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Search: id:A125650
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| A125650 |
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Numerator of n(n+3)/(4(n+1)(n+2)) = Sum[ 1/(k(k+1)(k+2)), {k,1,n} ]. |
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+0
6
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| 1, 5, 9, 7, 5, 27, 35, 11, 27, 65, 77, 45, 26, 119, 135, 38, 85, 189, 209, 115, 63, 275, 299, 81, 175, 377, 405, 217, 116, 495, 527, 140, 297, 629, 665, 351, 185, 779, 819, 215, 451, 945, 989, 517, 270, 1127, 1175, 306, 637, 1325, 1377, 715, 371, 1539, 1595, 413, 855
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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3^2 divides a(3k). p divides a(p) for an odd prime p. p divides a(p-3) for prime p>3. p^k divides a(p^k) for an odd prime p. a(n) = m^2 is a perfect square for n = {1,3,24,147,864,5043,29400,171363,...} = A125651(n). Corresponding numbers m such that m^2 = a[ A125651(n) ] are listed in A125652(n) = {1,3,9,105,306,3567,10395,121173,...}.
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FORMULA
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a(n) = Numerator[ n(n+3)/(4(n+1)(n+2)) ].
a(n)=n*(n+3)/2^min(3,valuation(n*(n+3),2)). a(n)=n*(n+3)/4 for n=1 or 4 (mod 8); a(n)=n*(n+3)/8 for n=0 or 5 (mod 8); a(n)=n*(n+3)/2 for n=2, 3, 6, or 7 (mod 8). - Max Alekseyev (maxale(AT)gmail.com), Jan 11 2007
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MATHEMATICA
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Table[Numerator[n(n+3)/(4(n+1)(n+2))], {n, 1, 100}]
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PROGRAM
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(PARI) a(n)=n*(n+3)/2^min(3, valuation(n*(n+3), 2)) - Max Alekseyev (maxale(AT)gmail.com), Jan 11 2007
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CROSSREFS
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Cf. A125651, A125652.
Sequence in context: A079459 A118309 A100106 this_sequence A140724 A086055 A077125
Adjacent sequences: A125647 A125648 A125649 this_sequence A125651 A125652 A125653
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KEYWORD
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nonn,frac
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AUTHOR
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Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 29 2006
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