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A169686
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a(n) = sqrt(T(k-1)*T(k)*T(k+1)) as k runs through the terms of A072221 and T(i)=i*(i+1)/2.
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0
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0, 30, 5850, 1157730, 229221540, 45384688830, 8985939059790, 1779170548525890, 352266782665431240, 69747043797185672190, 13809562405059974172930, 2734223609158076980818690, 541362465050894178032921580, 107187033856467889149087366750, 21222491341115591157198758976630
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OFFSET
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1,2
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COMMENTS
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It is known (see Beiler, p. 198) that the product of three consecutive triangular numbers, T(k-1)T(k)T(k+1), is a square if (and only if?) 2k+1 = 3a for a in A001541. The corresponding values of k are in A072221.
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REFERENCES
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A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1966.
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LINKS
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FORMULA
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Empirical g.f.: 30*x^2*(x^2-9*x+1) / ((x^2-198*x+1)*(x^2-6*x+1)). - Colin Barker, Jul 26 2013
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EXAMPLE
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sqrt (T(3)T(4)T(5)) = 30
sqrt (T(24)T(25)T(26)) = 5850
sqrt (T(147)T(148)T(149)) = 1157730
sqrt (T(864)T(865)T(866)) = 229221540
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane, Apr 13 2010, based on an email from Neven Juric, Mar 19 2010
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STATUS
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approved
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