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A276039
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Numbers using only digits 1 and 7.
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12
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1, 7, 11, 17, 71, 77, 111, 117, 171, 177, 711, 717, 771, 777, 1111, 1117, 1171, 1177, 1711, 1717, 1771, 1777, 7111, 7117, 7171, 7177, 7711, 7717, 7771, 7777, 11111, 11117, 11171, 11177, 11711, 11717, 11771, 11777, 17111, 17117, 17171, 17177, 17711, 17717, 17771, 17777
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OFFSET
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1,2
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COMMENTS
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Numbers k such that the product of digits of k is a power of 7.
There are no prime terms whose number of digits is divisible by 3: for every d that is a multiple of 3, every d-digit number j consisting of no digits other than 1's and 7's will have a digit sum divisible by 3, so j will also be divisible by 3. - Mikk Heidemaa, Mar 27 2021
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LINKS
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EXAMPLE
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7717 is in the sequence because 7*7*1*7 = 343 = 7^3.
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MATHEMATICA
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Select[Range[20000], IntegerQ[Log[7, Times@@(IntegerDigits[#])]] &] (* or *) Flatten[Table[FromDigits/@Tuples[{1, 7}, n], {n, 6}]]
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PROG
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(Magma) [n: n in [1..24000] | Set(Intseq(n)) subset {1, 7}];
(PARI) is(n) = my(d=digits(n), e=[0, 2, 3, 4, 5, 6, 8, 9]); if(#setintersect(Set(d), Set(e))==0, return(1), return(0)) \\ Felix Fröhlich, Aug 19 2016
(PARI) a(n) = { my(b = binary(n + 1)); b = b[^1]; b = apply(x -> 6*x + 1, b); fromdigits(b) } \\ David A. Corneth, Mar 27 2021
(Python)
def a(n):
b = bin(n+1)[3:]
return int("".join(b.replace("1", "7").replace("0", "1")))
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CROSSREFS
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Cf. similar sequences listed in A276037.
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KEYWORD
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nonn,easy,base
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AUTHOR
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STATUS
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approved
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