cp's OEIS Frontend

A253015 Sequence of determinants of matrices based on the digits of nonnegative integers.

%I A253015 #32 Oct 06 2018 03:55:36
%S A253015 0,1,2,3,4,5,6,7,8,9,-1,1,-1,-5,-11,-19,-29,-41,-55,-71,-4,-1,4,1,-4,
%T A253015 -11,-20,-31,-44,-59,-9,-5,1,9,5,-1,-9,-19,-31,-45,-16,-11,-4,5,16,11,
%U A253015 4,-5,-16
%N A253015 Sequence of determinants of matrices based on the digits of nonnegative integers.
%C A253015 A given nonnegative integer is transformed into a square matrix whose order equals the quantity of the number's digits. Each element of the main diagonal is a digit of this original number, while other elements are calculated from this diagonal. The determinant of this matrix is the element of the sequence.
%H A253015 Filipi R. de Oliveira, <a href="/A253015/b253015.txt">Table of n, a(n) for n = 0..999</a>
%F A253015 a(n) = det(B) where B is the n X n matrix with B(i,i) given by the i-th digit of n, B(i,j) = abs(B(i,j-1)-B(i+1,j)) if i < j and B(i,j) = B(i-1,j) + B(i,j+1) if i > j.
%e A253015 For n=124, a(124)=2, as follows:
%e A253015 B(1,1) = 1;
%e A253015 B(2,2) = 2;
%e A253015 B(3,3) = 4;
%e A253015 B(1,2) = abs(B(1,1) - B(2,2)) = abs(1-2) = 1;
%e A253015 B(2,3) = abs(B(2,2) - B(3,3)) = abs(2-4) = 2;
%e A253015 B(1,3) = abs(B(1,2) - B(2,3)) = abs(1-1) = 1;
%e A253015 B(2,1) = B(1,1) + B(2,2) = 1 + 2 = 3;
%e A253015 B(3,2) = B(2,2) + B(3,3) = 2 + 4 = 6;
%e A253015 B(3,1) = B(2,1) + B(3,2) = 3 + 6 = 9.
%e A253015 Thus,
%e A253015 _______|1 1 1|
%e A253015 B(124)=|3 2 2| --> det(B(124)) = a(124) = 2.
%e A253015 _______|9 6 4|
%Y A253015 See A227876, since the process of matrix construction is this so-called "pyramidalization".
%K A253015 sign,base,easy,dumb
%O A253015 0,3
%A A253015 _Filipi R. de Oliveira_, Dec 25 2014