cp's OEIS Frontend

The rows are of length 1, 3, 5, 7, ...

a(n,k) is also the number of independent rank n tensor operators to appear in the tensor product of two spaces each spanned by k+1 tensor operators of ranks 0 to k,

{Y_{l,m},l=0,1,...,k, m:-l,-l+1,...,l} times {Y'_{l'm'}, l'=0,1,...,k, m':-l,-l+1,...,l}.

Basis elements of the tensor product space are given by

psi^{l,l'}{p,q} = Sum{m,m'} C^{ll'p}{mm'q} Y{l,m}Y'_{l'm'}

for all l,l' = 0,1,...,k and where p = |l-l'|, |l-l'|+1, ..., l+l' is the rank, q=-p, -p+1,...,p and where C^{ll'p}_{mm'q} are the Clebsch-Gordon coefficients.

Sum_{k=0..2*n+1} a(n,k)*(2*k+1) = (n+1)^4. - L. Edson Jeffery, Oct 29 2012

Sum_{k=0..2*n+1} (a(n,k) - a(n-1,k))*(2*k+1) = n^4 - (n-1)^4 = A005917(n+1), for n > 0. - L. Edson Jeffery, Nov 02 2012

The length of row k is A005917, the rhombic dodecahedral numbers, (k+1)^4 - k^4. The triangle has rows beginning with 2^k and ending with 210^k.

This is the case k=2 of Sophie Germain's Identity n^4+(2*k^2)^2 = ((n-k)^2+k^2)*((n+k)^2+k^2).

An elementary square of type 2 is the smallest square that can be tiled with squares of two different sides a < b satisfying a^2+b^2 = c^2 and so that the numbers of small and large squares are equal.

Every term is an odd square >= 9 and each odd square is present infinitely many times.

Notation: s_p (resp. s_e) = side of a primitive (resp. elementary) tiled square, a = side of small squares and b = side of large squares used to tile a primitive square, and z_p (z_e) = number of small squares = number of large squares used to tile a primitive (resp. elementary) square.

A primitive square with side s_p = a*c/(c-b) is tiled with z_p small and z_p large squares with sides a and b, and z_p = (a/(c-b))^2.

Each elementary square with a side s_e = k*s_p, k>0, is tiled with z_e small and z_e large squares with sides k*a and k*b, and z_e = z_p = (a/(c-b))^2.

When an elementary side A344332(n) is a multiple of m distinct primitive sides s_p, then there are m different values T(n,1), ..., T(n,m) in the row n (see example).

A octachoral number is a centered figurate number that represents a octachoron, which is a four-dimensional regular polytope composed of 8 cells (also known as tesseract).

One of the 6 centered regular polichoral (centered pentachoral, centered hexadecachoral, centered octachoral, centered icositetrachoral, centered hexacosichoral and centered hecatonicosachoral) numbers.

a(n) <= A000583(n), which is the number of 2 X 2 matrices with entries in {1, ..., n}.

a(n) >= A005917(n), which is the number of 2 X 2 matrices with entries in {1, ..., n} that contain the element 1. All such matrices are not decomposable as a product of 2 X 2 positive integer matrices.

This definition is a generalization of the notion of prime numbers to the family of 2 X 2 positive integer matrices. Since the matrices do not contain 0, max(A*B) > max(A) and max(A*B) > max(B). Thus, for every matrix there is a finite number of possible decompositions to check.

Larger of pair of integers whose Pythagorean means are all integers.

The smaller of the pairs are: (A001844).

The arithmetic means are: (A007204)

The geometric means are: (A005917)

The harmonic means are: (A016754).

Subtracting terms in A016754 from A007204 gives complementary harmonics (A060300).

These squares with side = A344330(n) can be tiled with squares of two different sizes so that the numbers of large or small squares are equal.

Notation: s = side of the tiled squares, a = side of small squares, b = side of large squares, and z = number of small squares = number of large squares.

For any n, consecutive n-th powers will never share a divisor > 1, so now consider the second differences. Specifically, each m > 0, define the binary sequence {b(m)} as follows: b(m) = 1 if the first difference (m+1)^n - m^n and the second difference (m+2)^n - 2*(m+1)^n + m^n are coprime, 0 otherwise. I conjecture that {b(m)} is periodic with period a(n).

If m^n mod p == (m+1)^n mod p == (m+2)^n mod p, then p is in the prime factorization of a(n).

All primes p >= 5 belong to a prime factorization for a(n). p will always belong to the prime factorization of n=p-1 due to Fermat's Little Theorem.

I conjecture that the greatest prime factor for any prime n >= 5 is phi(2^n+1)/2 + 1 = Jacobsthal(n). n*A069925 + 1 = A001045(n).

I conjecture that all prime factors "f" are f=n*k+1, unless n is composite, in which case additionally all prime factors for any divisor of n will also be included in the prime factorization for a(n).