A381810 Array read by downward antidiagonals: A(n,k) is a generalization of odd columns of A125790 defined in Comments for n > 0, k >= 0.

2, 4, 4, 6, 16, 6, 8, 36, 20, 10, 10, 64, 42, 84, 14, 12, 100, 72, 286, 100, 20, 14, 144, 110, 680, 322, 120, 26, 16, 196, 156, 1330, 744, 364, 140, 36, 18, 256, 210, 2300, 1430, 816, 406, 656, 46, 20, 324, 272, 3654, 2444, 1540, 888, 3396, 740, 60, 22, 400, 342, 5456, 3850, 2600, 1650, 10816, 3682, 840, 74

Offset: 1

Author

Mikhail Kurkov, May 05 2025

Keywords

• nonn
• tabl

Comments

This is generalization in the sense that first column of A125790 is A000123(2^(n-1)) while in this square array column zero is conjecturally A000123(n).

A(n,k) = v_{A001511(n)} where we start with vector v of fixed length L(n) = A070939(n) with elements v_i = A125790(i,2*k+1), pre-calculate A078121 up to L(n)-th row, reserve t as an empty vector of fixed length L(n) and for i=1..A119387(n+1), for j=1..L(n)-i+1 apply t := v (at the beginning of each cycle for i) and also apply v_j := Sum_{k=1..j+1} A078121(j,k-1)*t_k if R(n,L(n)-i) = 1, otherwise v_j := Sum_{k=1..j+1} A078121(j,k-1)*t_k*(-1)^(j+k+1). Here R(n,k) = floor(n/(2^k)) mod 2 is the (k+1)-th bit in the binary expansion of n.

Conjecture: sequence A(n,k) for fixed n is a polynomial of degree A070939(n).

Examples

			Array begins:
===========================================================
n\k|  0    1     2      3      4      5       6       7 ...
---+-------------------------------------------------------
1  |  2,   4,    6,     8,    10,    12,     14,     16 ...
2  |  4,  16,   36,    64,   100,   144,    196,    256 ...
3  |  6,  20,   42,    72,   110,   156,    210,    272 ...
4  | 10,  84,  286,   680,  1330,  2300,   3654,   5456 ...
5  | 14, 100,  322,   744,  1430,  2444,   3850,   5712 ...
6  | 20, 120,  364,   816,  1540,  2600,   4060,   5984 ...
7  | 26, 140,  406,   888,  1650,  2756,   4270,   6256 ...
8  | 36, 656, 3396, 10816, 26500, 55056, 102116, 174336 ...
  ...
		

Crossrefs

Cf. A000123, A001511, A007814, A053645, A062383, A070939, A078121, A106400, A119387, A125790, A236206.

Programs

PARI

upto1(n) = my(v1); v1 = vector(n+1, i, vector(i, j, j==1 || j==i)); for(i=2, n, for(j=1, i-1, v1[i+1][j+1] = sum(k=j-1, i-1, v1[i][k+1]*v1[k+1][j]))); v1
A(n,m) = my(L = logint(n,2), A = valuation(n,2), B = logint(n>>A,2), v1, v2, v3); v1 = upto1(L+2); v2 = vector(L+2, i, vecsum(v1[i])); for(i=1, 2*m, v2 = vector(L+2, i, sum(j=1, i, v1[i][j]*v2[j]))); for(i=1, B, v3 = v2; for(j=1, L-i+1, v2[j+1] = sum(k=1, j+1, v1[j+1][k]*v3[k+1]*if(!bittest(n,L-i+1), (-1)^(j+k+1), 1)))); v2[A+2]

PARI

upto1(n) = my(v1); v1 = vector(n+1, i, vector(i, j, j==1 || j==i)); for(i=2, n, for(j=1, i-1, v1[i+1][j+1] = sum(k=j-1, i-1, v1[i][k+1]*v1[k+1][j]))); v1
upto2(n,m) = my(L = logint(n,2), A = valuation(n,2), B = logint(n>>A,2), v1, v2, v3, v4, v5); v1 = upto1(L+2); v2 = vector(L+2, i, 1); v3 = vector(m+1, i, 0); for(s=0, m, for(i=1, min(s+1,2), v2 = vector(L+2, i, sum(j=1, i, v1[i][j]*v2[j]))); v4 = v2; for(i=1, B, v5 = v4; for(j=1, L-i+1, v4[j+1] = sum(k=1, j+1, v1[j+1][k]*v5[k+1]*if(!bittest(n,L-i+1), (-1)^(j+k+1), 1)))); v3[s+1] = v4[A+2]); v3 \\ slightly modified version of the first program, some kind of memoization; generates A(n,k) for k=0..m

Formula

A(2^(n-1),k) = A125790(n,2*k+1) for n > 0, k >= 0.

Conjectured formulas: (Start)

A(n,0) = A000123(n) for n > 0.

A(n,k) = Sum_{j=0..k} A000123(A062383(n)*j+n)*A106400(k-j) for n > 0, k >= 0.

If we change v_i = A125790(i,2*k+1) to v_i = A125790(i,2*k) to get similar generalization of even columns, then for resulting array B(n,k) we have B(n,k) = Sum_{j=0..k} A000123(A062383(n)*j+A053645(n))*A106400(k-j) for n > 0, k >= 0.

2*(k+1) divides A(n,k) for n > 0 if (k+1) is a term of A236206.

G.f. for n-th row is f(A070939(n)+1,n) for n > 0 where f(n,k) = (Sum_{(c_0 + c_1 + ... + c_{n-1}) == 2*k (mod 2^n), 0 <= c_i < 2^n, 2^i divides c_i} x^((c_0 + c_1 + ... + c_{n-1} - 2*k)/2^n))/(1-x)^n for n > 0, k >= 0. Similarly, g.f. for n-th row of B(n,k) is f(A070939(n)+1,A053645(n)).

G.f. for n-th row is (Sum_{i=0..L(n)-1} x^i * Sum_{j=0..i} binomial(L(n)+1,j)*A(n,i-j)*(-1)^j)/(1-x)^(L(n)+1) for n > 0 where L(n) = A070939(n).

s(4*n+1) = 1 for n >= 0, s(4*n) = s(4*n+2) = 1 if A010060(n) = 1 for n > 0 where s(n) = A007814(Sum_{k=0..n-1} A(k+1,n-k-1)). (End)