A292625 - OEIS

%I #136 Aug 18 2024 23:03:02

%S 3,-29,14,509,-283,31,-17053,10104,-1306,64,1116637,-682005,94994,

%T -5466,129,-144570461,89619570,-12936231,800108,-22107,258,

%U 37221717341,-23243908815,3414230937,-218563987,6481607,-88413,515

%N Triangle read by rows: row n gives y transposed, where y is the solution to the matrix equation M*y=b, where the matrix M and vector b are defined by M(i,j) = ((2^(i+1) + 1)^(j-1) + 1)/2 and b(i) = ((2^(i+1)+1)^n + 1)/2 for 1 <= i,j <= n.

%C The matrix M is given by A266577.

%C The solution is unique and has an explicit formula as shown by _Max Alekseyev_, see the MathOverflow link.

%C 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

%H Alois P. Heinz, <a href="/A292625/b292625.txt">Rows n = 1..50, flattened</a>

%H Ahmad J. Masad, <a href="https://mathoverflow.net/q/281442">Conjecture that relates matrix systems with some polynomials of integer coefficients as solution sets</a>, MathOverflow, Sep 2017.

%e 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.

%e The first seven rows in the triangular array are:

%e             3;

%e           -29,           14;

%e           509,         -283,         31;

%e        -17053,        10104,      -1306,         64;

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

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

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

%e   ...

%o (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)~)~;

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

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

%o (Sage) 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

%Y Cf. A266577, A365450.

%K sign,tabl

%O 1,1

%A _Ahmad J. Masad_, Sep 21 2017

%E Edited by _Max Alekseyev_, Mar 20 2019