<HTML><BODY style="word-wrap: break-word; -khtml-nbsp-mode: space; -khtml-line-break: after-white-space; "><DIV><BLOCKQUOTE type="cite"><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; ">a(1) = 1. a(n) is the lowest positive integer such that the continued fraction [a(1);a(2),a(3),...,a(n)] is equal to a rational with a square numerator.</DIV></BLOCKQUOTE><BLOCKQUOTE type="cite"><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; ">The sequence begins: 1, 3, 2, 5,...</DIV></BLOCKQUOTE><DIV><BR class="khtml-block-placeholder"></DIV><DIV>Jack Brennen:</DIV><BR class="Apple-interchange-newline"><BLOCKQUOTE type="cite"><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; "><FONT face="Monaco" size="5" style="font: 16.0px Monaco">The next few terms of Leroy's sequence are:</FONT></DIV><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; "><FONT face="Monaco" size="5" style="font: 16.0px Monaco">1,3,2,5, 43, 522, 1104509, ...</FONT></DIV></BLOCKQUOTE><BR></DIV><DIV><BR class="khtml-block-placeholder"></DIV><DIV>... and here's my attempt at a(8)-a(12):</DIV><DIV><BR class="khtml-block-placeholder"></DIV><DIV>60248974744, </DIV><DIV>2075863890266492169136, </DIV><DIV>10942918579397694712648387271683911959312808, </DIV><DIV>30436613005235318097155473477154291219175029919236526500330140104415890363628017565032, </DIV><DIV>78393726730883984175796140914361366606308435898478368950014907168360536050435575630074874593765271306284234877366717354182102726789039077399421798511614768052156175104 </DIV><DIV><BR class="khtml-block-placeholder"></DIV><DIV><BR class="khtml-block-placeholder"></DIV><BLOCKQUOTE type="cite"><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; ">The continued fraction [1,3,2,5,43,522,1104509] corresponds to:</DIV><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; "> 1220041748025/946166591401</DIV><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; ">Where the numerator is 1104555 squared.</DIV><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; min-height: 14px; "><BR></DIV><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; ">I'm sure it's not a coincidence that the square root of the numerator</DIV><DIV style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; ">is numerically very close to the last term of the continued fraction</DIV></BLOCKQUOTE><BR><DIV><BR class="khtml-block-placeholder"></DIV><DIV>My take on this is that it's not a coincidence in the sense that there exists a term close to the square root of the resulting convergent's numerator that will satisfy the condition of the numerator being a square; however, i think it *is* a coincidence with the added stipulation (as is the case with Leroy's formulation) that the last term of the continued fraction be the *smallest* positive integer that satisfies that condition.</DIV><DIV><BR class="khtml-block-placeholder"></DIV><DIV><BR class="khtml-block-placeholder"></DIV></BODY></HTML>