A334636 -id:A334636 - cp's OEIS frontend

A333978 Numbers of the form b_1 * b_2 * ... * b_t, where b_1 = 1 and b_(i + 1) - b_i = 0 or 1.

1, 2, 4, 6, 8, 12, 16, 18, 24, 32, 36, 48, 54, 64, 72, 96, 108, 120, 128, 144, 162, 192, 216, 240, 256, 288, 324, 360, 384, 432, 480, 486, 512, 576, 600, 648, 720, 768, 864, 960, 972, 1024, 1080, 1152, 1200, 1296, 1440, 1458, 1536, 1728, 1800, 1920, 1944, 2048

Offset: 1

Peter Kagey, Sep 20 2020

This sequence gives the distinct values in A284001, sorted.
If m and k are in this sequence, then so is their product m*k.
If a prime p divides a(n), then so does p!.
A001013 is a subsequence.
Define a set S of polynomials by: (i) 1 is in S; (ii) if P is in S, then x*P and dP/dx are in S; (iii) if the repeated application of (i) and (ii) fails to prove that P is in S then P is not in S. This sequence enumerates the elements of S of degree 0. - Luc Rousseau, Aug 20 2022
Numbers k divisible by A102068(k)  (or in other words, numbers k divisible by h(k)! where h(k) is the largest prime factor of k). - David A. Corneth, Aug 20 2022

			The first 11 terms can be written as
   1 = 1
   2 = 1 * 2
   4 = 1 * 2 * 2
   6 = 1 * 2 * 3
   8 = 1 * 2 * 2 * 2
  12 = 1 * 2 * 2 * 3
  16 = 1 * 2 * 2 * 2 * 2
  18 = 1 * 2 * 3 * 3
  24 = 1 * 2 * 3 * 4 or 1 * 2 * 2 * 2 * 3
  32 = 1 * 2 * 2 * 2 * 2 * 2
  36 = 1 * 2 * 2 * 3 * 3

Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Peter Kagey)

Cf. A001013, A003586, A006530, A102068, A284001, A334636.

is(n) = if(n==1, return(1)); my(f = factor(n), p = f[#f~, 1]); n%p! == 0 \\ David A. Corneth, Sep 05 2022

import heapq
from math import factorial
from sympy import nextprime
from itertools import islice
def agen(): # generator of terms
    oldv, h, primes, nextp, nextfact = 0, [(1, 1)], [], 0, 0
    while True:
        v, maxp = heapq.heappop(h)
        if v != oldv:
            yield v; oldv = v
            while nextfact < v:
                nextp = nextprime(nextp); nextfact = factorial(nextp)
                primes.append(nextp); heapq.heappush(h, (nextfact, nextp))
            for p in primes:
                if p <= maxp: heapq.heappush(h, (v*p, max(maxp, p)))
                else: break
print(list(islice(agen(), 60))) # Michael S. Branicky, Aug 20 2022

cp's OEIS Frontend

A333978 Numbers of the form b_1 * b_2 * ... * b_t, where b_1 = 1 and b_(i + 1) - b_i = 0 or 1.

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