A274765 Cyclops numbers with circular digits {0,6,8,9}.

0, 606, 608, 609, 806, 808, 809, 906, 908, 909, 66066, 66068, 66069, 66086, 66088, 66089, 66096, 66098, 66099, 68066, 68068, 68069, 68086, 68088, 68089, 68096, 68098, 68099, 69066, 69068, 69069, 69086, 69088, 69089, 69096, 69098, 69099, 86066, 86068, 86069, 86086, 86088, 86089, 86096, 86098, 86099

Offset

1,2

Comments

Intersection of A001743 and A134808.

Links

Kenny Lau, Table of n, a(n) for n = 1..20000

Example

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

Mathematica

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 *)

Prog

(Python)
import sys
f = open('b274765.txt', 'w')
i = 1
n = 0
a = [""]
while True:
    for x in a:
        for y in a:
            f.write(str(i)+" "+x+"0"+y+"\n")
            i += 1
            if i>20000:
                f.close()
                sys.exit()
    a = sum([[x+"6", x+"8", x+"9"] for x in a], [])
# Kenny Lau, Jul 05 2016
(PARI) is_a001633(n) = #Str(n)%2==1
is_a001743(n) = #setintersect([1, 2, 3, 4, 5, 7], Set(digits(n)))==0
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))))
is(n) = is_a001633(n) && is_a001743(n) && is_a134808(n) \\ Felix Fröhlich, Jul 05 2016

Crossrefs

Cf. A001633, A001742, A134808, A160561.

Sequence in context: A250969 A265244 A228036 * A249524 A064250 A186818

Adjacent sequences: A274762 A274763 A274764 * A274766 A274767 A274768

Keyword

nonn,base,easy,changed

Author

Kenny Lau, Jul 05 2016

Status

approved