A046413 Numbers k such that the repunit of length k (11...11, with k 1's) has exactly 2 prime factors.
3, 4, 5, 7, 11, 17, 47, 59, 71, 139, 211, 251, 311, 347, 457, 461
Offset: 1
Examples
7 is a term because 1111111 = 239*4649.
References
- Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
Links
- Patrick De Geest, Repunits prime factors.
- Shyam Sunder Gupta, Repunit Numbers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 11, 327-352.
- Makoto Kamada, Factorizations of 11...11 (Repunit).
- Yousuke Kiode, Factorizations of Repunit Numbers.
- Eric Weisstein's World of Mathematics, Repunit.
- Status of (10^509-1)/9 in factordb.com.
- Status of (10^557-1)/9 in factordb.com.
- Status of (10^647-1)/9 in factordb.com.
Programs
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Mathematica
Select[Range[60],PrimeOmega[FromDigits[PadRight[{},#,1]]]==2&] (* The program generates the first 8 terms of the sequence. *) (* Harvey P. Dale, Aug 26 2024 *)
Extensions
More terms from Rick L. Shepherd, Mar 11 2003
a(13)-a(16) from Robert G. Wilson v, Apr 26 2010
Comments