A307251 Numbers k for which A319100(k) sets a new record.

1, 3, 7, 21, 56, 63, 168, 252, 504, 819, 2184, 3276, 6552, 15561, 32760, 62244, 124488, 482391, 622440, 1929564, 3859128, 17848467, 19295640, 71393868, 142787736, 713938680, 3069936324, 6139872648, 30699363240, 187266115764, 337692995640, 374532231528

Offset: 1

Author

Jianing Song, Mar 31 2019

Keywords

• nonn

Comments

Let b = A319100 and v(k, p) be the p-adic valuation of k. Note that:

- if k is an odd number, then b(2*k) = b(k), b(4*k) = 2*b(k), b(2^e*k) = 4*b(k) for e >= 3;

- if k is not divisible by 3, then b(3*k) = 2*b(k), b(3^e*k) = 6*b(k) for e >= 2;

- for all primes p > 3, if k is not divisible by p, then b(p^e*k) = b(p*k).

As a result, every term k of this sequence satisfies: v(k, 2) = 0, 2 or 3, v(k, 3) <= 2 and v(k, p) <= 1 for all primes p > 3.

All terms k such that v(k, 3) = 1 are k = 3, 21, 168 and 2184. Proof:

- if k has a prime factor q == 5 (mod 6) or v(k, 2) = 2 (let q = 4), then b(9/(3*q)*k) = 6*b(k/(3*q)) = (3/2)*b(k), but 9/(3*q) < 1;

- if k has a prime factor p == 1 (mod 6) and p >= 19, then b((9*5)/(3*p)*k) = b(k), but (9*5)/(3*p) < 1.

So all terms k such that v(k, 3) = 1 are of the form k = 3*8^e*7^f*13^g, where 0 <= e <= f <= 1, 0 <= g <= f <= 1, but note that 3*7*13 = 273 is not a term because b(252) = b(273).

It is easy to see that all the other terms are of the form F, where F(i,j) = Product_{s=1..i} (p_s)*Product_{t=1..j} (q_t), where p_1 = 7, p_2 = 9, p_s = A002476(s-1) for s >= 3; q_1 = 4, q_2 = 2, q_t = A007528(t-2) for t >= 3. This is because F(i,j) is the smallest number k such that v(k, 3) != 1 and that b(k) = 6^i*2^j. Other than 4, 28 and 2520, the number F(i,j) is a term if and only if for all i', j' such that F(i',j') < F(i,j) we have 6^i'*2^j' < 6^i*2^j. (Note that 4, 28 and 2520 are of the form F and they satisfy this but they are not terms.)

Equivalently, a number k other than 3, 21, 168, 2184 and 4, 28, 2520 is a term if and only if k is of the form F and: (a) for any u, v such that u <= i and 6^u < 2^v, Product_{s=i-u+1..i} (p_s) < Product_{t=j+1..j+v} (q_t); (b) for any u, v such that v <= j and 6^u > 2^v, Product_{s=i+1..i+u} (p_s) > Product_{t=j-v+1..j} (q_t).

Specially, if a number k other than 3, 21, 168, 2184 and 4, 28, 2520 is a term, then k is of the form F and: (a') p_i < (q_(j+1))*(q_(j+2))*(q_(j+3)); (b') p_(i+1) > (q_(j-1))*(q_j). From this we can see that for every prime r, there are only finitely many terms that are not divisible by r (the largest term not divisible by 5 is (7*9*13*19*...*919)*(4*2) = 1.1832*10^190, nevertheless). But note that these are not sufficient. For example, k = (7*9*13*19*...*55819*55831*55837*55849)*(4*2*5*11*17*23) is not a term because although 55849 < 29*41*47, 55831*55837*55849 > 29*41*...*89 so k' = (7*9*13*19*...*55819)*(4*2*5*11*...*89) have k' < k but b(k') = (256/216)*b(k). Similarly, k = (7*9*13*19*...*643*661)*(4*2*5*11*17*23*29) is not a term because although 23*29 < 673, 5*11*17*23*29 > 673*691 so k' = (7*9*13*19*...*643*661*673*691)*(4*2) have k' < k but b(k') = (36/32)*b(k).

Examples

			A319100(504) = 144 which is larger than A319100(i) for i < 504, so 144 is a term.
		

Crossrefs

Cf. A319100, A002476, A007528.

For the records see A307252.

Programs

PARI

P(n) = if(!n, 1, if(n==1, 7, my(i=0,N=9); forprime(p=7, oo, if(p%3==1, i++; N*=p); if(i==n-1, return(N)))))
Q(n) = if(!n, 1, if(n==1, 4, my(i=0,N=4); forprime(p=2, oo, if(p%3==2, i++; N*=p); if(i==n-1, return(N)))))
v = []; for(i=0, 15, for(j=0, 15, if(P(i)*Q(j) < min(P(16), Q(16)), v=concat(v, [P(i)*Q(j)])))); v=vecsort(v);
u = []; for(i=1, #v, if(sum(j=1, i-1, A319100(v[j]) >= A319100(v[i]))==0, u=concat(u, [v[i]])));
vecsort(concat(select(i->(i!=4&&i!=28&&i!=2520), u), [3, 21, 168, 2184])) \\ See A319100 for its program