A365450 -id:A365450 - cp's OEIS frontend

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.

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

Offset: 1

Ahmad J. Masad, Sep 21 2017

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

			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;
  ...

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.

Cf. A266577, A365450.

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)R. J. Cano, Oct 01 2017

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

Edited by Max Alekseyev, Mar 20 2019

A374258 Square array: T(n,k) = ((3^(n+1) + 1)^(k-1) + 2)/3, read by descending antidiagonals.

Original entry on oeis.org

1, 4, 1, 34, 10, 1, 334, 262, 28, 1, 3334, 7318, 2242, 82, 1, 33334, 204886, 183790, 19846, 244, 1, 333334, 5736790, 15070726, 4842262, 177634, 730, 1, 3333334, 160630102, 1235799478, 1181511766, 129672334, 1595782, 2188, 1, 33333334, 4497642838, 101335557142, 288288870742, 94660803334, 3491569558, 14353282, 6562, 1

Offset: 1

Ahmad J. Masad, Jul 01 2024

This sequence gives the matrix M in the definition of A365450. Similar to A266577.
Conjecture: For each natural number n, the digits of the product of any (n+1) not necessarily distinct terms of the n-th row in the base (3^(n+1)+1) numeral system appear in nondecreasing order.
Proof of the conjecture. Let b := 3^(n+1)+1. The product of any (n+1) terms of the n-th row has the form p/(b-1), where p is the product of (n+1) numbers of the form b^k+2. Let p = (dm, ..., d1, d0)b, and we have d0+d1+...+dm = 3^(n+1) = b-1. Then p/(b-1) = (dm, ..., d2+...+dm, 1+d1+...+dm)_b, which do form a nondecreasing sequence. - _Max Alekseyev, Jul 03 2024
The preceding result is similar to the property of the nondecreasing products mentioned in A237424. Specifically; the first row of this array is A093137, which is a subsequence of A237424.  - Ahmad J. Masad, Jul 30 2024
More generally: Let r and n be positive integers and S be a sequence of all numbers of the form (b^c(1)+b^c(2)+...+b^c(r)+1)/(r+1), where c(1),...,c(r) are nonnegative integers. Then in the numeral system base b := (r+1)^(n+1)+1, the digits of the product of any n+1 (possibly equal) terms of S appear in nondecreasing order. Proof is similar. - Ahmad J. Masad, Jul 30 2024; edited by Max Alekseyev, Aug 01 2024

			The array begins:
  1    4    34   334   3334
  1   10   262  7318
  1   28  2242
  1   82
  1
Example of the conjecture: Take 5 terms from the 4th row and find their product in base 244 numeral system (since 3^(4+1)+1=244) as follows: 82,19846 twice and 4842262 twice, the product is equal to 82*19846*19846*4842262*4842262 = 757279838666167487626528 = (1, 3, 7, 19, 31, 55, 91, 115, 163, 195, 212)_244 which is in agreement with the conjecture since the digits in 244 base numeral system are in nondecreasing order.
Example of the general property: Take r=3 and n=4, then b=4^5+1=1025. The sequence S is the sequence of the numbers of the form (1025^b(1)+1025^b(2)+1025^b(3)+1)/4. Let's multiply 5 terms of the sequence S, say ((1025^0+1025^0+1025^1+1)/4)*(((1025^0+1025^1+1025^1+1)/4)^2)*(((1025^3+1025^4+1025^4+1)/4)^2) = 257*513^2*552175667969^2 = 20621601208620337073958261562113 = (16,112,308,488,580,680,832,936,964,984,1013)_1025. The digits of the product in base 1025 are in nondecreasing order.

Michel Marcus, Table of n, a(n) for n = 1..820

Cf. A093137, A266577, A365450, A237424.

T[n_, k_] := ((3^(n+1) + 1)^(k-1) + 2)/3; Table[T[k, n-k+1], {n, 1, 9}, {k, 1, n}] // Flatten (* Amiram Eldar, Jul 02 2024 *)

T(n,k) = ((3^(n+1) + 1)^(k-1) + 2)/3 \\ Andrew Howroyd, Jul 01 2024

cp's OEIS Frontend

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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