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User: Boas Bakker

Boas Bakker has authored 2 sequences.

A383397 Numbers in whose canonical prime factorization the powers of the primes form a strictly increasing sequence.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 46, 47, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 97, 98, 99, 100, 101

Offset: 1

Boas Bakker, Apr 26 2025

Alternative name: positive integers with canonical prime factorization p_1 ^ e_1 * p_2 ^ e_2 * ... * p_k ^ e_k which satisfy p_1 ^ e_1 < p_2 ^ e_2 < ... < p_k ^ e_k.

The asymptotic density of this sequence seems to be about 0.84.

			18 = 2^1 * 3^2 is in the sequence as 2^1 < 3^2.
12 is not in the sequence because 12 = 2^2 * 3^1 and 4>3.

Boas Bakker, Table of n, a(n) for n = 1..100000

Complement of A140831.

Cf. A005117.

Select[Range[100], Less @@ Power @@@ FactorInteger[#] &] (* Amiram Eldar, Apr 26 2025 *)

is(n) = {my(f = factor(n), r = 0); for(i = 1, #f~, c = f[i,1]^f[i,2]; if(c > r, r = c, return(0))); 1} \\ David A. Corneth, Apr 26 2025

A383038 Positive integers which can be expressed in the form b^r - b^s where b, r, and s are positive integers.

Original entry on oeis.org

2, 4, 6, 8, 12, 14, 16, 18, 20, 24, 28, 30, 32, 42, 48, 54, 56, 60, 62, 64, 72, 78, 90, 96, 100, 110, 112, 120, 124, 126, 128, 132, 156, 162, 180, 182, 192, 210, 216, 224, 234, 240, 248, 252, 254, 256, 272, 294, 306, 336, 342, 380, 384, 420, 448, 462, 480, 486, 496, 500

Offset: 1

Boas Bakker, Apr 13 2025

nonn

Terms are all even because b^r - b^s == b - b == 0 (mod 2).
From David A. Corneth, Apr 26 2025: (Start)
By definition, b, s, r and b^r - b^s is positive. Therefore, r > s. If b = 1 then b^r - b^s = 1^r - 1^s = 0 which is excluded by default so b > 1. Suppose we look for terms <= U. Then b is maximal when (s, r) = (1, 2) i.e. b^2 - b <= U. Solving for b gives (sqrt(4*U + 1) + 1) / 2.
As r > s >= 1, r >= 2.
Once b is fixed, b^r - b^s is minimal when s = r-1, enabling the largest r.
So we'd have b^r - b^(r-1) = (b - 1)*(b^(r-1)) <= U. Solving for r gives r <= log(U / (b - 1)) / log(b) + 1.
Once b and r are fixed we have
b^r - b^s <= U so b^r - U <= b^s. Solving for s gives log(b^r - U) / log(b) <= s.
Summarizing we have
2 <= b <= (sqrt(4*U + 1) + 1) / 2,
2 <= r <= log(U / (b - 1)) / log(b) + 1,
log(b^r - U) / log(b) <= s <= r-1. (End)

			a(6) = 14 = 16 - 2 = 2^4 - 2^1.
a(9) = 20 = 25 - 5 = 5^2 - 5^1.
As _David A. Corneth_ said, we know r > s > 0 and b > 1, so r > 1, and the smallest value of b is 2. So b^r - b^s >= b^(r-1) >= 2^(r-1). So to prove 10 is not in the sequence, we only need to check up up to r=4, because for r=5, 2^(s-1) = 2^4 = 16 > 10. This means there are 6 combinations of (r, s) we need to check. We also know b^r - b^s == 0 (mod b) because s > 0, so we only need to check divisors of 10. So with b = 2 and s < r < 5, we get the terms {2, 4, 6, 8, 12, 14}. With b=5 the smallest term is 5^2 - 5^1 = 20, which is bigger than 10. For b>5, b^r - b^s >= b^2 - b > 5^2 - 5, so we don't need to check those values of b.

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

The terms of the case b=2 are in A043569.

function A383038List(limit::Int)
    res = Set{Int}()
    for b in 2:floor(Int, sqrt(limit)) + 2
        for r in 2:limit
            valr = b^(r-1) * (b-1)
            valr > limit && break
            br = b^r
            for s in 1:r-1
                val = br - b^s
                val > 0 && val <= limit && push!(res, val)
    end end end
    sort(collect(res)) end
println(A383038List(500))  # Peter Luschny, Aug 17 2025

upto(n) = {n++; my(res = List());
    maxb = ceil((sqrtint(4*n + 1) + 1) / 2);
    for(b = 2, maxb,
        maxr = logint(n\(b-1), b) + 1;
        for(r = 2, maxr,
            mins = max(1, ceil(log(max(b^r - n, 1)) / log(b)));
            mins = min(mins, r-1); \\ min() to fix messed up rounding when b^r - n is a power of b. Also solved by increasing n by 1 initially (n++).
            forstep(s = r-1, mins, -1,
                c = b^r - b^s;
                listput(res, c);
            );
        );
    );
    Set(res)
} \\ David A. Corneth, Apr 26 2025

from sympy import integer_nthroot
aupto = 500
b_max, A383038 = (i := integer_nthroot(aupto,2))[0] + 2 - i[1], set()
for b in range(2, b_max):
    r, s = 2, 1
    while (br:=b**r) - b**s <= aupto:
        while s > 0 and (res := br - b**s) <= aupto: A383038.add(res); s -= 1
        r += 1
        s = r - 1
print(sorted(A383038)) # Karl-Heinz Hofmann, Apr 29 2025

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User: Boas Bakker

A383397 Numbers in whose canonical prime factorization the powers of the primes form a strictly increasing sequence.

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