cp's OEIS Frontend

Here we store the exponent e_i of p_i (p1=2, p2=3, p3=5, ...) in the factorization of n to the bit positions given by the column i-1 of A001477 viewed as a table (the exponent of 2 is thus stored to bit positions 0, 2, 5, 9, 14, 20, ..., exponent of 3 to 1, 4, 8, 13, 19, ..., exponent of 5 to 3, 7, 12, 18, 25, ...) using unary system, i.e. we actually store 2^(e_i) - 1 in binary.

This injective mapping from N to N offers an example of the proof given in Cameron's book that any distributive lattice can be represented as a sublattice of the power-set lattice P(X) of some set X. With this we can implement GCD (A003989) with bitwise AND (A004198) and LCM (A003990) with bitwise OR (A003986). Also, to test whether x divides y, it is enough to check that ((a(x) OR a(y)) XOR a(y)) = A003987(A003986(a(x),a(y)),a(y)) is zero.

Rows alternate between all 1's and alternating 1's and 0's. A 'mixed' sequence array: rows alternate between the rows of the sequence array for the all 1's sequence and the sequence array for the sequence 1,0,1,0,...

Column 2*k has g.f. x^(2*k)/(1-x); column 2*k+1 has g.f. x^(2*k+1)/(1-x^2).

Row sums are A029578(n+2). Antidiagonal sums are A106466.

This triangle is the Kronecker product of an infinite lower triangular matrix filled with 1's with a 2 X 2 lower triangular matrix of 1's. - Christopher Cormier, Sep 24 2017

From Peter Bala, Aug 21 2021: (Start)

Using the notation of Davenport et al.:

This is the double Riordan array ( 1/(1 - x); x/(1 + x), x*(1 + x) ).

The inverse array equals ( (1 - x)*(1 - x^2); x*(1 - x), x*(1 + x) ).

They are examples of double Riordan arrays of the form (g(x); x*f_1(x), x*f_2(x)), where f_1(x)*f_2(x) = 1. Arrays of this type form a group under matrix multiplication. For the group law see the Bala link. (End)

The array is read by descending antidiagonals as A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.

With interpolated 0's, this is Sum_{k=0..n} gcd(n-k+1,k+1)*(-1)^k.

Can also be viewed as a triangular table T(n,k) (n>=1, 1<=k<=n) read by rows: T(1,1); T(2,1), T(2,2); T(3,1), T(3,2), T(3,3); T(4,1), T(4,2), T(4,3), T(4,4); ... where T(n,k) gives the least value v>0 such that v*k = 0 modulo n+1, i.e., in other words, T(n,k) = (n+1)/gcd(n+1,k).

The fractions A053382(n,m)/A053383(n,m) give the triangle of the coefficients of the Bernoulli polynomials:

1;

-1/2, 1;

1/6, -1, 1;

0, 1/2, -3/2, 1;

-1/30, 0, 1, -2, 1;

0, -1/6, 0, 5/3, -5/2, 1;

1/42, 0, -1/2, 0, 5/2, -3, 1;

The matrix inverse of this triangle defines coefficients of the umbral inverse Bernoulli polynomials B^{-1}(n,x) in row n:

1;

1/2, 1;

1/3, 1, 1;

1/4, 1, 3/2, 1;

1/5, 1, 2, 2, 1;

1/6, 1, 5/2, 10/3, 5/2, 1;

1/7, 1, 3, 5, 5, 3, 1;

1/8, 1, 7/2, 7, 35/4, 7, 7/2, 1;

1/9, 1, 4, 28/3, 14, 14, 28/3, 4, 1;

1/10, 1, 9/2, 12, 21, 126/5, 21, 12, 9/2, 1;

1/11, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1;

The current triangle T(n,m) is the numerator of the entry in row n and column m.

In the majority of cases, T(n,m) = A050169(n,m), but since we use the numerators of the reduced fractions, an integer factor may be missing in this equation.

Umbral composition (e.g., B(.,x)^k = B(k,x)) gives B^(-1)(n,B(.,x)) = x^n = B(n,B^(-1)(.,x)). - Tom Copeland, Aug 25 2015

The gcd is given in A003989, and u is given in A345415. Minimal means minimize u^2+v^2. We follow Maple, PARI, etc., in setting u=0 and v=1 when m=n. If we ignore the diagonal, the v table is the transpose of the u table.

The indices run as A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), etc. The array is symmetric.