%I #46 Feb 04 2024 18:45:47

%S 1,2,4,6,12,24,36,48,60,120,180,240,300,360,420,840,1680,2520,5040,

%T 7560,10080,12600,15120,17640,20160,22680,25200,27720,55440,83160,

%U 110880,138600,166320,194040,221760,249480,277200,304920,332640,360360,720720,1441440

%N Once a number in this sequence is divisible by all numbers 1 to m, subsequent terms are constrained to have the same property; choose the smallest permissible number that is greater than the previous term.

%C All terms from A095848 belong to this sequence.

%H David A. Corneth, <a href="/A242298/b242298.txt">Table of n, a(n) for n = 1..10000</a>

%F a(1) = 1. If n > 1, a(n) = a(n-1) + A368777(a(n-1)). - _Hal M. Switkay_, Jan 26 2024

%e After 6, none of 7,8,9,10 or 11 are in the sequence since they are not divisible by 1,2 and 3 as 6 is. 12 is a term, but is now divisible by 1,2,3 and 4, adding a new constraint on subsequent terms.

%e After 24, 30 is not in the sequence because 24 is divisible by all numbers from 1 to 4 and 30 is not divisible by 4. But 36, which is divisible by all of 1 through 4, is a term.

%e As an irregular table, the n-th row consists of all numbers divisible by A051451(n) but not by A051451(n+1). - _Tom Edgar_, May 22 2014

%o (PARI) consecd(a) = {d = divisors(a); for (i=2, #d, if (d[i] - d[i-1] > 1, return(i-1));); return(a);}

%o findnext(a) = {nconsd = consecd(a); na = a + 1; while (consecd(na) < nconsd, na++); na;}

%o lista(nn) = {a = 1; print1(a, ", "); for (n=1, nn, a = findnext(a); print1(a, ", "););} \\ _Michel Marcus_, May 11 2014

%o (PARI)

%o first(n) = {

%o my(res = vector(n), step = 1, oldm = 1, newm = 1);

%o res[1] = 1;

%o for(i = 2, n,

%o while(res[i-1] % (newm+1) == 0,

%o newm++;

%o );

%o if(newm > oldm,

%o step = lcm([step, lcm([oldm..newm])]);

%o oldm = newm

%o );

%o res[i] = res[i-1]+step

%o );

%o res

%o } \\ _David A. Corneth_, Jan 28 2024

%Y Cf. A003418, A051451, A080765, A368777.

%K nonn

%O 1,2

%A _J. Lowell_, May 10 2014

%E More terms from _Michel Marcus_, May 11 2014