|169 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |119 |120 |169 |- !8 |577 |408 |288 |289 |204 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !9 |1393 |985 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |696 |697 |985 |- !10 |3363 |2378 |1681 |1682 |1189 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !11 |8119 |5741 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |4059 |4060 |5741 |- !12 |19601 |13860 |9800 |9801 |6930 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |} Definitions The half-companion Pell numbers Hn and the Pell numbers Pn can be derived in a number of easily equivalent ways. Raising to powers From this it follows that there are closed forms: and Paired recurrences Reciprocal recurrence formulas Let n be at least 2. ; . Matrix formulations So Approximations The difference between Hn and Pn is which goes rapidly to zero. So is extremely close to 2Hn. From this last observation it follows that the integer ratios rapidly approach ; and and rapidly approach 1 + . H2 − 2P2 = ±1 Since is irrational, we cannot have = , i.e., The best we can achieve is either The (non-negative) solutions to are exactly the pairs with n even, and the solutions to are exactly the pairs with n odd. To see this, note first that so that these differences, starting with , are alternately 1 and −1. Then note that every positive solution comes in this way from a solution with smaller integers since The smaller solution also has positive integers, with the one exception: which comes from H0 = 1 and P0 = 0. Square triangular numbers The required equation is equivalent to which becomes with the substitutions H = 2t + 1 and P = 2s. Hence the n-th solution is Observe that t and t + 1 are relatively prime, so that = s2 happens exactly when they are adjacent integers, one a square H&hairsp;&hairsp;2 and the other twice a square 2P&hairsp;&hairsp;2. Since we know all solutions of that equation, we also have and This alternate expression is seen in the next table. {| class="wikitable" style="text-align:center" |- !n !Hn !Pn !t !t + 1 !s !a !b !c |- !0 |1 |0 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !1 |1 |1 |1 |2 |1 |3 |4 |5 |- !2 |3 |2 |8 |9 |6 |20 |21 |29 |- !3 |7 |5