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The offset is 1 since A049802(k) = 0 for infinitely many values (when k = 2^r, r >= 0).

Every m > 0 in A049802 has a finite multiplicity, since, except for n = 2, the range of numbers for which A049802(k) = s is bounded above by 2^s+1 (see Englezou link).

The tuple of summands (x_1, ..., x_t) for m in A049802 can also be seen as a finite subset of an infinite tuple which is the representation of m as a profinite integer isomorphic to the normalized 2-adic series of m. This is because m is an element of the inverse limit of the finite rings Z/(2^i)Z, which is a profinite group isomorphic to the ring of 2-adic integers. In the infinite tuple (x_1, x_2, ...), x_i = m for every i such that m < 2^i. For example, for m = 29, we have the tuple (1, 1, 5, 13, 29, 29, 29, ...). See the Wikipedia link for more information.

From a combinatorial perspective, the tuple of summands (x_1, ..., x_t) mentioned above can be seen as a set of t counters, where the j-th counter cycles through 0 to 2^j-1. The natural question 'which m in A049802 appear k times?' becomes a question about how this cycling condition restricts the number of tuples which sum to m. For example, for n <= 100, when n = 1, 3, 5, 9, 15, 23, 35, 63, 65, and 67 there is only one m such that the tuple of summands sums to n (a trivial tuple consisting of n 1s, trivial because there is such a tuple for every n >= 1, i.e. for every m = 2^n+1).

From Petros Hadjicostas, Dec 11 2019: (Start)

Conjecture: For b >= 2, consider the function s(n,b) = Sum_{1 <= b^j <= n} (n mod b^j) from p. 8 in Dearden et al. (2011). Then s(b*n + r, b) = b*s(n,b) + r*N(n,b) for 0 <= r <= b-1, where N(n,b) = floor(log_b(n)) + 1 is the number of digits in the base-b representation of n. As initial conditions, we have s(n,b) = 0 for 1 <= n <= b. (This is a generalization of a result by Robert Israel in A049802.)

Here b = 3 and a(n) = s(n,3).

We have N(n,2) = A070939(n), N(n,3) = A081604(n), N(n,4) = A110591(n), and N(n,5) = A110592(n).

If A_b(x) = Sum_{n >= 1} s(n,b)*x^n is the g.f. of the sequence (s(n,b): n >= 1) and the above conjecture is correct, then it can be proved that A_b(x) = b * A_b(x^b) * (1-x^b)/(1-x) + x * ((b-1)*x^b - b*x^(b-1) + 1)/((1-x)^2 * (1-x^b)) * Sum_{k >= 1} x^(b^k). (End)

From Petros Hadjicostas, Dec 11 2019: (Start)

Conjecture: For b >= 2, consider the function s(n,b) = Sum_{1 <= b^j <= n} (n mod b^j) from p. 8 in Dearden et al. (2011). Then s(b*n + r, b) = b*s(n,b) + r*N(n,b) for 0 <= r <= b-1, where N(n,b) = floor(log_b(n)) + 1 is the number of digits in the base-b representation of n. As initial conditions, we have s(n,b) = 0 for 1 <= n <= b. (This is a generalization of a result by Robert Israel in A049802.)

Here b = 4 and a(n) = s(n,4).

We have N(n,2) = A070939(n), N(n,3) = A081604(n), N(n,4) = A110591(n), and N(n,5) = A110592(n).

If A_b(x) = Sum_{n >= 1} s(n,b)*x^n is the g.f. of the sequence (s(n,b): n >= 1) and the above conjecture is correct, then it can be proved that A_b(x) = b * A_b(x^b) * (1-x^b)/(1-x) + x * ((b-1)*x^b - b*x^(b-1) + 1)/((1-x)^2 * (1-x^b)) * Sum_{k >= 1} x^(b^k). (End)

Conjecture: For b >= 2, consider the function s(n,b) = Sum_{1 <= b^j <= n} (n mod b^j) from p. 8 in Dearden et al. (2011). Then s(b*n + r, b) = b*s(n,b) + r*N(n,b) for 0 <= r <= b-1, where N(n,b) = floor(log_b(n)) + 1 is the number of digits in the base-b representation of n. As initial conditions, we have s(n,b) = 0 for 1 <= n <= b. (This is a generalization of a result by Robert Israel in A049802.)

Here b = 5 and a(n) = s(n,5).

We have N(n,2) = A070939(n), N(n,3) = A081604(n), N(n,4) = A110591(n), and N(n,5) = A110592(n).

If A_b(x) = Sum_{n >= 1} s(n,b)*x^n is the g.f. of the sequence (s(n,b): n >= 1) and the above conjecture is correct, then it can be proved that A_b(x) = b * A_b(x^b) * (1-x^b)/(1-x) + x * ((b-1)*x^b - b*x^(b-1) + 1)/((1-x)^2 * (1-x^b)) * Sum_{k >= 1} x^(b^k).

Every k in A024934 is the sum of a tuple (x_1, ..., x_t) for prime(t) <= k < prime(t+1), where x_i = k mod prime(i). The tuples can be seen combinatorially as sets of t counters where the i-th counter cycles through 0 to prime(i)-1.

Since A024934(n) > n for n > 21, the set of numbers k for which A024934(k) = n is bounded above by n for those n, though smaller bounds are possible.

It is interesting to compare A024934 to A049802 (and likewise the current sequence with A383327). Every m in A049802 is also the sum of a tuple of congruences except the moduli are ascending powers of 2. Ordered by divisibility, the set of moduli in A049802 therefore form a chain (i.e., they are totally ordered) whilst the set of moduli in A024934 form an antichain (no one modulus divides any other). These opposed orders mean the two sequences behave quite differently.

Comparison of A049802 and A383327 vs. A024934 and {a(n)}:

A049802 and A383327:

- 0 appears infinitely many times in A049802, for every m = 2^n. Therefore 0 is not a term of A383327.

- A049802(2^n+1) = n. Therefore every n appears at least once in A049802.

- It is likely that every n appears at least once in A383327, however this is currently conjectural.

- If A049802(k) = m for 2^r-2^(r-2) <= k < 2^r, and if the rightmost summand in the tuple of m is x, then for s >= 0, A049802(k+2^((r-1)+s)) = m + x*(s+1).

A024934 and {a(n)}:

- 0 appears 3 times in A024934, for n = 0, 1, 2. Therefore a(0) = 3.

- It is not true that every n appears at least once in A024934 (e.g., 2 and 5 are not terms), and this is likely to be the case for infinitely many n, meaning it is likely that a(n) = 0 for infinitely many n.

- It appears to be unlikely that a(k) = n for every n: for 0 < n < 3500, a(n) <= 3 (and a(n) = 3 only for n = 0, 1, 8, 37, 781).

Numbers k such that there is only one m such that Sum_{i=1..t} m mod 2^i for 2^t <= m < 2^(t+1) is equal to k (see A049802). Every m = 2^k+1 (see A383327).

All terms are odd since for every even r there are at least two numbers u, v where the congruences sum to r (for u = 2^r+1 and v = 2^(r/2+1)+2).