ; Length[f] == 1 && IntegerQ[Log[2, f[[1, 2]]]]) &, n]]; pp = First /@ fa; mo = Last /@ fa; k = fa[[-1, 1]]; While[ bad[k, pp, mo], k++]; k]; Array[a,15] (* Giovanni Resta, Apr 11 2014 *)

a(5)-a(23) from Giovanni Resta, Apr 11 2014

a(24)-a(33) from Hiroaki Yamanouchi, Oct 01 2014

A240755 Smallest prime which occurs in prime power factorization of A240751(n)! with exponent n.

Original entry on oeis.org

2, 3, 2, 2, 3, 3, 2, 2, 3, 2, 2, 5, 3, 3, 2, 2, 3, 2, 2, 5, 3, 2, 2, 5, 2, 2, 3, 3, 7, 3, 2, 2, 5, 2, 2, 3, 5, 2, 2, 3, 2, 2, 5, 3, 3, 2, 2, 3, 2, 2, 5, 5, 2, 2, 3, 2, 2, 3, 3, 7, 3, 3, 2, 2, 5, 2, 2, 3, 5, 2, 2, 3, 2, 2, 3, 3, 5, 2, 2, 3, 2, 2, 5, 3, 2, 2, 5

Offset: 1

Vladimir Shevelev, Apr 12 2014

nonn

			A240751(4)=6, since 6!=2^4*3^2*5. Here, only the prime 2 is in power 4, thus A240755(4)=2.

Peter J. C. Moses, Table of n, a(n) for n = 1..1000

Cf. A240537, A240606, A240619, A240620, A240668, A240669, A240670, A240672, A240695, A240751.

a(n)^n || A240751(n)!.

More terms from Peter J. C. Moses, Apr 12 2014

A240764 Least k such that prime power factorization of A240751(k)! contains p^k when the smallest such p equals prime(n), or a(n)=0 if there is no such k.

Original entry on oeis.org

1, 2, 12, 29, 186, 2865, 3265, 379852, 7172525

Offset: 1

Vladimir Shevelev, Apr 12 2014

The first position k in A240755 in which A240755(k) = prime(n), or a(n)=0 if prime(n) does not occur in A240755.

Conjecture: all a(n)>0.

			A240751(a(3))! = A240751(12)! = 50!. 50! is the least factorial having exponent 12 in its prime factorization. That exponent denotes the multiplicity of prime(3) = 5. - _David A. Corneth_, Mar 27 2017

David A. Corneth, A240751; a(n) is the smallest k such that in the prime power factorization of k! there exists at least one exponent n., Seqfan, (Mar 25 2017)

Cf. A240537, A240606, A240619, A240620, A240668, A240669, A240670, A240672, A240695, A240751, A240755.

a(5)-a(7) from Peter J. C. Moses, Apr 14 2014

a(8)-a(9) from David A. Corneth, Mar 27 2017

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A240619 Positions of records in A240606.

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A240751 a(n) is the smallest k such that in the prime power factorization of k! there exists at least one exponent n.

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A240620 a(n) is the smallest k such that in the prime power factorization of k! at least the first n positive exponents are even.

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A240668 Number of the first odd exponents in the prime power factorization of (2*n)!.

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A240669 Number of the first odious exponents (A000069) in the prime power factorization of (2n)!.

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A240670 Numbers n for which all exponents in the prime power factorization of (2*n)! are odious (A000069).

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A240672 Number of the first evil exponents (A001969) in the prime power factorization of (2n)!.

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