This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
SequencePosition[Table[If[SequenceCount[IntegerDigits[n,8],{1,3}]>0 || SequenceCount[ IntegerDigits[n,8],{3,1}]>0,1,0],{n,1600}],{0,1}][[All,2]] (* Harvey P. Dale, Nov 11 2022 *)
def has13or31(n): o = oct(n); return "13" in o or "31" in o def ok(n): return has13or31(n) and not has13or31(n-1) print([k for k in range(1600) if ok(k)]) # Michael S. Branicky, Nov 25 2021
SequencePosition[Table[If[SequenceCount[IntegerDigits[n,8],{1,3}]>0 || SequenceCount[IntegerDigits[n,8],{3,1}]>0,1,0],{n,1400}],{1,0}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Sep 07 2020 *)
def has13or31(n): o = oct(n); return "13" in o or "31" in o def ok(n): return has13or31(n) and not has13or31(n+1) print([k for k in range(1600) if ok(k)]) # Michael S. Branicky, Nov 25 2021
'42' might be read as an octal number, but '48' could not be and so belongs to the sequence.
Select[Range[200],Max[IntegerDigits[#]]>7&] (* Harvey P. Dale, May 26 2018 *)
isok(n) = vecmax(digits(n)) > 7; \\ Michel Marcus, Feb 18 2016
def ok(n): return {'8', '9'} & set(str(n)) != set()
print(list(filter(ok, range(189)))) # Michael S. Branicky, Aug 09 2021
For n=16, we see that a(15) = 8, '10' in octal, and '0' has occurred just twice in the octal representations of terms a(0) .. a(15), namely in a(0) = 0 (which is also zero when read in octal base) and a(15), thus a(16) = 2.
6 is a term as 6_10 = 6_7 = 6_8 = 6_9, so it has three other base representations where the digit product also equals 6. 12563124891 is a term as 12563124891_10 = 5434343211123_6 = 623216541162_7 = 135464411233_8, so it has three other base representations where the digit product also equals 103680. 115233863842 is a term as 115233863842_10 = 124534313342214_6 = 11216413452466_7 = 1532436234242_8, so it has three other base representations where the digit product also equals 829440.
110 is in the sequence because 110^2 = 12100 = 27504_8 and no octal digit occurs more than once.
Select[Range[2^12], DuplicateFreeQ[IntegerDigits[#^2, 8]] &] (* Michael De Vlieger, Oct 12 2024 *)
for k in range(1, 2**12):
octal = format(k**2, "o")
if len(octal) == len(set(octal)): print(k, end=",")
[Seqint(Intseq(n^2, 8)): n in [1..60]]; // Vincenzo Librandi, Oct 16 2015
Table[BaseForm[n^2,8],{n,5!}] (* Vladimir Joseph Stephan Orlovsky, May 12 2010 *)
Table[FromDigits[IntegerDigits[n^2, 8]], {n, 60}] (* Vincenzo Librandi, Oct 15 2015 *)
a(n)=fromdigits(digits(n^2,8)) \\ Charles R Greathouse IV, Apr 25 2016
[Seqint(Intseq(n^3, 8)): n in [1..40]]; // Vincenzo Librandi, Oct 15 2015
Table[FromDigits[IntegerDigits[n^3, 8]], {n, 40}] (* Vincenzo Librandi, Oct 15 2015 *)
for(n=1,40, print1(fromdigits(digits(n^3, 8)), ", ")) \\ G. C. Greubel, Sep 10 2018
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