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Challenges Fibonacci without consecutive digits

Python 3, 149 bytes g=lambda r:r>1and g(r-1)+g(r-2)or r lambda n:[i for(i)in map(g,range(2*n+2))if(lambda*l:all(abs(n-i)!=1for(n,i)in zip(l,l[1:])))(*map(int,str(i)))][n] Try it online! ...

posted 10mo ago by ok12d‭ · edited 10mo ago by ok12d‭

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#2: Post edited by

ok12d‭ · 2024-03-03T14:43:07Z (10 months ago)
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# [Python 3], 149 bytes
g=lambda r:r>1and g(r-1)+g(r-2)or r
lambda n:[i for(i)in map(g,range(2*n+2))if(lambda*l:all(abs(n-i)!=1for(n,i)in zip(l,l[1:])))(*map(int,str(i)))][n]
[Try it online!][TIO-ltbli504]
Based on @isaacg‭'s idea of mapping to twice the input, I had a similar thought, but I wasn't sure about the growth rate of the non-adjacent sequence...
Here's a mashed up version of my previous attempt:
~~while len(l:=[*filter(g,[*map(lambda i:[*map(int,str(i))],map(h,range(j:=j+1)))])])<n+1:0~~
print(l[:-1])
[Python 3]: https://docs.python.org/3/
[TIO-ltbli504]: https://tio.run/##NY9NboMwEIX3nGK6wpNMJGz@kdyLEBaEGIJKHGS7UtsoZ6eGUs9iPPJ7n@fN3@720PEyyKm9X64tmMq881ZfYWDmxPG4NoEPAybo5XnZVbqqR@gfho04ari3MxvItHpQTBz0USCOPfuTHqaqnSbWXizTpxHfJF9tmjbjzziziaaaVw0issMKGrUj61YyYlPrZnHKOit3OsdAfc2qc@oq64i4L0ExpVQQ9y0lniQU5znxtMxJpEVCCS@8LM85pySLs4LyNBJeWGb@hVKeCFF6WxlHcepZZVQUPCuboLup7kMZkMDC86fIky4k2G48DjHwMcCRgj3GtiXB/3JYBeCP9e6eOdyG2fhsrA@f7gUVPO0LnvsftZLSNi9PXX4B "Python 3 – Try It Online"

# [Python 3], 149 bytes
g=lambda r:r>1and g(r-1)+g(r-2)or r
lambda n:[i for(i)in map(g,range(2*n+2))if(lambda*l:all(abs(n-i)!=1for(n,i)in zip(l,l[1:])))(*map(int,str(i)))][n]
[Try it online!][TIO-ltbli504]
Based on @isaacg‭'s idea of mapping to twice the input, I had a similar thought, but I wasn't sure about the growth rate of the non-adjacent sequence...
Here's a mashed up version of my previous attempt:
j=n
while len(l:=[*filter(lambda*l:all(abs(n-i)!=1for(n,i)in zip(l,l[1:])),[*map(lambda i:[*map(int,str(i))],map(g,range(j:=j+1)))])])<n+1:0
print(l[:-1])
[Python 3]: https://docs.python.org/3/
[TIO-ltbli504]: https://tio.run/##NY9NboMwEIX3nGK6wpNMJGz@kdyLEBaEGIJKHGS7UtsoZ6eGUs9iPPJ7n@fN3@720PEyyKm9X64tmMq881ZfYWDmxPG4NoEPAybo5XnZVbqqR@gfho04ari3MxvItHpQTBz0USCOPfuTHqaqnSbWXizTpxHfJF9tmjbjzziziaaaVw0issMKGrUj61YyYlPrZnHKOit3OsdAfc2qc@oq64i4L0ExpVQQ9y0lniQU5znxtMxJpEVCCS@8LM85pySLs4LyNBJeWGb@hVKeCFF6WxlHcepZZVQUPCuboLup7kMZkMDC86fIky4k2G48DjHwMcCRgj3GtiXB/3JYBeCP9e6eOdyG2fhsrA@f7gUVPO0LnvsftZLSNi9PXX4B "Python 3 – Try It Online"

#1: Initial revision by

ok12d‭ · 2024-03-03T14:38:40Z (10 months ago)

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# [Python 3], 149 bytes

<!-- language-all: lang-python -->
    g=lambda r:r>1and g(r-1)+g(r-2)or r
    lambda n:[i for(i)in map(g,range(2*n+2))if(lambda*l:all(abs(n-i)!=1for(n,i)in zip(l,l[1:])))(*map(int,str(i)))][n]

[Try it online!][TIO-ltbli504]

Based on @isaacg‭'s idea of mapping to twice the input, I had a similar thought, but I wasn't sure about the growth rate of the non-adjacent sequence...

Here's a mashed up version of my previous attempt:

    while len(l:=[*filter(g,[*map(lambda i:[*map(int,str(i))],map(h,range(j:=j+1)))])])<n+1:0
    print(l[:-1])

[Python 3]: https://docs.python.org/3/
[TIO-ltbli504]: https://tio.run/##NY9NboMwEIX3nGK6wpNMJGz@kdyLEBaEGIJKHGS7UtsoZ6eGUs9iPPJ7n@fN3@720PEyyKm9X64tmMq881ZfYWDmxPG4NoEPAybo5XnZVbqqR@gfho04ari3MxvItHpQTBz0USCOPfuTHqaqnSbWXizTpxHfJF9tmjbjzziziaaaVw0issMKGrUj61YyYlPrZnHKOit3OsdAfc2qc@oq64i4L0ExpVQQ9y0lniQU5znxtMxJpEVCCS@8LM85pySLs4LyNBJeWGb@hVKeCFF6WxlHcepZZVQUPCuboLup7kMZkMDC86fIky4k2G48DjHwMcCRgj3GtiXB/3JYBeCP9e6eOdyG2fhsrA@f7gUVPO0LnvsftZLSNi9PXX4B "Python 3 – Try It Online"

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