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

Author

Vincenzo Librandi, Aug 19 2016

Keywords

• nonn
• easy
• base

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

Examples

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

Links

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

Crossrefs

Cf. similar sequences listed in A276037.

Cf. A020455, A002275, A002281, A034054.

Programs

Magma

[n: n in [1..24000] | Set(Intseq(n)) subset {1,7}];

Mathematica

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

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

Python

def A276039(n): return 6*int(bin(n+1)[3:])+(10**((n+1).bit_length()-1)-1)//9 # Chai Wah Wu, Jun 28 2025