cp's OEIS Frontend

Number of digits in A007090(n).

In terms of the repetition convolution operator #, where (sequence A) # (sequence B) = the sequence consisting of A(n) copies of B(n), this sequence is the repetition convolution A110594 # n. Over the set of positive infinite integer sequences, # gives a nonassociative noncommutative groupoid (magma) with a left identity (A000012) but no right identity, where the left identity is also a right nullifier and idempotent. For any positive integer constant c, the sequence c*A000012 = (c,c,c,c,...) is also a right nullifier; for c = 1, this is A000012; for c = 3 this is A010701.

For n > 0, first differences of A022331. - Michel Marcus, Oct 03 2013

In terms of the repetition convolution operator #, where (sequence A) # (sequence B) = the sequence consisting of A(n) copies of B(n), then this sequence is the repetition convolution A110595 # n. Over the set of positive infinite integer sequences, # gives a nonassociative noncommutative groupoid (magma) with a left identity (A000012) but no right identity, where the left identity is also a right nullifier and idempotent. For any positive integer constant c, the sequence c*A000012 = (c,c,c,c,...) is also a right nullifier; for c = 1, this is A000012; for c = 3 this is A010701.

A003462 is a subsequence; A171960(a(n)) >= a(n). - Reinhard Zumkeller, Jan 20 2010

A157671 is a subsequence. - Reinhard Zumkeller, Jan 20 2010

It is known that there is exactly one bad coin, which is heavier than the others. No weights are used in the weighings.

0 appears once, 1 twice, 2 6 times, 3 18 times, 4 54 times, ... which is the same as the number of base-3 numbers of length n; see A007089. - Jonathan Vos Post, Apr 20 2011

Records appear at positions 3^n+1 (=A034472(n)). - Robert G. Wilson v, Aug 06 2012

The "Heavy Marble" section of "Brainteaser Problems" in the Mongan et al. reference describes the n = 8 case in detail and then derives the general formula given below. Of course this sequence applies also when the single, unlike object is lighter than all the others. If the unlike object is only known to have a different weight (that is, to be lighter than all the others or heavier than all the others), use A064099. - Rick L. Shepherd, Sep 05 2013

If it is unknown whether the bad coin is heavier or lighter, then the minimum number of weighings is A029837(n) and the number of coins that must be used in the first weighing is A004526(n), for n > 2. - Ivan N. Ianakiev, Apr 13 2017

From A.H.M. Smeets, Dec 14 2019: (Start)

a(n)-a(n-1) >= 2 due to the fact that n = 10_n, so there is an increment of at least 2. If n can be written as a perfect power m^s, an additional +1 comes to it for the representation of n in each base m.

For instance, for n = 729 we have 729 = 3^6 = 9^3 = 27^2, so there is an additional increment of 3. For n = 1296 we have 1296 = 6^4 = 36^2, so there is an additional increment of 2. For n = 4096 we have 4096 = 2^12 = 4^6 = 8^4 = 16^3= 64^2, so there is an additional increment of 5. (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 = 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)