1,1

The matrix M is given by A266577.

The solution is unique and has an explicit formula as shown by Max Alekseyev, see the MathOverflow link.

Conjecture: as m approaches infinity, the point continuation of the inverse hyperbolic sine scatterplot of the first m*(m+1)/2 terms of this sequence approaches a perfect circular sector with an angle equal to 2*Pi/9. See the last scatterplot in the graph section. - Ahmad J. Masad, Jun 02 2022

Alois P. Heinz, Rows n = 1..50, flattened

Ahmad J. Masad, Conjecture that relates matrix systems with some polynomials of integer coefficients as solution sets, MathOverflow, Sep 2017.

The first row contains a single term, the solution x=3; the second row contains the solution of the system { x+3y=13, x+5y=41 }, which is x=-29 and y=14; the third row contains the solution of the system { x+3y+13z=63, x+5y+41z=365, x+9y+145z=2457 }, which is x=509, y=-283 and z=31; and so on.

The first seven rows in the triangular array are:

3;

-29, 14;

509, -283, 31;

-17053, 10104, -1306, 64;

1116637, -682005, 94994, -5466, 129;

-144570461, 89619570, -12936231, 800108, -22107, 258;

37221717341, -23243908815, 3414230937, -218563987, 6481607, -88413, 515;

...

(PARI) tblRow(k)=matsolve(matrix(k, k, i, j, ((2^(i+1)+1)^(j-1) + 1)/2), vector(k, l, ((2^(l+1)+1)^k + 1)/2)~)~;

firstTerms(r)={my(ans=[], t); while(t++<=r, ans=concat(ans, tblRow(t))); return(ans)}

a(n)={my(u); while(binomial(u+1, 2)<n, u++); firstTerms(u)[n]} \\ R. J. Cano, Oct 01 2017

(SageMath) def A292625row(n): return tuple([(-1)^(n+1) * ( product(2^(i+2)+1 for i in range(n)) - 2^(n*(n+3)/2-1) )]) + tuple( (-1)^(n+k) * SymmetricFunctions(QQ).e()[n+1-k].expand(n)( tuple(2^(i+2)+1 for i in range(n)) ) for k in range(2, n+1) ) # Max Alekseyev, Mar 20 2019

Cf. A266577, A365450.

Sequence in context: A143234 A378147 A071195 * A030274 A249518 A194392

Adjacent sequences: A292622 A292623 A292624 * A292626 A292627 A292628

sign,tabl

Ahmad J. Masad, Sep 21 2017

Edited by Max Alekseyev, Mar 20 2019

approved