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A088912
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a(n) = smallest m such that sigma(m) = (n+1/2)*m.
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4
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OFFSET
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1,1
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COMMENTS
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2 is the only number m such that sigma(m)=1.5*m.
A direct consequence of Robin's theorem is that a(6)>5E16, a(7)>1.898E29, a(8)>2.144E51, a(9)>9.877E89 and a(10)>6.023E157. - Washington Bomfim, Oct 30 2008
If the Riemann hypothesis (RH) is true then Robin's theorem (Guy Robin, 1984) implies that the n-th term of this sequence is greater than exp(exp((n+1/2)/exp(gamma))) where gamma=0.5772156649... is the Euler-Mascheroni constant (A001620). For the 6th term (which is actually 1.7*10^44) this lower bound is 5.0*10^16. Similarly, if RH is true, the next term (7th term) is at least 1.9*10^29 (and is probably more than 10^90 or so). - Gerard P. Michon, Jun 10 2009
An upper bound for a(7) is provided by a 97-digit integer of abundancy 15/2 (5.71379...10^96) discovered by Michel Marcus on July 4, 2009. The factorization of that number is: 2^53 3^15 5^6 7^6 11^3 13 17 19^3 23 29 31 37 41 43 61 73 79 97 181 193 199 257 263 4733 11939 19531 21803 87211 262657.
Similarly, an upper bound for a(8) is provided by a 286-digit integer of abundancy 17/2 (3.30181...10^285) equal to x/17, where x is the smallest known number of abundancy 9 (a 287-digit integer discovered by Fred W. Helenius in 1995). This is so because 17 happen to occur with multiplicity 1 in the factorization of x. (End)
A new upper bound for a(7) was found on Aug 15 2009 by Michel Marcus, who broke his own record by finding two "small" multiples of 2^35*3^20*5^5*7^6*11^2*13^2*17 that are of abundancy 15/2. The lower one (1.27494722...10^88) has only 89 digits. - Gerard P. Michon, Aug 15 2009
These are the least hemiperfects of abundancy n + 1/2. - Walter Nissen, Aug 17 2010
On Jul 24 2010, Michel Marcus found a 191-digit integer of abundancy 17/2 (2.7172904...10^190) whose factorization starts with 2^81 3^29 5^9 7^10 11^4 13^3 17^2 19 23^2... This is the best upper bound to a(8) known so far. - Gerard P. Michon, Aug 22 2010
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REFERENCES
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Guy Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187-213. [From Gerard P. Michon, Jun 10 2009]
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LINKS
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EXAMPLE
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a(2)=24 because 1+2+3+4+6+8+12+24=2.5*24 and 24 is the earliest m such that sigma(m)=2.5*m.
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MATHEMATICA
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a[n_] := (For[m=1, DivisorSigma[1, m]!=(n+1/2)m, m++ ]; m); Do[Print[a[n]], {n, 4}]
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CROSSREFS
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KEYWORD
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hard,more,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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