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A227771
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Antiharmonic numbers that are not squares.
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6
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20, 50, 117, 180, 200, 242, 325, 450, 468, 500, 578, 605, 650, 800, 968, 980, 1025, 1058, 1280, 1300, 1445, 1476, 1620, 1682, 1700, 1800, 1872, 2178, 2312, 2340, 2420, 2450, 2600, 2645, 2925, 3200, 3362, 3380, 3757, 3872, 4050, 4100, 4205, 4232, 4352, 4418, 4500, 4693, 5200
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OFFSET
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1,1
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COMMENTS
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Given prime factorization m = product (p_i^e_i), the antiharmonic (or contraharmonic) mean of the divisors of m is sigma_2(m)/sigma_1(m) = product (p_i^(e_i+1)+1)/(p_i+1). If this is an integer, then m is called antiharmonic.
All squares are trivially antiharmonic, since (p^(2*e+1)+1)/(p+1) = p^(2*e) - p^(2*e-1) + p^(2*e-2) - ... + 1 is an integer. Sequence gives the nontrivial antiharmonic numbers.
The antiharmonic means of their divisors are A227986.
Sequence is infinite, since if n is in the sequence and gcd(n, k) = 1 then nk^2 is also in the sequence. - Charles R Greathouse IV, Aug 02 2013
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REFERENCES
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R. Guy, Unsolved Problems in Number Theory, B2 (see harmonic number).
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LINKS
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FORMULA
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EXAMPLE
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sigma_2(20)/sigma_1(20) = (1^2 + 2^2 + 4^2 + 5^2 + 10^2 + 20^2)/(1 + 2 + 4 + 5 + 10 + 20) = 546/42 = 13 is an integer, 20 is not a square, and no smaller number has these properties, so a(1) = 20.
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PROG
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(PARI) is(n)=if(issquare(n), return(0)); my(f=factor(n)); denominator(prod(i=1, #f~, (f[i, 1]^(f[i, 2]+1)+1)/(f[i, 1]+1)))==1 \\ Charles R Greathouse IV, Aug 02 2013
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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