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 A307251 Numbers k for which A319100(k) sets a new record. 1

%I

%S 1,3,7,21,56,63,168,252,504,819,2184,3276,6552,15561,32760,62244,

%T 124488,482391,622440,1929564,3859128,17848467,19295640,71393868,

%U 142787736,713938680,3069936324,6139872648,30699363240,187266115764,337692995640,374532231528

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

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

%C - 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;

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

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

%C 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.

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

%C - 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;

%C - 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.

%C 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).

%C 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.)

%C 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).

%C 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).

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

%o (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)))))

%o 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)))))

%o 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);

%o u = []; for(i=1, #v, if(sum(j=1, i-1, A319100(v[j]) >= A319100(v[i]))==0, u=concat(u, [v[i]])));

%o vecsort(concat(select(i->(i!=4&&i!=28&&i!=2520), u), [3, 21, 168, 2184])) \\ See A319100 for its program

%Y Cf. A319100, A002476, A007528.

%Y For the records see A307252.

%K nonn

%O 1,2

%A _Jianing Song_, Mar 31 2019

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Last modified May 28 17:37 EDT 2020. Contains 334684 sequences. (Running on oeis4.)