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<DIV>I can't actually prove it, but this does lead to some interesting areas.</DIV>
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<DIV>I'm focussing on the (s[n]-1)/8 from Jack's sequence. These are:</DIV>
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<DIV>0,1,6,35,204,1189,6930,40391,...</DIV>
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<DIV>These are A001109, and certainly any solutions are in this sequence. Call this sequence a(n).</DIV>
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<DIV>Note, by the way, that these numbers are one half of the bisection of the Pell numbers (A000129). I have submitted a comment to this effect. Since the other half of the Pell numbers are odd, the assertion that these are the only triangular numbers with triangular squares is equivalent to the assertion that the only Pell numbers of the form n^2+n are 0, 2, and 12.</DIV>
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<DIV>The members of A001109 can be represented in a regular way as the difference of triangular numbers. Specifically, let b(n) be the sequence:</DIV>
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<DIV>0,0,1,3,8,20,49,119,288,...</DIV>
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<DIV>This is A048739 (partial sums of Pell numbers) with two initial zeros. Then a(n) = b(n+1)*(b(n+1)+1)/2 - b(n-1)*(b(n-1)+1)/2 (with b(-1)=0).</DIV>
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<DIV>Also, the bisection of the b(n)'s 0,1,8,49,288,1681,... gives A001108, the numbers whose triangular is the squares of the a(n)'s. The other bisection of the b(n)'s is A001652, the numbers whose triangular is of the form n^2+n.</DIV></DIV>
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<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV>
<DIV> </DIV> <BR>-----Original Message-----<BR>From: Jack Brennen <jb@brennen.net><BR>To: seqfan@ext.jussieu.fr<BR>Sent: Fri, 6 Jan 2006 20:24:46 -0500 (EST)<BR>Subject: Re: Triangular numbers with triangular squares<BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_7bb094f2-8807-4725-84e4-0c915866b4fd><PRE><TT>
I'm pretty sure that 0, 1, and 6 are the only triangular numbers
with a triangular square.
If I'm not mistaken, all solutions can be isolated to this
recursive sequence:
s[0] = 1
s[1] = 9
s[n] = 6*s[n-1] - s[n-2] - 4
The sequence goes:
1, 9, 49, 281, 1633, 9513, 55441, 323129, ...
If a number X of this sequence is a perfect square, then (X-1)/8
is a triangular number. Note that due to the form of the
sequence, ((X-1)/8)^2 is always a triangular number.
This sequence grows very quickly and did not produce any more
squares even searching up to 10000 digit numbers.
Jack
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