For the series: 1,1,2,3,4,6,9,13...
The rule is: F(n+3)=F(n+2)+F(n) with starting conditions F(0)=F(1)=F(2)=1.
I found a closed form expression using a variation on a matrix based proof used for the Fibonacci series and came up with the following: $$\left[ {\begin{array}{*{20}{c}}{{F_{2t + 2}}}\\{{F_{2t + 1}}}\\{{F_{2t}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&0&1\\1&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{F_{2t}}}\\{{F_{2t - 1}}}\\{{F_{2t - 2}}}\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakeaaqaaaaaaaaaWdbmaadmaapaqaauaabe % qadeaaaeaapeGaamOra8aadaWgaaWcbaWdbiaaikdacaWG0bGaey4k % aSIaaGOmaaWdaeqaaaGcbaWdbiaadAeapaWaaSbaaSqaa8qacaaIYa % GaamiDaiabgUcaRiaaigdaa8aabeaaaOqaa8qacaWGgbWdamaaBaaa % leaapeGaaGOmaiaadshaa8aabeaaaaaak8qacaGLBbGaayzxaaGaey % ypa0ZaamWaa8aabaqbaeqabmWaaaqaa8qacaaIXaaapaqaa8qacaaI % Xaaapaqaa8qacaaIXaaapaqaa8qacaaIXaaapaqaa8qacaaIWaaapa % qaa8qacaaIXaaapaqaa8qacaaIXaaapaqaa8qacaaIWaaapaqaa8qa % caaIWaaaaaGaay5waiaaw2faamaadmaapaqaauaabeqadeaaaeaape % GaamOra8aadaWgaaWcbaGaaGOmaiaadshaaeqaaaGcbaWdbiaadAea % paWaaSbaaSqaa8qacaaIYaGaamiDaiabgkHiTiaaigdaa8aabeaaaO % qaa8qacaWGgbWdamaaBaaaleaapeGaaGOmaiaadshacqGHsislcaaI % YaaapaqabaaaaaGcpeGaay5waiaaw2faaaaa!58FC!$$ $$ = {\left[ {\begin{array}{*{20}{c}}1&1&1\\1&0&1\\1&0&0\end{array}} \right]^t}\left[ {\begin{array}{*{20}{c}}{{F_2}}\\{{F_1}}\\{{F_0}}\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakeaacqGH9aqpqaaaaaaaaaWdbmaadmaapa % qaauaabeqadmaaaeaapeGaaGymaaWdaeaapeGaaGymaaWdaeaapeGa % aGymaaWdaeaapeGaaGymaaWdaeaapeGaaGimaaWdaeaapeGaaGymaa % WdaeaapeGaaGymaaWdaeaapeGaaGimaaWdaeaapeGaaGimaaaaaiaa % wUfacaGLDbaadaahaaWcbeqaaiaadshaaaGcdaWadaWdaeaafaqabe % WabaaabaWdbiaadAeapaWaaSbaaSqaaiaaikdaaeqaaaGcbaWdbiaa % dAeapaWaaSbaaSqaaiaaigdaaeqaaaGcbaWdbiaadAeapaWaaSbaaS % qaaiaaicdaaeqaaaaaaOWdbiaawUfacaGLDbaaaaa!4578! $$ A notable difference from the Fibonacci proof, which I did not immediately notice, is that the F vector advances 2 numbers for each activation of the matrix.
Further steps were very similar: $$\left| {\begin{array}{*{20}{c}}{1 - \lambda }&1&1\\1&{ - \lambda }&1\\1&0&{ - \lambda }\end{array}} \right| = 0 \Rightarrow % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakeaacqGHshI3aaa!341E! \left\{ {\scriptstyle{\lambda _1} = 2.1479\atop{\scriptstyle{\lambda _2} = - 0.5739 - 0.3690i\atop\scriptstyle{\lambda _3} = - 0.5739 + 0.3690i}} \right. % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakeaalmaaceaaeaqabeaaqaaaaaaaaaWdbi % abeU7aSnaaBaaabaGaaGymaaqabaGaeyypa0ZdaiaaikdacaGGUaGa % aGymaiaaisdacaaI3aGaaGyoaaqaa8qacqaH7oaBdaWgaaqaaiaaik % daaeqaaiabg2da98aacqGHsislcaaIWaGaaiOlaiaaiwdacaaI3aGa % aG4maiaaiMdacqGHsislcaaIWaGaaiOlaiaaiodacaaI2aGaaGyoai % aaicdacaWGPbaabaWdbiabeU7aSnaaBaaabaGaaG4maaqabaGaeyyp % a0ZdaiabgkHiTiaaicdacaGGUaGaaGynaiaaiEdacaaIZaGaaGyoai % abgUcaRiaaicdacaGGUaGaaG4maiaaiAdacaaI5aGaaGimaiaadMga % aaGaay5Eaaaaaa!59B4! $$ Eigenvectors were all of the form: $$\left[ {\begin{array}{*{20}{c}}{{\textstyle{{ - \left( {1 + \lambda + {\lambda ^2}} \right)} \over {1 - {\lambda ^2}}}}}\\{{\textstyle{{{\lambda ^2}} \over {1 + \lambda }}}}\\1\end{array}} \right] \Rightarrow V = % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakeaadaWadaqaauaabeqadeaaaeaadaWcba % WcbaGaeyOeI0YaaeWaaeaacaaIXaGaey4kaSIaeq4UdWMaey4kaSIa % eq4UdW2aaWbaaWqabeaacaaIYaaaaaWccaGLOaGaayzkaaaabaGaaG % ymaiabgkHiTiabeU7aSnaaCaaameqabaGaaGOmaaaaaaaakeaadaWc % baWcbaGaeq4UdW2aaWbaaWqabeaacaaIYaaaaaWcbaGaaGymaiabgU % caRiabeU7aSbaaaOqaaiaaigdaaaaacaGLBbGaayzxaaaaaa!485B! \left[ {\begin{array}{*{20}{c}}{{\textstyle{{ - \left( {1 + {\lambda _1} + {\lambda _1}^2} \right)} \over {1 - {\lambda _1}^2}}}}&{{\textstyle{{ - \left( {1 + {\lambda _2} + {\lambda _2}^2} \right)} \over {1 - {\lambda _2}^2}}}}&{{\textstyle{{ - \left( {1 + {\lambda _3} + {\lambda _3}^2} \right)} \over {1 - {\lambda _3}^2}}}}\\{{\textstyle{{{\lambda _1}^2} \over {1 + {\lambda _1}}}}}&{{\textstyle{{{\lambda _2}^2} \over {1 + {\lambda _2}}}}}&{{\textstyle{{{\lambda _3}^2} \over {1 + {\lambda _3}}}}}\\1&1&1\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakeaadaWadaqaauaabeqadmaaaeaadaWcba % WcbaGaeyOeI0YaaeWaaeaacaaIXaGaey4kaSIaeq4UdW2aaSbaaWqa % aiaaigdaaeqaaSGaey4kaSIaeq4UdW2aaSbaaWqaaiaaigdaaeqaaS % WaaWbaaWqabeaacaaIYaaaaaWccaGLOaGaayzkaaaabaGaaGymaiab % gkHiTiabeU7aSnaaBaaameaacaaIXaaabeaalmaaCaaameqabaGaaG % OmaaaaaaaakeaadaWcbaWcbaGaeyOeI0YaaeWaaeaacaaIXaGaey4k % aSIaeq4UdW2aaSbaaWqaaiaaikdaaeqaaSGaey4kaSIaeq4UdW2aaS % baaWqaaiaaikdaaeqaaSWaaWbaaWqabeaacaaIYaaaaaWccaGLOaGa % ayzkaaaabaGaaGymaiabgkHiTiabeU7aSnaaBaaameaacaaIYaaabe % aalmaaCaaameqabaGaaGOmaaaaaaaakeaadaWcbaWcbaGaeyOeI0Ya % aeWaaeaacaaIXaGaey4kaSIaeq4UdW2aaSbaaWqaaiaaiodaaeqaaS % Gaey4kaSIaeq4UdW2aaSbaaWqaaiaaiodaaeqaaSWaaWbaaWqabeaa % caaIYaaaaaWccaGLOaGaayzkaaaabaGaaGymaiabgkHiTiabeU7aSn % aaBaaameaacaaIZaaabeaalmaaCaaameqabaGaaGOmaaaaaaaakeaa % daWcbaWcbaGaeq4UdW2aaSbaaWqaaiaaigdaaeqaaSWaaWbaaWqabe % aacaaIYaaaaaWcbaGaaGymaiabgUcaRiabeU7aSnaaBaaameaacaaI % XaaabeaaaaaakeaadaWcbaWcbaGaeq4UdW2aaSbaaWqaaiaaikdaae % qaaSWaaWbaaWqabeaacaaIYaaaaaWcbaGaaGymaiabgUcaRiabeU7a % SnaaBaaameaacaaIYaaabeaaaaaakeaadaWcbaWcbaGaeq4UdW2aaS % baaWqaaiaaiodaaeqaaSWaaWbaaWqabeaacaaIYaaaaaWcbaGaaGym % aiabgUcaRiabeU7aSnaaBaaameaacaaIZaaabeaaaaaakeaacaaIXa % aabaGaaGymaaqaaiaaigdaaaaacaGLBbGaayzxaaaaaa!7FC0! $$ Leading to the expression: $$\left[ {\begin{array}{*{20}{c}}{{F_{2t + 2}}}\\{{F_{2t + 1}}}\\{{F_{2t}}}\end{array}} \right] = V\left[ {\begin{array}{*{20}{c}}{{\lambda _1}^t}&0&0\\0&{{\lambda _2}^t}&0\\0&0&{{\lambda _3}^t}\end{array}} \right]{V^{ - 1}}\left[ {\begin{array}{*{20}{c}}{{F_2}}\\{{F_1}}\\{{F_0}}\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakeaaqaaaaaaaaaWdbmaadmaapaqaauaabe % qadeaaaeaapeGaamOra8aadaWgaaWcbaWdbiaaikdacaWG0bGaey4k % aSIaaGOmaaWdaeqaaaGcbaWdbiaadAeapaWaaSbaaSqaa8qacaaIYa % GaamiDaiabgUcaRiaaigdaa8aabeaaaOqaa8qacaWGgbWdamaaBaaa % leaapeGaaGOmaiaadshaa8aabeaaaaaak8qacaGLBbGaayzxaaGaey % ypa0ZdamaadmaabaqbaeqabmWaaaqaamaaleaaleaacqGHsisldaqa % daqaaiaaigdacqGHRaWkcqaH7oaBdaWgaaadbaGaaGymaaqabaWccq % GHRaWkcqaH7oaBdaWgaaadbaGaaGymaaqabaWcdaahaaadbeqaaiaa % ikdaaaaaliaawIcacaGLPaaaaeaacaaIXaGaeyOeI0Iaeq4UdW2aaS % baaWqaaiaaigdaaeqaaSWaaWbaaWqabeaacaaIYaaaaaaaaOqaamaa % leaaleaacqGHsisldaqadaqaaiaaigdacqGHRaWkcqaH7oaBdaWgaa % adbaGaaGOmaaqabaWccqGHRaWkcqaH7oaBdaWgaaadbaGaaGOmaaqa % baWcdaahaaadbeqaaiaaikdaaaaaliaawIcacaGLPaaaaeaacaaIXa % GaeyOeI0Iaeq4UdW2aaSbaaWqaaiaaikdaaeqaaSWaaWbaaWqabeaa % caaIYaaaaaaaaOqaamaaleaaleaacqGHsisldaqadaqaaiaaigdacq % GHRaWkcqaH7oaBdaWgaaadbaGaaG4maaqabaWccqGHRaWkcqaH7oaB % daWgaaadbaGaaG4maaqabaWcdaahaaadbeqaaiaaikdaaaaaliaawI % cacaGLPaaaaeaacaaIXaGaeyOeI0Iaeq4UdW2aaSbaaWqaaiaaioda % aeqaaSWaaWbaaWqabeaacaaIYaaaaaaaaOqaamaaleaaleaacqaH7o % aBdaWgaaadbaGaaGymaaqabaWcdaahaaadbeqaaiaaikdaaaaaleaa % caaIXaGaey4kaSIaeq4UdW2aaSbaaWqaaiaaigdaaeqaaaaaaOqaam % aaleaaleaacqaH7oaBdaWgaaadbaGaaGOmaaqabaWcdaahaaadbeqa % aiaaikdaaaaaleaacaaIXaGaey4kaSIaeq4UdW2aaSbaaWqaaiaaik % daaeqaaaaaaOqaamaaleaaleaacqaH7oaBdaWgaaadbaGaaG4maaqa % baWcdaahaaadbeqaaiaaikdaaaaaleaacaaIXaGaey4kaSIaeq4UdW % 2aaSbaaWqaaiaaiodaaeqaaaaaaOqaaiaaigdaaeaacaaIXaaabaGa % aGymaaaaaiaawUfacaGLDbaadaWadaqaauaabeqadmaaaeaacqaH7o % aBdaWgaaWcbaGaaGymaaqabaGcdaahaaWcbeqaaiaadshaaaaakeaa % caaIWaaabaGaaGimaaqaaiaaicdaaeaacqaH7oaBdaWgaaWcbaGaaG % OmaaqabaGcdaahaaWcbeqaaiaadshaaaaakeaacaaIWaaabaGaaGim % aaqaaiaaicdaaeaacqaH7oaBdaWgaaWcbaGaaG4maaqabaGcdaahaa % WcbeqaaiaadshaaaaaaaGccaGLBbGaayzxaaWaamWaaeaafaqabeWa % daaabaWaaSqaaSqaaiabgkHiTmaabmaabaGaaGymaiabgUcaRiabeU % 7aSnaaBaaameaacaaIXaaabeaaliabgUcaRiabeU7aSnaaBaaameaa % caaIXaaabeaalmaaCaaameqabaGaaGOmaaaaaSGaayjkaiaawMcaaa % qaaiaaigdacqGHsislcqaH7oaBdaWgaaadbaGaaGymaaqabaWcdaah % aaadbeqaaiaaikdaaaaaaaGcbaWaaSqaaSqaaiabgkHiTmaabmaaba % GaaGymaiabgUcaRiabeU7aSnaaBaaameaacaaIYaaabeaaliabgUca % RiabeU7aSnaaBaaameaacaaIYaaabeaalmaaCaaameqabaGaaGOmaa % aaaSGaayjkaiaawMcaaaqaaiaaigdacqGHsislcqaH7oaBdaWgaaad % baGaaGOmaaqabaWcdaahaaadbeqaaiaaikdaaaaaaaGcbaWaaSqaaS % qaaiabgkHiTmaabmaabaGaaGymaiabgUcaRiabeU7aSnaaBaaameaa % caaIZaaabeaaliabgUcaRiabeU7aSnaaBaaameaacaaIZaaabeaalm % aaCaaameqabaGaaGOmaaaaaSGaayjkaiaawMcaaaqaaiaaigdacqGH % sislcqaH7oaBdaWgaaadbaGaaG4maaqabaWcdaahaaadbeqaaiaaik % daaaaaaaGcbaWaaSqaaSqaaiabeU7aSnaaBaaameaacaaIXaaabeaa % lmaaCaaameqabaGaaGOmaaaaaSqaaiaaigdacqGHRaWkcqaH7oaBda % WgaaadbaGaaGymaaqabaaaaaGcbaWaaSqaaSqaaiabeU7aSnaaBaaa % meaacaaIYaaabeaalmaaCaaameqabaGaaGOmaaaaaSqaaiaaigdacq % GHRaWkcqaH7oaBdaWgaaadbaGaaGOmaaqabaaaaaGcbaWaaSqaaSqa % aiabeU7aSnaaBaaameaacaaIZaaabeaalmaaCaaameqabaGaaGOmaa % aaaSqaaiaaigdacqGHRaWkcqaH7oaBdaWgaaadbaGaaG4maaqabaaa % aaGcbaGaaGymaaqaaiaaigdaaeaacaaIXaaaaaGaay5waiaaw2faam % aaCaaaleqabaGaeyOeI0IaaGymaaaak8qadaWadaWdaeaafaqabeWa % baaabaWdbiaadAeapaWaaSbaaSqaaiaaikdaaeqaaaGcbaWdbiaadA % eapaWaaSbaaSqaa8qacaaIXaaapaqabaaakeaapeGaamOra8aadaWg % aaWcbaGaaGimaaqabaaaaaGcpeGaay5waiaaw2faaaaa!F8D1! $$ At this point I let matlab do the work and find a few numbers to make sure this works well. $$\left[ {\begin{array}{*{20}{c}}{{F_{10}}}\\{{F_9}}\\{{F_8}}\end{array}} \right] = V{D^4}{V^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{28}\\{19}\\{13}\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakabaaaaaaaaapeqaamaadmaapaqaauaabe % qadeaaaeaapeGaamOra8aadaWgaaWcbaWdbiaaigdacaaIWaaapaqa % baaakeaapeGaamOra8aadaWgaaWcbaGaaGyoaaqabaaakeaapeGaam % Ora8aadaWgaaWcbaGaaGioaaqabaaaaaGcpeGaay5waiaaw2faaiab % g2da9iaadAfacaWGebWaaWbaaSqabeaacaaI0aaaaOGaamOvamaaCa % aaleqabaGaeyOeI0IaaGymaaaakiabg2da9maadmaapaqaauaabeqa % deaaaeaacaaIYaGaaGioaaqaaiaaigdacaaI5aaabaGaaGymaiaaio % daaaaapeGaay5waiaaw2faaaaa!486D! $$ $$\left[ {\begin{array}{*{20}{c}}{{F_{16}}}\\{{F_{15}}}\\{{F_{14}}}\end{array}} \right] = V{D^7}{V^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{277}\\{189}\\{129}\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakabaaaaaaaaapeqaamaadmaapaqaauaabe % qadeaaaeaapeGaamOra8aadaWgaaWcbaWdbiaaigdacaaI2aaapaqa % baaakeaapeGaamOra8aadaWgaaWcbaGaaGymaiaaiwdaaeqaaaGcba % WdbiaadAeapaWaaSbaaSqaaiaaigdacaaI0aaabeaaaaaak8qacaGL % BbGaayzxaaGaeyypa0JaamOvaiaadseadaahaaWcbeqaaiaaiEdaaa % GccaWGwbWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaeyypa0ZaamWa % a8aabaqbaeqabmqaaaqaaiaaikdacaaI3aGaaG4naaqaaiaaigdaca % aI4aGaaGyoaaqaaiaaigdacaaIYaGaaGyoaaaaa8qacaGLBbGaayzx % aaaaaa!4C28! $$ $$\left[ {\begin{array}{*{20}{c}}{{F_{26}}}\\{{F_{25}}}\\{{F_{24}}}\end{array}} \right] = V{D^{12}}{V^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{12,664}\\{8,641}\\{5,896}\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakabaaaaaaaaapeqaamaadmaapaqaauaabe % qadeaaaeaapeGaamOra8aadaWgaaWcbaWdbiaaikdacaaI2aaapaqa % baaakeaapeGaamOra8aadaWgaaWcbaGaaGOmaiaaiwdaaeqaaaGcba % WdbiaadAeapaWaaSbaaSqaaiaaikdacaaI0aaabeaaaaaak8qacaGL % BbGaayzxaaGaeyypa0JaamOvaiaadseadaahaaWcbeqaaiaaigdaca % aIYaaaaOGaamOvamaaCaaaleqabaGaeyOeI0IaaGymaaaakiabg2da % 9maadmaapaqaauaabeqadeaaaeaacaaIXaGaaGOmaiaacYcacaaI2a % GaaGOnaiaaisdaaeaacaaI4aGaaiilaiaaiAdacaaI0aGaaGymaaqa % aiaaiwdacaGGSaGaaGioaiaaiMdacaaI2aaaaaWdbiaawUfacaGLDb % aaaaa!51ED! $$ I am not entirely certain how this process would be completed in the event where the geometric multiplicity of the eigenvalues falls short and would appreciate a note on how, for example, the following proposition is arrived at: $$\begin{array}{l}{a_n} = 4{a_{n - 1}} - 4{a_{n - 2}} \Rightarrow {\lambda _{1,2}} = 2\\ \Rightarrow {a_n} = {c_1}{\lambda ^n} + {c_2}n{\lambda ^n}\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaGadeWadaaakqaabeaeaaaaaaaaa8qabaGaamyyamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaisdacaWGHbWaaSbaaSqaaiaa % d6gacqGHsislcaaIXaaabeaakiabgkHiTiaaisdacaWGHbWaaSbaaS % qaaiaad6gacqGHsislcaaIYaaabeaakiabgkDiElabeU7aSnaaBaaa % leaacaaIXaGaaiilaiaaikdaaeqaaOGaeyypa0JaaGOmaaqaaiabgk % DiElaadggadaWgaaWcbaGaamOBaaqabaGccqGH9aqpcaWGJbWaaSba % aSqaaiaaigdaaeqaaOGaeq4UdW2aaWbaaSqabeaacaWGUbaaaOGaey % 4kaSIaam4yamaaBaaaleaacaaIYaaabeaakiaad6gacqaH7oaBdaah % aaWcbeqaaiaad6gaaaaaaaa!57B1! $$
The general approach is to reduce the matrix to Jordan canonical form using generalized eigenvectors. Each Jordan block will give you terms in the final recurrence solution of the form you presented ($\lambda^n$ times a polynomial in $n$, whose order is one less than the eigenvalue multiplicity). I'll run through your example below, and hope that you can infer how this generalizes to higher order recurrences.
Using the more standard form where you encode the recurrence in the top line of the matrix and fill the rest with "shifts-by-one" (thanks Peter), the example you want to know more about is represented by the system $$ \left[ \begin{array}{c} a_{n} \\ a_{n-1} \end{array} \right] = \left[ \begin{array}{cc} 4 & -4 \\ 1 & 0\end{array} \right] \left[ \begin{array}{c} a_{n-1} \\ a_{n-2} \end{array} \right] $$
If that matrix is called $A$, it is clear that applying $A$ to the initial conditions $t$ times will get you the $a_t$ and $a_{t+1}$ terms of your recurrence. When the eigenvalues are unique, there is a nice closed form for the operation $A^t$ in terms of exponentiation of a diagonal matrix, which you worked out in your original example. However, when there are degeneracies, we have to find generalized eigenvectors.
The associated eigenvalue problem $Av=\lambda v$ boils down to the characteristic equation $$ (4-\lambda)(-\lambda)+4 = 0 $$ $$ \lambda^2-4\lambda+4 = 0 $$ And indeed there is a single root of this equation $$ \lambda=2 $$ with normalized eigenvector $$ v_{1}=\frac{1}{\sqrt{5}}\left[ \begin{array}{c} 2 \\ 1 \end{array} \right] $$
The fact that the eigenvalue was degenerate tells us that the matrix $A$ isn't diagonalizable, so we continue on to find a generalized eigenvector for $\lambda$, which solves $$ (A-\lambda I)v_2 = v_1 $$ $$ \left[ \begin{array}{cc} 2 & -4 \\ 1 & -2\end{array} \right] \left[ \begin{array}{c} v_{21} \\ v_{22} \end{array} \right] = \frac{1}{\sqrt{5}}\left[ \begin{array}{c} 2 \\ 1 \end{array} \right] $$ There's a lot of freedom in solving this, but a simple solution is $$ v_2 = \frac{1}{\sqrt{5}}\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] $$ Now we form $$ V = \left[ \begin{array}{c} v_1 & v_2 \end{array} \right]= \left[ \begin{array}{cc} 2/\sqrt{5} & 1/\sqrt{5}\\ 1/\sqrt{5} & 0 \end{array} \right] $$ Now our $D$ isn't really diagonal (it'll be in Jordan form), but let's form it anyway $$ D = V^{-1}AV = \left[ \begin{array}{cc} 2 & 1\\ 0 & 2 \end{array} \right] $$ Again, this isn't diagonal, but it's "close". In fact, this is a $2\times 2$ Jordan block. Powers of a diagonal matrix are really easy, but powers of Jordan blocks aren't that much harder. Notice that it has the eigenvalue on the diagonal, and a single value on the first upper diagonal: $$ D = V^{-1}AV = \lambda I + N $$ Where $N=\left[\begin{array}{c}0 & 1\\0&0\end{array}\right]$. Notice that $N^2=N^3=\ldots=0$. Together, this gives us a closed form for $D^t$ (which you can prove by induction): $$ D^t = \lambda^t I + t \lambda^{t-1} N $$ All that remains is to believe that $A^t=VD^tV^{-1}$ for Jordan matrices (it is), and change notation / absorb some constants into each other to get the form of the solution to this problem which you presented. For general (larger order) recurrences, you'll have a block Jordan matrix, where there will be Jordan blocks on the matrix diagonal. There will be one block per unique eigenvalue, and the size of the block is equal to the eigenvalue multiplicity. The power of a block diagonal matrix is the block matrix with each block raised to the same power. Also, blocks always have the form of $\lambda I + N$, but computing the powers of the block in closed form for larger block sizes is a slightly more tedious exercise since for a $m\times m$ block, the $N^m$ terms and higher powers vanish, but not the lower order powers of $N$.