A363063 Positive integers k such that the largest power of p dividing k is larger than or equal to the largest power of q dividing k (i.e., A305720(k,p) >= A305720(k,q)) for all primes p and q with p < q.
1, 2, 4, 8, 12, 16, 24, 32, 48, 64, 96, 128, 144, 192, 256, 288, 384, 512, 576, 720, 768, 864, 1024, 1152, 1440, 1536, 1728, 2048, 2304, 2880, 3072, 3456, 4096, 4320, 4608, 5760, 6144, 6912, 8192, 8640, 9216, 10368, 11520, 12288, 13824, 16384, 17280, 18432
Offset: 1
Keywords
Examples
151200 = 2^5 * 3^3 * 5^2 * 7 is a term, because 2^5 >= 3^3 >= 5^2 >= 7. 72 = 2^3 * 3^2 is not a term, because 2^3 < 3^2. 40 = 2^3 * 3^0 * 5 is not a term, because 3^0 < 5. From _Michael De Vlieger_, May 19 2023: (Start) Sequence read as an irregular triangle delimited by appearance of 2^m: 1 2 4 8 12 16 24 32 48 64 96 128 144 192 256 288 384 512 576 720 768 864 1024 1152 1440 1536 1728 2048 2304 2880 3072 3456 4096 4320 4608 5760 6144 6912 8192 8640 9216 10368 11520 12288 13824 ... (End)
Links
- Pontus von Brömssen, Table of n, a(n) for n = 1..10000
- Michael De Vlieger, Plot p^e | a(n) at (x,y) = (n, pi(p)), n = 1..1024, showing multiplicity e with a color function such that e = 1 is black, e = 2 is red, e = 3 is orange, etc., 12X vertical exaggeration. On the bottom, a color code represents a(n) is empty product (black), prime (red), composite prime power (gold), neither squarefree nor prime power (blue).
- Michael De Vlieger, Plot multiplicities e in a(n) = Product p^e at (x,y) = (e, -n) for n = 1..1024, 8X horizontal exaggeration.
Crossrefs
Programs
-
Mathematica
Select[Range[20000], # == 1 || PrimePi[(f = FactorInteger[#])[[-1, 1]]] == Length[f] && Greater @@ (Power @@@ f) &] (* Amiram Eldar, May 16 2023 *)
-
Python
from sympy import nextprime primes = [2] # global list of first primes def f(kmax,pi,ppmax): # Generate numbers up to kmax with nonincreasing prime-powers <= ppmax, starting at the (pi+1)-st prime. if len(primes) <= pi: primes.append(nextprime(primes[-1])) p0 = primes[pi] ppmax = min(ppmax,kmax) if ppmax < p0: yield 1 return pp = 1 while pp <= ppmax: for x in f(kmax//pp,pi+1,pp): yield pp*x pp *= p0 def A363063_list(kmax): return sorted(f(kmax,0,kmax))
Comments