A333978 - OEIS

%I #90 Sep 24 2022 15:45:17

%S 1,2,4,6,8,12,16,18,24,32,36,48,54,64,72,96,108,120,128,144,162,192,

%T 216,240,256,288,324,360,384,432,480,486,512,576,600,648,720,768,864,

%U 960,972,1024,1080,1152,1200,1296,1440,1458,1536,1728,1800,1920,1944,2048

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

%C This sequence gives the distinct values in A284001, sorted.

%C If m and k are in this sequence, then so is their product m*k.

%C If a prime p divides a(n), then so does p!.

%C A001013 is a subsequence.

%C 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

%C 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

%H Michael S. Branicky, <a href="/A333978/b333978.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from Peter Kagey)

%e The first 11 terms can be written as

%e    1 = 1

%e    2 = 1 * 2

%e    4 = 1 * 2 * 2

%e    6 = 1 * 2 * 3

%e    8 = 1 * 2 * 2 * 2

%e   12 = 1 * 2 * 2 * 3

%e   16 = 1 * 2 * 2 * 2 * 2

%e   18 = 1 * 2 * 3 * 3

%e   24 = 1 * 2 * 3 * 4 or 1 * 2 * 2 * 2 * 3

%e   32 = 1 * 2 * 2 * 2 * 2 * 2

%e   36 = 1 * 2 * 2 * 3 * 3

%o (SWI-Prolog)

%o main :- iter(1).

%o iter(K) :-

%o     (legal(K * x ^ 0) -> (maplist(write, [K, ', ']), flush_output) ; true),

%o     KK is K + 1, iter(KK).

%o legal(1 * x ^ 0).

%o legal(K * x ^ N) :-

%o     NN is N + 1, 0 is K mod NN, KK is K / NN,

%o     legal(KK * x ^ NN).

%o legal(K * x ^ N) :-

%o     ((K = 1, N = 1) ; (N > 1)), NN is N - 1,

%o     legal(K * x ^ NN).

%o % _Luc Rousseau_, Aug 20 2022

%o (Python)

%o import heapq

%o from math import factorial

%o from sympy import nextprime

%o from itertools import islice

%o def agen(): # generator of terms

%o     oldv, h, primes, nextp, nextfact = 0, [(1, 1)], [], 0, 0

%o     while True:

%o         v, maxp = heapq.heappop(h)

%o         if v != oldv:

%o             yield v; oldv = v

%o             while nextfact < v:

%o                 nextp = nextprime(nextp); nextfact = factorial(nextp)

%o                 primes.append(nextp); heapq.heappush(h, (nextfact, nextp))

%o             for p in primes:

%o                 if p <= maxp: heapq.heappush(h, (v*p, max(maxp, p)))

%o                 else: break

%o print(list(islice(agen(), 60))) # _Michael S. Branicky_, Aug 20 2022

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

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

%K nonn,easy

%O 1,2

%A _Peter Kagey_, Sep 20 2020