Write $ n=2^b + c$ where $ 0\leq c < 2^b$ We get, following the definition of $ a_n$ that: $ a_{2^b+j}=2^b+j+b$ (for $ 0\leq j \leq 2^b-b-1$) $ a_{2^b+j}=2^b+j+b+1$ (for $ 2^b-b \leq j \leq 2^b-1$) (This is valid for $ b>0$) $ a_{2^{b-1}+j}=2^{b-1}+j+b$ (for $ 2^{b-1}-(b-1) \leq j \leq 2^{b-1}-1$) The image of this set is all integers $ d$ that satisfy: $ 2^b+1 \leq d \leq 2^b +b-1$ The image of the set $ a_{2^b+j}=2^b+j+b$ (for $ 0\leq j \leq 2^b-b-1$) is all integers $ d$ satisfying: $ 2^b+b \leq d \leq2^{b+1}-1$ Thus the union of these disjoint sets is all numbers which are not a power of two (and the number $ 1$, for $ b=0$) Then the answer of the problem is all numbers of the form $ 2^x$ with $ x$ a natural number (Remark: there are some details to be given but I think the main idea is ok and the rest is easy to understand)

Do sombody know about the othe 2 problems? That solution seem to be the same as mine, but ther is no proof of the anoying parts I hope have time to post some of them But here is time to sleep

Can you post the rest of the problems? Thank you

Here are the links: Prob 1 http://www.mathlinks.ro/Forum/viewtopic.php?t=302378 Prob 3 http://www.mathlinks.ro/Forum/viewtopic.php?p=1636964

Yeah I've got some time now As you wish I'm gonna writ the solution down here. So we must define $ f(n)=n\times 2^n$ and $ r(m)=2^{2^m}$, where $ f(n)$ is obviously crescent in positives. Lemma 1: $ a< 2^a$ for $ a\geq 0$ (integer) Proof: Case $ a=0$ we get that $ 0<1$ witch is obviously true. So in the inductive step we have $ 2^{a+1}=2^a+2^a<a+a\leq a+1$ as we want. So by the Lemma we know for: $ m-1\leq 2^{m-1}\Leftrightarrow (m-1) \times 2^{2^m-(m-1)}\leq 2^{2^m}\Leftrightarrow$ $ \Leftrightarrow m\times 2^{2^m-(m-1)}+2^{2^m}\leq 2^{2^m}+2^{2^m}+2^{2^m-(m-1)}\Leftrightarrow$ $ \Leftrightarrow 2^{2^m}\leq 2^{2^m+1}+ m\times 2^{2^m-(m-1)}+2^{2^m-(m-1)}=(2^m-(m-1))\times 2^{2^m-(m-1)}\Leftrightarrow r(m)\leq f(2^m-(m-1))$ (1) And we can notice that for $ m>0$: $ m\times 2^{2^m-m}>0\Leftrightarrow 2^{2^m}-m\times 2^{2^m-m}<2^{2^m}\Leftrightarrow (2^m-m)\times 2^{2^m-m}<2^{2^m}\Leftrightarrow f(2^m-m)<r(m)$ (2) Define the set $ A_{m}=\{a_n| a_n-n=m\}$ so $ a_n\in A_{m}$ iif $ r(m)\leq f(n)<r(m+1)$, but for $ n=2^m-m, 2^m-(m-1), 2^{m+1}-(m+1), 2^{m+1}-m$ we get $ f(n)<r(m)$, $ r(m)\leq f(n)$, $ f(n)< r(m+1)$ and $ r(m+1)<f(n)$ by inequality (1), (2), (1), (2) respectively. ecause f is a inceassing function in positives the set $ A_m=\{a_{2^m-(m-1)}, \cdots, a_{2^{m+1}-(m+1)}\}=\{2^m+1, \cdots, 2^{m+1}-1\}$ exept for the case that $ m=0$. So we get that Only the powers of 2 are not in the sets $ A_i$, $ i>0$ so they must be in the set $ A_0$ or they don't exist in the succession. But $ A_0=\{a_1\}=\{1\}$ because $ f(2)=8>2=r(0)$. So the positive powers of 2 are the ones that doesn't occur in the succession. Yeah is this correct?

Raúl wrote: Yeah I've got some time now As you wish I'm gonna writ the solution down here. So we must define $ f(n) = n\times 2^n$ and $ r(m) = 2^{2^m}$, where $ f(n)$ is obviously crescent in positives. Lemma 1: $ a < 2^a$ for $ a\geq 0$ (integer) Proof: Case $ a = 0$ we get that $ 0 < 1$ witch is obviously true. So in the inductive step we have $ 2^{a + 1} = 2^a + 2^a < a + a\leq a + 1$ as we want. So by the Lemma we know for: $ m - 1\leq 2^{m - 1}\Leftrightarrow (m - 1) \times 2^{2^m - (m - 1)}\leq 2^{2^m}\Leftrightarrow$ $ \Leftrightarrow m\times 2^{2^m - (m - 1)} + 2^{2^m}\leq 2^{2^m} + 2^{2^m} + 2^{2^m - (m - 1)}\Leftrightarrow$ $ \Leftrightarrow 2^{2^m}\leq 2^{2^m + 1} + m\times 2^{2^m - (m - 1)} + 2^{2^m - (m - 1)} = (2^m - (m - 1))\times 2^{2^m - (m - 1)}\Leftrightarrow r(m)\leq f(2^m - (m - 1))$ (1) And we can notice that for $ m > 0$: $ m\times 2^{2^m - m} > 0\Leftrightarrow 2^{2^m} - m\times 2^{2^m - m} < 2^{2^m}\Leftrightarrow (2^m - m)\times 2^{2^m - m} < 2^{2^m}\Leftrightarrow f(2^m - m) < r(m)$ (2) Define the set $ A_{m} = \{a_n| a_n - n = m\}$ so $ a_n\in A_{m}$ iif $ r(m)\leq f(n) < r(m + 1)$, but for $ n = 2^m - m, 2^m - (m - 1), 2^{m + 1} - (m + 1), 2^{m + 1} - m$ we get $ f(n) < r(m)$, $ r(m)\leq f(n)$, $ f(n) < r(m + 1)$ and $ r(m + 1) < f(n)$ by inequality (1), (2), (1), (2) respectively. ecause f is a inceassing function in positives the set $ A_m = \{a_{2^m - (m - 1)}, \cdots, a_{2^{m + 1} - (m + 1)}\} = \{2^m + 1, \cdots, 2^{m + 1} - 1\}$ exept for the case that $ m = 0$. So we get that Only the powers of 2 are not in the sets $ A_i$, $ i > 0$ so they must be in the set $ A_0$ or they don't exist in the succession. But $ A_0 = \{a_1\} = \{1\}$ because $ f(2) = 8 > 2 = r(0)$. So the positive powers of 2 are the ones that doesn't occur in the succession. Yeah is this correct? Hello, I saw an error in your lemma's prove: you want to prove that $ 2^a>a, \ a \ge 0$ (integer), but in the prove you wrote $ 2^{a + 1} = 2^a + 2^a < a + a\leq a + 1$.

I'm very glad to see that sombody read my solution Raúl wrote: Lemma 1: $ a < 2^a$ for $ a\geq 0$ (integer) Proof: Case $ a = 0$ we get that $ 0 < 1$ witch is obviously true. So in the inductive step we have $ 2^{a + 1} = 2^a + 2^a < a + a\leq a + 1$ as we want. So the right Lemma is this: $ a < 2^a$ for $ a\geq 0$ Proof: Case $ a = 0$ and $ a = 1$ we get that $ 0 < 1$ and $ 1 < 2$ witch is obviously true. So in the inductive step we have $ 2^{a + 1} = 2^a + 2^a > a + a\geq a + 1$ as we want. I notice some other typos: I want to meam Raúl wrote: So by the Lemma we know for: $ 2^{m - 1}\Leftrightarrow (m - 1) \times 2^{2^m - (m - 1)}\leq 2^{2^m}\Leftrightarrow m\times 2^{2^m - (m - 1)} + 2^{2^m}\leq 2^{2^m} + 2^{2^m} + 2^{2^m - (m - 1)}\Leftrightarrow 2^{2^m}\leq 2^{2^m + 1} - m\times 2^{2^m - (m - 1)} + 2^{2^m - (m - 1)} = (2^m - (m - 1))\times 2^{2^m - (m - 1)}\Leftrightarrow r(m)\leq f(2^m - (m - 1))$ (1) Thanks a lot

Raúl wrote: I'm very glad to see that sombody read my solution Raúl wrote: Lemma 1: $ a < 2^a$ for $ a\geq 0$ (integer) Proof: Case $ a = 0$ we get that $ 0 < 1$ witch is obviously true. So in the inductive step we have $ 2^{a + 1} = 2^a + 2^a < a + a\leq a + 1$ as we want. So the right Lemma is this: $ a < 2^a$ for $ a\geq 0$ Proof: Case $ a = 0$ and $ a = 1$ we get that $ 0 < 1$ and $ 1 < 2$ witch is obviously true. So in the inductive step we have $ 2^{a + 1} = 2^a + 2^a > a + a\geq a + 1$ as we want. I notice some other typos: I want to meam Raúl wrote: So by the Lemma we know for: $ 2^{m - 1}\Leftrightarrow (m - 1) \times 2^{2^m - (m - 1)}\leq 2^{2^m}\Leftrightarrow m\times 2^{2^m - (m - 1)} + 2^{2^m}\leq 2^{2^m} + 2^{2^m} + 2^{2^m - (m - 1)}\Leftrightarrow 2^{2^m}\leq 2^{2^m + 1} - m\times 2^{2^m - (m - 1)} + 2^{2^m - (m - 1)} = (2^m - (m - 1))\times 2^{2^m - (m - 1)}\Leftrightarrow r(m)\leq f(2^m - (m - 1))$ (1) Thanks a lot Your solution is very nice, yeah... beacuse it's ingenious