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 A169686 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. 0

%I #10 Dec 26 2023 12:19:31

%S 0,30,5850,1157730,229221540,45384688830,8985939059790,

%T 1779170548525890,352266782665431240,69747043797185672190,

%U 13809562405059974172930,2734223609158076980818690,541362465050894178032921580,107187033856467889149087366750,21222491341115591157198758976630

%N 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.

%C 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.

%D A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1966.

%F 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

%e sqrt (T(3)T(4)T(5)) = 30

%e sqrt (T(24)T(25)T(26)) = 5850

%e sqrt (T(147)T(148)T(149)) = 1157730

%e sqrt (T(864)T(865)T(866)) = 229221540

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Apr 13 2010, based on an email from Neven Juric, Mar 19 2010

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Last modified August 13 20:02 EDT 2024. Contains 375144 sequences. (Running on oeis4.)