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Cyclops numbers with circular digits {0,6,8,9}.
2

%I #22 Sep 24 2024 03:14:27

%S 0,606,608,609,806,808,809,906,908,909,66066,66068,66069,66086,66088,

%T 66089,66096,66098,66099,68066,68068,68069,68086,68088,68089,68096,

%U 68098,68099,69066,69068,69069,69086,69088,69089,69096,69098,69099,86066,86068,86069,86086,86088,86089,86096,86098,86099

%N Cyclops numbers with circular digits {0,6,8,9}.

%C Intersection of A001743 and A134808.

%H Kenny Lau, <a href="/A274765/b274765.txt">Table of n, a(n) for n = 1..20000</a>

%e 86069 is a member because it is cyclops (A134808) and each digit contains at least one loop (A001743).

%t cncdQ[n_]:=Module[{idn=IntegerDigits[n]},OddQ[Length[idn]]&&Count[idn,0] == 1&&idn[[(Length[idn]+1)/2]]==0&&SubsetQ[{0,6,8,9},idn]]; Select[ Range[ 0,90000],cncdQ] (* _Harvey P. Dale_, Jan 06 2022 *)

%o (Python)

%o import sys

%o f = open('b274765.txt', 'w')

%o i = 1

%o n = 0

%o a = [""]

%o while True:

%o for x in a:

%o for y in a:

%o f.write(str(i)+" "+x+"0"+y+"\n")

%o i += 1

%o if i>20000:

%o f.close()

%o sys.exit()

%o a = sum([[x+"6", x+"8", x+"9"] for x in a], [])

%o # _Kenny Lau_, Jul 05 2016

%o (PARI) is_a001633(n) = #Str(n)%2==1

%o is_a001743(n) = #setintersect([1, 2, 3, 4, 5, 7], Set(digits(n)))==0

%o is_a134808(n) = if(n==0, return(1), if(n < 10, return(0), my(d=digits(n), x=1, y=#d); while(x < #d, if(d[x]==0, break); x++); while(y > 1, if(d[y]==0, break); y--); if(x==y && x==ceil(#Str(n)/2), return(1), return(0))))

%o is(n) = is_a001633(n) && is_a001743(n) && is_a134808(n) \\ _Felix Fröhlich_, Jul 05 2016

%Y Cf. A001633, A001742, A134808, A160561.

%K nonn,base,easy

%O 1,2

%A _Kenny Lau_, Jul 05 2016