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%I #15 Feb 24 2023 11:54:28
%S 0,1,3,3,7,4,12,13,13,8,24,26,58,51,53,53,117,116,244,240,250,235,491,
%T 488,488,457,459,451,963,964,1988,1989,2007,1942,1946,1946,3994,3867,
%U 3897,3900,7996,7991,16183,16167,16163,15906,32290,32288,32288,32289,32355
%N a(n) = Sum_{p | A055204(n)} 2^(pi(p) - 1).
%C All terms of A055204 are squarefree by definition, therefore we can compress the terms of A055204 by interpreting the terms of reverse(A067255(A055204(n))) as a binary number and converted to decimal.
%H Michael De Vlieger, <a href="/A336510/b336510.txt">Table of n, a(n) for n = 1..10000</a>
%H Michael De Vlieger, <a href="/A336510/a336510.png">Plot of the bits of a(n)</a> with (x,y) = (n, a(n)) for 1 <= n <= 2^14.
%e A055204(1) = 1, the empty product; by convention a(1) = 0.
%e 5! = 120 = 2^3 * 3 * 5, therefore 2 * 3 * 5 = 30 is the squarefree part, which we write "111", a 1 in the first three places to signify a product of the first three primes. Interpreting "111" as a binary number yields 8. Thus a(5) = 8.
%e 13! = 6227020800 = 2^10 * 3^5 * 5^2 * 7 * 11 * 13; its squarefree part is 3 * 7 * 11 * 13 = 3003, a product of the 2nd, 4th, 5th, and 6th primes. Therefore we write "111010", which, interpreted as a binary number and converted to decimal, is 58. Thus a(13) = 58.
%e Table illustrating the first terms of this sequence, with b(n) = A055204(n):
%e Multiplicities of p|b(n)
%e n b(n) 2 3 5 7 11 13 17 -> Binary a(n)
%e --------------------------------------------------
%e 1 1 . . . . . . . 0 0
%e 2 2 1 . . . . . . 1 1
%e 3 6 1 1 . . . . . 11 3
%e 4 6 1 1 . . . . . 11 3
%e 5 30 1 1 1 . . . . 111 7
%e 6 5 . . 1 . . . . 100 4
%e 7 35 . . 1 1 . . . 1100 12
%e 8 70 1 . 1 1 . . . 1101 13
%e 9 70 1 . 1 1 . . . 1101 13
%e 10 7 . . . 1 . . . 1000 8
%e 11 77 . . . 1 1 . . 11000 24
%e 12 231 . 1 . 1 1 . . 11010 26
%e 13 3003 . 1 . 1 1 1 . 111010 58
%e 14 858 1 1 . . 1 1 . 110011 51
%e 15 1430 1 . 1 . 1 1 . 110101 53
%e 16 1430 1 . 1 . 1 1 . 110101 53
%e 17 24310 1 . 1 . 1 1 1 1110101 117
%e 18 12155 . . 1 . 1 1 1 1110100 116
%e ...
%t Block[{nn = 51, k, p}, k = PrimePi@ nn; Array[Set[p[Prime@ #], 0] &, k]; {0}~Join~Reap[Do[Map[Set[p[#1], Mod[p[#1] + Mod[#2, 2], 2]] & @@ # &, FactorInteger@ i]; Sow[FromDigits[Array[p[Prime[k - # + 1]] &, k], 2]], {i, 2, nn}]][[-1, 1]]] (* or *)
%t Block[{nn = 51, k = 1}, Reap[Do[Map[If[Mod[k, #] == 0, k /= #, k *= #] &, Flatten[ConstantArray[#1, #2] & @@@ FactorInteger@ i]]; Sow[If[k == 1, 0, Total@ Map[2^(PrimePi[#] - 1) &, FactorInteger[k][[All, 1]] ] ] ], {i, nn}]][[-1, 1]]]
%Y Cf. A055204, A067255, A240504.
%K nonn,easy
%O 1,3
%A _Michael De Vlieger_, Sep 18 2020
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