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A262868
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Number of squarefree numbers appearing among the larger parts of the partitions of n into two parts.
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6
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0, 1, 1, 2, 1, 2, 2, 3, 3, 3, 3, 4, 3, 4, 4, 5, 5, 6, 6, 7, 6, 7, 7, 8, 8, 8, 8, 8, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 14, 14, 14, 15, 15, 15, 15, 16, 15, 16, 16, 17, 17, 18, 18, 19, 18, 19, 19, 19, 19, 20, 20, 21, 20, 21, 21, 22, 22
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OFFSET
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1,4
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COMMENTS
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Number of distinct rectangles with squarefree length and integer width such that L + W = n, W <= L. For example, a(14) = 4; the rectangles are 1 X 13, 3 X 11, 4 X 10 and 7 X 7. - Wesley Ivan Hurt, Nov 02 2017
a(10) = 3, a(100) = 30, a(10^3) = 302, a(10^4) = 3041, a(10^5) = 30393, a(10^6) = 303968, a(10^7) = 3039658, a(10^8) = 30396350, a(10^9) = 303963598, a(10^10) = 3039635373, a(10^11) = 30396355273, a(10^12) = 303963551068, a(10^13) = 3039635509338, a(10^14) = 30396355094469, a(10^15) = 303963550926043, a(10^16) = 3039635509271763, a(10^17) = 30396355092700721, and a(10^18) = 303963550927014110. The limit of a(n)/n is 3/Pi^2. - Charles R Greathouse IV, Nov 04 2017
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LINKS
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FORMULA
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a(n) = Sum_{i=1..floor(n/2)} mu(n-i)^2, where mu is the Möbius function A008683.
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EXAMPLE
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a(4)=2; there are two partitions of 4 into two parts: (3,1) and (2,2). Both of the larger parts are squarefree, thus a(4)=2.
a(5)=1; there are two partitions of 5 into two parts: (4,1) and (3,2). Among the larger parts, only 3 is squarefree, thus a(5)=1.
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MAPLE
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with(numtheory): A262868:=n->add(mobius(n-i)^2, i=1..floor(n/2)): seq(A262868(n), n=1..100);
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MATHEMATICA
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Table[Sum[MoebiusMu[n - i]^2, {i, Floor[n/2]}], {n, 100}]
Table[Count[IntegerPartitions[n, {2}][[All, 1]], _?SquareFreeQ], {n, 80}] (* Harvey P. Dale, Jan 03 2022 *)
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PROG
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(PARI) a(n) = sum(i=1, n\2, moebius(n-i)^2); \\ Michel Marcus, Oct 04 2015
(PARI) f(n)=my(s); forfactored(k=1, sqrtint(n), s+=n\k[1]^2*moebius(k)); s
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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