cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A292625 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.

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%I A292625 #136 Aug 18 2024 23:03:02
%S A292625 3,-29,14,509,-283,31,-17053,10104,-1306,64,1116637,-682005,94994,
%T A292625 -5466,129,-144570461,89619570,-12936231,800108,-22107,258,
%U A292625 37221717341,-23243908815,3414230937,-218563987,6481607,-88413,515
%N A292625 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 A292625 The matrix M is given by A266577.
%C A292625 The solution is unique and has an explicit formula as shown by _Max Alekseyev_, see the MathOverflow link.
%C A292625 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 A292625 Alois P. Heinz, <a href="/A292625/b292625.txt">Rows n = 1..50, flattened</a>
%H A292625 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 A292625 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 A292625 The first seven rows in the triangular array are:
%e A292625             3;
%e A292625           -29,           14;
%e A292625           509,         -283,         31;
%e A292625        -17053,        10104,      -1306,         64;
%e A292625       1116637,      -682005,      94994,      -5466,     129;
%e A292625    -144570461,     89619570,  -12936231,     800108,  -22107,    258;
%e A292625   37221717341, -23243908815, 3414230937, -218563987, 6481607, -88413, 515;
%e A292625   ...
%o A292625 (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 A292625 firstTerms(r)={my(ans=[],t);while(t++<=r,ans=concat(ans,tblRow(t)));return(ans)}
%o A292625 a(n)={my(u);while(binomial(u+1,2)<n,u++);firstTerms(u)[n]} \\ _R. J. Cano_, Oct 01 2017
%o A292625 (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 A292625 Cf. A266577, A365450.
%K A292625 sign,tabl
%O A292625 1,1
%A A292625 _Ahmad J. Masad_, Sep 21 2017
%E A292625 Edited by _Max Alekseyev_, Mar 20 2019