A078121 Infinite lower triangular matrix, M, that satisfies [M^2](i,j) = M(i+1,j+1) for all i,j>=0 where [M^n](i,j) denotes the element at row i, column j, of the n-th power of matrix M, with M(0,k)=1 and M(k,k)=1 for all k>=0.
1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 10, 16, 8, 1, 1, 36, 84, 64, 16, 1, 1, 202, 656, 680, 256, 32, 1, 1, 1828, 8148, 10816, 5456, 1024, 64, 1, 1, 27338, 167568, 274856, 174336, 43680, 4096, 128, 1, 1, 692004, 5866452, 11622976, 8909648, 2794496, 349504, 16384, 256, 1
Offset: 0
Examples
The square of the matrix is the same matrix excluding the first row and column: [1, 0, 0, 0, 0]^2 = [ 1, 0, 0, 0, 0] [1, 1, 0, 0, 0] [ 2, 1, 0, 0, 0] [1, 2, 1, 0, 0] [ 4, 4, 1, 0, 0] [1, 4, 4, 1, 0] [10,16, 8, 1, 0] [1,10,16, 8, 1] [36,84,64,16, 1]
Links
- Alois P. Heinz, Rows n = 0..80, flattened
- MathOverflow, Closed form for diagonals of A078121, (2025).
Programs
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Maple
M:= proc(i, j) option remember; `if`(j=0 or i=j, 1, add(M(i-1, k)*M(k, j-1), k=0..i-1)) end: seq(seq(M(n,k), k=0..n), n=0..10); # Alois P. Heinz, Feb 27 2015 -
Mathematica
rows = 10; M[k_] := Table[ Which[j == 1, 1, i == j, 1, 1 < j < i, m[i, j], True, 0], {i, 1, k}, {j, 1, k}]; m2[i_, j_] := m[i+1, j+1]; M2[k_] := Table[ Which[j -
PARI
rows_upto(n) = my(A, v1); v1 = vector(n+1, i, vector(i, j, 0)); v1[1][1] = 1; for(i=1, n, v1[i+1][1] = 1; v1[i+1][i+1] = 1); for(i=2, n, for(j=1, i-1, A = (i+j+1)%2; v1[i+1][j+1] = 2*sum(k=0, (i-j-1)\2, v1[i-j+1][2*k+A+1]*v1[j+2*k+A+1][j]))); v1 \\ Mikhail Kurkov, Aug 27 2025
Formula
M(1,j) = A002577(j) (partitions of 2^j into powers of 2), M(j+1,j) = 2^j, M(j+2,j) = 4^j, M(j+3,j) = A016131(j).
M(n,k) = the coefficient of x^(2^n - 2^(n-k)) in the power series expansion of 1/Product_{j=0..n-k} (1-x^(2^j)) whenever 0<=k0 (conjecture).
M(n,k) = Sum_{j=0..n-k-1} M(n-k,j)*M(k+j,k-1)*(1+(-1)^(n+k+j+1)) for 0 < k < n with M(n,0) = M(n,n) = 1. - Mikhail Kurkov, Jun 01 2025
From Mikhail Kurkov, Jul 01 2025: (Start)
Conjecture 1: let R(n,x) be the n-th row polynomial, then R(n,x) = x*R(n-1,x) + Sum_{k=1..n-1} M(n-1,k-1)*R(k,x)*(-1)^(n+k+1) = R(n-1,x) + x*Sum_{k=1..n-1} (M(n-1,k) - M(n-2,k))*R(k,x) for n > 1 with R(0,x) = 1, R(1,x) = x + 1.
Conjecture 2: M(n+m,n) ~ 2^(m*(2*n+m-1)/2)/m! as n -> oo. More generally, it also looks like that M(n+m,n) for m > 0 can be represented as (Sum_{j=0..flooor((m-1)/2)} 2^((m-2*j)*(2*(n-j)+m-1)/2)*P(m,j)*(-1)^j)/m! where P(m,j) are some positive integer coefficients. (End)
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