A069353 Numbers of form 2^i*3^j - 1 with i, j >= 0.

0, 1, 2, 3, 5, 7, 8, 11, 15, 17, 23, 26, 31, 35, 47, 53, 63, 71, 80, 95, 107, 127, 143, 161, 191, 215, 242, 255, 287, 323, 383, 431, 485, 511, 575, 647, 728, 767, 863, 971, 1023, 1151, 1295, 1457, 1535, 1727, 1943, 2047, 2186, 2303, 2591, 2915, 3071, 3455, 3887

Offset: 1

Author

Reinhard Zumkeller, Mar 18 2002

Keywords

• nonn

Comments

Are there infinitely many primes in this sequence? See A005105.

If m is a term then also 2*m + 1 and 3*m + 2.

Links

• Graham Everest, Peter Rogers, and Thomas Ward, A higher-rank Mersenne problem, Algorithmic Number Theory: 5th International Symposium, ANTS-V Sydney, Australia, July 7-12, 2002 Proceedings 5, Lect. Notes Computer Sci. 2369, Springer Berlin Heidelberg, 2002, pp. 95-107.

Crossrefs

Cf. A003586, A055600, A069355, A005105.

Programs

Mathematica

With[{max = 4000}, Sort[Flatten[Table[2^i*3^j - 1, {i, 0, Log2[max]}, {j, 0, Log[3, max/2^i]}]]]] (* Amiram Eldar, Jul 13 2023 *)

Python

from sympy import integer_log
def A069353(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum(((x+1)//3**i).bit_length() for i in range(integer_log(x+1,3)[0]+1))
    return bisection(f,n-1,n-1) # Chai Wah Wu, Mar 31 2025

Formula

a(n) = A003586(n)-1.