cp's OEIS Frontend

Conjecture: a(n) is the height of the bottom region in column n of the Conant gasket in A328078.

If we consider each square k as having a continued fraction expansion c of all zeros after c(0) = sqrt(k)-1, then the continued fraction expansion of sqrt(k) for each square is trivially multiplicative.

For nonsquares, c(1) must be 1 and so k must satisfy m + 1/2 < sqrt(k) <= m+1, for some integer m.

A000034 = 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ... = continued fraction for (sqrt(3)+1)/2 (cf. A040001) = base 3 digital root of n+1. In general, the ascending descending base exponent transform of any simple periodic sequence can be written as a periodic set of interleaved sequences.

Take two consecutive triangular numbers t1 and t2 and create a triangle using (0,0), (t1,t2) and (t2,t1). The area of this triangle will be ((n+1)^3)/2 for t1 = n*(n+1)/2. - J. M. Bergot, May 08 2012

Multiplicative because both A000578 and A040001 are. - Andrew Howroyd, Jul 26 2018

Numbers m such that Ad(m) = A000203(m)/A000005(m) = A057020(m)/A057021(m) and Ah(m) = A023896(m)/A000010(m) = A026741(m)/A040001(m-1) are both integers.

{a(n)} = 1 and the even arithmetic numbers from A003601.

Subsequence of A003601.

Union of {a(n)} and A175679 = A003601 (arithmetic numbers).

Numbers m such that Ad(m) = A000203(m) / A000005(m) = A057020(m) / A057021(m) and Ak(m) = A000217(m) / A000027(m) = A145051(m) / A040001(m) are both integer.

Subsequence of A003601: {a(n)} = odd arithmetic numbers from A003601.

{a(n)} union A175678 = A003601 (arithmetic numbers).

From Robert G. Wilson v, Aug 09 2010: (Start)

All terms are odd because the second criterion is equivalent to n|T(n), where T(n) is the n-th triangular number, A000217(n).

Terms that are not prime are 1, 15, 21, 27, 33, 35, 39, 45, 49, 51, 55, 57, 65, 69, 77, 85, ..., .

Odd integers that are not terms: 9, 25, 63, 75, 81, 117, 121, 171, 175, 225, 243, 279, 289, ..., . (End)

There is more complicated way of defining the sequence: consider the sequence of modified Bernoulli numbers EVB(n) = A176327(n)/A176289(n) and its inverse binomial transform IEVB(n) = A176328(n)/A176591(n). Then a(n) is the numerator of the difference EVB(n)-IEVB(n). The denominator of the difference is 1 if n=0, else A040001(n-1).

A particularity of EVB(n) is: its (forward) binomial transform is 1, 1, 7/6, 3/2, 59/30,.. = (-1)^n*IEVB(n).

Note that A026741 is related to the Rydberg-Ritz spectrum of the hydrogen atom.

For Bernoulli numbers, B(1) excluded.

B(n) is calculated via

B(0) = 1;

B(0) + 0 = 1;

B(0) + 0 + 3*B(2) = 3/2;

B(0) + 0 + 6*B(2) + 4*B(3) = 2;

etc.

The diagonal is A026741(n+1)/A040001(n).

Row sums: 1, 1, 4, 11, 26, 57, ..., essentially Euler numbers A000295. See A130103, A008292 and A173018.

There is an infinitude of Bernoulli number sequences. They are of the form

B(n,q) = 1, q, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, ... .

Chronologically, the first, and the most regular, is, for q=1/2, A164555(n)/A027642(n), from Jacob Bernoulli (1654-1705), published in Ars Conjectandi in 1713 and(?) Seko Kowa (1642-1708) in 1712. See A159688. The second is, for q=-1/2, B(n,-1/2) = A027641(n)/A027642(n), from B(n,1/2) via Pascal's triangle. We could choose Be(n,q) instead of B(n,q) to avoid confusion with Sloane's B(n,p) for A027641(n)/A027642(n) (p=-1), A164555(n)/A027642(n) (p=1), A164558(n)/A027642(n) (p=2), A157809(n)/A027642(n) (p=3), ..., successive binomial transforms of the previous sequence.

This motivates the proposal of the (independent of q) sequence Bernoulli(n+2):

B(n+2) = 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, ... and its inverse binomial transform. See A190339.