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A075036
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Smaller of two smallest consecutive numbers with 2n divisors.
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8
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2, 14, 44, 104, 2511, 735, 29888, 2295, 6075, 5264, 2200933376, 5984, 689278976, 156735, 180224, 21735, 2035980763136, 223244, 9399153082499072, 458864, 41680575, 701443071, 2503092614937444351, 201824, 2707370000, 29785673727, 46977524, 5475519, 1737797404898095794225152
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OFFSET
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1,1
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COMMENTS
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There cannot be two consecutive numbers with the same odd number of divisors as both cannot be squares.
These numbers have the property that a(n) * (a(n) + 1) has 4*n^2 divisors. - David A. Corneth, Jun 24 2016
Conjecture: if a term k is even, the highest p-adic order of k (the maximum may be attained by several p's) occurs at p=2 and the highest p-adic order of k+1 occurs at p=3. If a term k is odd, the highest p-adic order of k occurs at p=3 and the highest p-adic order of k+1 occurs at p=2. - Chai Wah Wu, Mar 12 2019
a(49) = 378401464109375, a(58) = 79921490583489592950783. - Jon E. Schoenfield, May 07 2022
a(51) = 34210814718574592, a(55) = 2481402804069375, a(57) = 394311388855795712. - Jon E. Schoenfield, Nov 06 2023 - Nov 08 2023
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LINKS
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FORMULA
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a(n) <= A215199(n-1) for n > 1. Conjecture: if p is prime, then a(p) = A215199(p-1). This conjecture is true if the conjecture in A215199 is true. The b-file of A215199 thus shows that a(p) = A215199(p-1) for prime p < 1279. - Chai Wah Wu, Mar 12 2019
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EXAMPLE
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a(4) = 104 as tau(104) = tau(105) = 8.
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MATHEMATICA
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a[n_] := (For[k=1, ! (DivisorSigma[0, k] == 2*n && DivisorSigma[0, k+1] == 2*n), k++]; k); Array[a, 10] (* Giovanni Resta, Jun 24 2016 *)
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PROG
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(PARI) a(n) = my(k=1); while(numdiv(k)!=2*n || numdiv(k+1)!=2*n, k++); k \\ Felix Fröhlich, Jun 24 2016
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CROSSREFS
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KEYWORD
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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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