18:I[96096,["/_next/static/chunks/c6db19d9c5fa184a.js","/_next/static/chunks/bfe171a9b29c6141.js","/_next/static/chunks/215a462e6e3d5704.js","/_next/static/chunks/8e5a4ef2cd34cd2e.js","/_next/static/chunks/7f58f759df62ad2a.js","/_next/static/chunks/3d80c91f0baaa7e1.js","/_next/static/chunks/69017dabd5a7bdb1.js","/_next/static/chunks/16e35d7ca69704d5.js","/_next/static/chunks/1e99a352826083c0.js","/_next/static/chunks/d500812f89b1de2c.js","/_next/static/chunks/276dde09bcd6a58a.js","/_next/static/chunks/05f60f5d8542f370.js","/_next/static/chunks/a3977a4b8be0f95f.js"],"Recent"] 1a:I[93966,["/_next/static/chunks/c6db19d9c5fa184a.js","/_next/static/chunks/bfe171a9b29c6141.js","/_next/static/chunks/215a462e6e3d5704.js","/_next/static/chunks/8e5a4ef2cd34cd2e.js","/_next/static/chunks/7f58f759df62ad2a.js","/_next/static/chunks/3d80c91f0baaa7e1.js","/_next/static/chunks/69017dabd5a7bdb1.js","/_next/static/chunks/16e35d7ca69704d5.js","/_next/static/chunks/1e99a352826083c0.js","/_next/static/chunks/d500812f89b1de2c.js","/_next/static/chunks/276dde09bcd6a58a.js","/_next/static/chunks/05f60f5d8542f370.js","/_next/static/chunks/a3977a4b8be0f95f.js"],"default"] 1b:I[57442,["/_next/static/chunks/c6db19d9c5fa184a.js","/_next/static/chunks/bfe171a9b29c6141.js","/_next/static/chunks/215a462e6e3d5704.js","/_next/static/chunks/8e5a4ef2cd34cd2e.js","/_next/static/chunks/7f58f759df62ad2a.js","/_next/static/chunks/3d80c91f0baaa7e1.js","/_next/static/chunks/69017dabd5a7bdb1.js","/_next/static/chunks/16e35d7ca69704d5.js","/_next/static/chunks/1e99a352826083c0.js","/_next/static/chunks/d500812f89b1de2c.js","/_next/static/chunks/276dde09bcd6a58a.js","/_next/static/chunks/05f60f5d8542f370.js","/_next/static/chunks/a3977a4b8be0f95f.js"],"default"] 1c:I[56378,["/_next/static/chunks/c6db19d9c5fa184a.js","/_next/static/chunks/bfe171a9b29c6141.js","/_next/static/chunks/215a462e6e3d5704.js","/_next/static/chunks/8e5a4ef2cd34cd2e.js","/_next/static/chunks/7f58f759df62ad2a.js","/_next/static/chunks/3d80c91f0baaa7e1.js","/_next/static/chunks/69017dabd5a7bdb1.js","/_next/static/chunks/16e35d7ca69704d5.js","/_next/static/chunks/1e99a352826083c0.js","/_next/static/chunks/d500812f89b1de2c.js","/_next/static/chunks/276dde09bcd6a58a.js","/_next/static/chunks/05f60f5d8542f370.js","/_next/static/chunks/a3977a4b8be0f95f.js"],"default"] 1f:I[69035,["/_next/static/chunks/c6db19d9c5fa184a.js","/_next/static/chunks/bfe171a9b29c6141.js","/_next/static/chunks/215a462e6e3d5704.js","/_next/static/chunks/8e5a4ef2cd34cd2e.js","/_next/static/chunks/7f58f759df62ad2a.js","/_next/static/chunks/3d80c91f0baaa7e1.js","/_next/static/chunks/69017dabd5a7bdb1.js","/_next/static/chunks/16e35d7ca69704d5.js","/_next/static/chunks/1e99a352826083c0.js","/_next/static/chunks/d500812f89b1de2c.js","/_next/static/chunks/276dde09bcd6a58a.js","/_next/static/chunks/05f60f5d8542f370.js","/_next/static/chunks/a3977a4b8be0f95f.js"],"Separator"] 19:T49e,Sea $\phi=\frac{1+\sqrt{5}}{2}$, de modo que tenemos las identidades $\phi+1=\phi^2$ y $\phi-1=\frac{1}{\phi}$. Entonces $f(n)=\lfloor\phi n\rfloor$. Probamos esto por inducción fuerte. El caso base está dado. Supongamos entonces que $f(n)=\lfloor\phi n\rfloor$ para todo $n\leq k$, de modo que $f(f(k)-k+1)=f(\lfloor\phi k\rfloor -k+1)=\lfloor\phi(\lfloor\phi k\rfloor-k+1)\rfloor$. Es fácil probar que este valor es o bien $k$ o bien $k+1$. Para establecer la hipótesis inductiva, si $f(f(k)-k+1)=k$ (de modo que $f(k+1)=\lfloor\phi k\rfloor+2$) entonces $k<\phi(\lfloor\phi k\rfloor-k+1)

Olimpiada Nacional de Irán 2000 Problema 3

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