A276039 Numbers using only digits 1 and 7.

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

Offset

1,2

Comments

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

Links

David A. Corneth, Table of n, a(n) for n = 1..10000

Example

7717 is in the sequence because 7*7*1*7 = 343 = 7^3.

Mathematica

Select[Range[20000], IntegerQ[Log[7, Times@@(IntegerDigits[#])]] &] (* or *) Flatten[Table[FromDigits/@Tuples[{1, 7}, n], {n, 6}]]

Prog

(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")))
print([a(n) for n in range(1, 47)]) # Michael S. Branicky, Mar 27 2021

Crossrefs

Cf. similar sequences listed in A276037.

Cf. A020455, A002275, A002281, A034054.

Sequence in context: A094104 A167513 A260891 * A199327 A260892 A020455

Adjacent sequences: A276036 A276037 A276038 * A276040 A276041 A276042

Keyword

nonn,easy,base

Author

Vincenzo Librandi, Aug 19 2016

Status

approved