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A377486 a(n) = product of {p^k : p | n, k = 1..floor(log n/log p)}, a(1) = 1.

%I A377486 #6 Nov 01 2024 11:11:57
%S A377486 1,2,3,8,5,24,7,64,27,320,11,1728,13,448,135,1024,17,27648,19,5120,
%T A377486 189,11264,23,27648,125,13312,729,7168,29,93312000,31,32768,8019,
%U A377486 557056,875,23887872,37,622592,9477,4096000,41,167215104,43,360448,91125,753664,47,23887872
%N A377486 a(n) = product of {p^k : p | n, k = 1..floor(log n/log p)}, a(1) = 1.
%C A377486 Compare with A064446, where A064446(n) = Product_{p|n} p^floor(log n / log p).
%H A377486 Michael De Vlieger, <a href="/A377486/b377486.txt">Table of n, a(n) for n = 1..10000</a>
%H A377486 Michael De Vlieger, <a href="/A377486/a377486.png">Log log scatterplot of a(n)</a>, n = 1..2^20.
%F A377486 a(p) = p for prime p.
%F A377486 a(p^k) = Product_{j=1..k} p^j = p^(k*(k+1)/2) = p^A000217(k).
%F A377486 a(n) = Product_{p|n} p^(k*(k+1)/2), where k = floor(log n / log p).
%F A377486 Product of row n of A377485.
%e A377486 Let S(n) = row n of A377485 = { p^k : p | n, p^k <= n, k > 0 }.
%e A377486 a(4) = 8 since S(4) = {2, 4} and the product of these is 8.
%e A377486 a(6) = 24 since S(6) = {2, 3, 4} and the product of these is 24.
%e A377486 a(12) = 1728 since S(12) = {2, 3, 4, 8, 9}, etc.
%t A377486 {1}~Join~Table[Times @@ Flatten@ Map[#^Range[Floor@ Log[#, n]] &, FactorInteger[n][[All, 1]]], {n, 2, 120}]
%Y A377486 Cf. A064446, A162306, A246655, A377485.
%K A377486 nonn,easy
%O A377486 1,2
%A A377486 _Michael De Vlieger_, Oct 29 2024