A369099 - OEIS

%I #15 Jan 20 2024 09:35:31

%S 1,2,4,6,16,12,64,30,36,48,65536,60,4096,192,144,210,

%T 18446744073709551616,180,1073741824,240,576,196608,68719476736,420,

%U 1296,12288,900,960,281474976710656,720

%N Index of first occurrence of n in A369015; smallest number whose prime tower factorization tree has Matula-Göbel number n.

%C After a(30), the sequence continues 2^65536, 2*3*5*7*11, 2^16*3^2, 2^64*3, 2^6*3^4, 2^2*3^2*5*7, 2^60, 2^30*3, 2^12*9, 2^4*3*5*7, 2^4096, ... .

%C a(n) can be determined recursively as follows. Let n = Product_{i>=1} p_i^e_i, where p_i is the i-th prime. Take f_1 >= f_2 >= ... >= f_k so that the number a(i) occurs e_i times for i >= 1. Then a(n) = Product_{i>=1} p_i^f_i.

%C All terms are in A025487 (products of primorials).

%H <a href="/index/Mat#matula">Index entries for sequences related to Matula-Göbel numbers</a>.

%F a(prime(n)) = 2^a(n). As a consequence, a(A007097(n)) = A014221(n).

%F a(2^n) = A002110(n).

%F a(n) = A284456(A369137(n)) = A025487(A369138(A369137(n))).

%e Using the method described in the comments for n = 20 = p(1)^2*p(3)^1, the exponents f_i shall include the term a(1)=1 twice and the term a(3)=4 once, i.e., (f_1, f_2, f_3) = (4, 1, 1), so a(20) = p(1)^4*p(2)^1*p(3) = 240.

%o (Python)

%o from sympy import factorint,nextprime,primepi

%o def A369099(n):

%o     f = {A369099(primepi(p)):e for p,e in factorint(n).items()}

%o     a = p = 1

%o     for k in sorted(f,reverse=True):

%o         for i in range(f[k]):

%o             p = nextprime(p)

%o             a *= p**k

%o     return a

%Y Cf. A002110, A007097, A014221, A025487, A369015, A369137, A369138.

%Y A permutation of A284456.

%K nonn

%O 1,2

%A _Pontus von Brömssen_, Jan 13 2024