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I would appreciate if somebody could help me with the following problem:

Q: How to define this in Mathematica.

RecurrenceTable[{a[2 n] == a[n] + 1, a[2 n + 1] == a[n] + 3, a[1] == 1},a,{n,1,10}]
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  • $\begingroup$ You should be getting an error which tells you what the problem is: RecurrenceTable::overdet: There are fewer dependent variables than equations, so the system is overdetermined. $\endgroup$
    – flinty
    Jan 4, 2022 at 9:56

1 Answer 1

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RSolve and RecurrenceTable can't seem to handle this.

You can at least get values from the sequence like this though:

sequence = Module[{a,n=1,seq = CreateDataStructure["LinkedList"]},
   seq["Append", 1];
   While[n < 100,
    a = seq["Part", n];
    seq["Append", a + 1];
    seq["Append", a + 3];
    ++n;]; Normal@seq
   ];

ListPlot[sequence, Filling -> Axis, AxesLabel -> {"n", "a[n]"}]

enter image description here

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