cp's OEIS Frontend

The level 0 Sierpinski carpet graph is a single vertex. The level n Sierpinski carpet graph is formed from 8 copies of level n-1 by joining boundary vertices between adjacent copies.

Interleaving of A001018 and A083233 without initial term 1.

Binomial transform is A164544. Third binomial transform is A038761.

The level 0 Menger sponge graph is a single vertex. The level n Menger sponge graph is formed from 20 copies of level n-1 in the shape of a cube with middle faces removed by joining boundary vertices between adjacent copies.

The level 0 Menger sponge graph is a single vertex. The level n Menger sponge graph is formed from 20 copies of level n-1 in the shape of a cube with middle faces removed by joining boundary vertices between adjacent copies.

The level 0 Menger sponge graph is a single vertex. The level n Menger sponge graph is formed from 20 copies of level n-1 in the shape of a cube with middle faces removed by joining boundary vertices between adjacent copies.

The level 0 Menger sponge graph is a single vertex. The level n Menger sponge graph is formed from 20 copies of level n-1 in the shape of a cube with middle faces removed by joining boundary vertices between adjacent copies.

The level 0 Menger sponge graph is a single vertex. The level n Menger sponge graph is formed from 20 copies of level n-1 in the shape of a cube with middle faces removed by joining boundary vertices between adjacent copies.

Binomial transform of A053524 (without leading zero).

The n-th row is {T(n,0),T(n,1),...,T(n,n-1)}.

Let m denote the prime power p^e.

T(m,0) = A020478(m) = (p^(e+1) + p^e-1)*p^(2*e-1).

T(m,1) = A000056(m) = (p^2-1)*p^(3*e-2).

T(prime(n),1) = A127917(n).

Sum_{k=1..n-1} T(n,k) = A005353(n).

T(n,1) = n*A007434(n) for n>=1 because A000056(n) = n*Jordan_Function_J_2(n).

T(2^n,1) = A083233(n) = A164640(2n) for n>=1. Proof: a(n):=T(2^n,1); a(1)=6, a(n)=8*a(n-1); A083233(1)=6 and A083233(n) is a geometric series with ratio 8 (because of its g.f.), too; A164640 = {b(1)=1, b(2)=6, b(n)=8*b(n-2)}.

T(2^n,0) = A165148(n) for n>=0, because 2*T(2^n,0) = (3*2^n-1)*4^n.

T(2^e,2) = A003951(e) for 2 <= e. Proof: T(2^e,2) = 9*8^(e-1) is a series with ratio 8 and initial term 72, as A003951(2...inf) is.

Working with consecutive powers of a prime p, we need a definition (0 <= i < e):

N(p^e,i):=#{k: 0 < k < p^e, gcd(k,p^e) = p^i} = (p-1)*p^(e-1-i). We say that these k's belong to i (respect to p^e). Note that N(p^e,0) = EulerPhi(p^e), and if 0 < k < p^e then gcd(k,p^e) = gcd(k,p^(e+1)). Let T(p^e,[i]) denote the common value of T(p^e,k)'s, where k's belong to i (q.v.PROGRAM); for example, T(p^e,[0]) = T(p^e,1). The number of the 2 X 2 matrices over Z(p^e), T(p^e,0) + Sum_{i=0..e-1} T(p^e,[i])*N(p^e,i) = p^(4e) will be useful.

On the hexagon property: Let prime p be given and let T(p^e,[0]), T(p^e,[1]), T(p^e,[2]), ..., T(p^e,[e-2]), T(p^e,[e-1]) form the e-th row of a Pascal-like triangle, e>=1. Let denote X(r,s) an element of the triangle and its value T(p^r,[s]). Let positive integers a and b given, so that the entries A(m-a,n-b), B(m-a,n), C(m,n+a), D(m+b,n+a), E(m+b,n), F(m,n-b) of the triangle form a hexagon spaced around T(p^m,[n]); if a=b=1 then they surround it. If A*C*E = B*D*F, then we say that the triangle T(.,.) has the "hexagon property". (In the case of binomial coefficients X(r,s) = COMB(r,s), the "hexagon property" holds (see [Gupta]) and moreover gcd(A,C,E) = gcd(B,D,F) (see [Hitotumatu & Sato]).)

Corollary 2.2 in [Brent & McKay] says that, for the d X d matrices over Z(p^e), (mutatis mutandis) T_d(p^e,0) = K*(1-P(d+e-1)/P(e-1)) and T_d(p^e,[i]) = K*(q^e)*((1-q^d)/(1-q))*P(d+i-1)/P(i), where q=1/p, K=(p^e)^(d^2), P(t) = Product_{j=1..t} (1-q^j), P(0):=1. (For the case d=2, we have T(p^e,[i]) = (p+1)*(p^(i+1)-1)*p^(3*e-i-2).) Due to [Brent & McKay], it can be simply proved that for d X d matrices the "hexagon property" is true. The formulation implies an obvious generalization: For the entries A(r,u), B(r,v), C(s,w), D(t,w), E(t,v), F(s,u) of the T_d(.,.)-triangle, a hexagon-like property A*C*E = B*D*F holds. This is false in general for the COMB(.,.)-triangle.

Another (rotated-hexagon-like) property: for the entries A(m-b1,n), B(m-a1,n+c2), C(m+a2,n+c2), D(m+b2,n), E(m+a2,n-c1), F(m-a1,n-c1) of the T_d(.,.)-triangle, the property A*C*E = B*D*F holds, if and only if 2*(a1 + a2) = b1 + b2. This is also in general false for COMB(.,.)-triangle.

Each term of the sequence encodes a line of the inverse of a Jacobi matrix that has 1s on its lower, main, and upper diagonals in GF(2). The associated inverse matrix column values come from the binary representation of that base-10 number, being a bit per column. These matrices have ascending and consecutive multiples of 3 sizes. If the binary number has fewer bits than the number of columns, it must be zero-padded to the left. To obtain the inverse matrices in real numbers instead of GF(2), alternate between + and - between the 1s in a row. If a row is a multiple of 3, alternate between - and +. The determinants of these 3m x 3m Jacobi matrices are 1 in GF(2), as proven by Sutner (1989), and alternate between -1 and 1 in R if m is odd or even, as proven by Melo (1987).

The recurrence, in line 3, uses the Iverson notation as presented in Graham, Knuth, and Patashnik (2002).

The proof of the correctness of that sequence of inverses is done by induction.