|
%I #20 Nov 04 2022 10:47:18
%S 1,2,3,4,5,6,7,6,6,10,11,12,13,14,15,8,17,12,19,20,21,22,23,18,10,26,
%T 9,28,29,30,31,10,33,34,35,24,37,38,39,30,41,42,43,44,30,46,47,24,14,
%U 20,51,52,53,18,55,42,57,58,59,60,61,62,42,12,65,66,67,68
%N Multiplicative with a(p^k) = p*a(k) for any prime p and k>0.
%C To compute a(n): multiply (with multiplicity) each prime number appearing in the prime tower factorization of n (see A182318 for definition).
%C a(n) = n if n is squarefree.
%C a(n) <= A000026(n) for any n>0.
%C First differs from A000026 at n=256: a(256)=12 and A000026(256)=16.
%C If n = p_1 * p_2 * ... * p_k with p_1, p_2, ..., p_k primes, then a(p_1 ^ p_2 ^ ... ^ p_k) = n.
%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.33 Hall-Montgomery Constant, p. 207.
%H Rémy Sigrist, <a href="/A279513/b279513.txt">Table of n, a(n) for n = 1..10000</a>
%H Richard Alan Gilman and Rainer Tschiersch, <a href="https://www.jstor.org/stable/2324471">Problem 6660</a>, The American Mathematical Monthly, Vol. 100, No. 3 (1993), pp. 296-298.
%F Sum_{k=1..n} a(k) ~ (1/2) * c * n^2, where c = Product_{p prime} (1 - 1/p^2 + (p-1)*Sum_{k>=2} a(k)/p^(2*k)) = 0.8351076361... (Gilman and Tschiersch, 1993; Finch, 2003; the constant was calculated by Kevin Ford). - _Amiram Eldar_, Nov 04 2022
%e a(6!) = a(2^(2^2)*3^2*5) = 2*2*2*3*2*5 = 240.
%p a:= proc(n) option remember; `if`(n=1, 1,
%p mul(i[1]*a(i[2]), i=ifactors(n)[2]))
%p end:
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Aug 22 2020
%t a[n_] := a[n] = If[n==1, 1, Times @@ (#[[1]] a[#[[2]]]& /@ FactorInteger[n] )]; Array[a, 256] (* _Jean-François Alcover_, Mar 31 2017 *)
%o (PARI) a(n) = my (f=factor(n)); return (prod(i=1, #f~, f[i,1]*a(f[i,2])))
%Y Cf. A000026, A182318, A279510.
%K nonn,mult
%O 1,2
%A _Rémy Sigrist_, Dec 13 2016
|
