A242298 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.
1, 2, 4, 6, 12, 24, 36, 48, 60, 120, 180, 240, 300, 360, 420, 840, 1680, 2520, 5040, 7560, 10080, 12600, 15120, 17640, 20160, 22680, 25200, 27720, 55440, 83160, 110880, 138600, 166320, 194040, 221760, 249480, 277200, 304920, 332640, 360360, 720720, 1441440
Offset: 1
Keywords
Examples
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. 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. 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
Links
- David A. Corneth, Table of n, a(n) for n = 1..10000
Programs
-
PARI
consecd(a) = {d = divisors(a); for (i=2, #d, if (d[i] - d[i-1] > 1, return(i-1));); return(a);} findnext(a) = {nconsd = consecd(a); na = a + 1; while (consecd(na) < nconsd, na++); na;} lista(nn) = {a = 1; print1(a, ", "); for (n=1, nn, a = findnext(a); print1(a, ", "););} \\ Michel Marcus, May 11 2014 -
PARI
first(n) = { my(res = vector(n), step = 1, oldm = 1, newm = 1); res[1] = 1; for(i = 2, n, while(res[i-1] % (newm+1) == 0, newm++; ); if(newm > oldm, step = lcm([step, lcm([oldm..newm])]); oldm = newm ); res[i] = res[i-1]+step ); res } \\ David A. Corneth, Jan 28 2024
Formula
a(1) = 1. If n > 1, a(n) = a(n-1) + A368777(a(n-1)). - Hal M. Switkay, Jan 26 2024
Extensions
More terms from Michel Marcus, May 11 2014
Comments