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%I #12 Jun 05 2021 02:32:08
%S 1,2,3,3,6,5,4,9,10,7,5,12,15,14,9,6,15,20,21,18,11,7,18,25,28,27,22,
%T 13,8,21,30,35,36,33,26,15,9,24,35,42,45,44,39,30,17,10,27,40,49,54,
%U 55,52,45,34,19,11,30,45,56,63,66,65,60,51,38,21,12,33,50,63,72,77,78,75,68
%N Triangle read by rows: T(n,k) = (2*k+1)*(n+1-k), 0 <= k < n.
%C The triangle is generated from the product of matrix A and matrix B, i.e., A * B where A = the infinite lower triangular matrix:
%C 1 0 0 0 0 ...
%C 1 1 0 0 0 ...
%C 1 1 1 0 0 ...
%C 1 1 1 1 0 ...
%C 1 1 1 1 1 ...
%C ... and B = the infinite lower triangular matrix:
%C 1 0 0 0 0 ...
%C 1 3 0 0 0 ...
%C 1 3 5 0 0 ...
%C 1 3 5 7 0 ...
%C 1 3 5 7 9 ...
%C ...
%C Row sums give the square pyramidal numbers A000330.
%C T(n+0,0)=1*n=A000027(n+1); T(n+1,1)=3*n=A008585(n); T(n+2,2)=5*n=A008587(n); T(n+3,3)=7*n=A008589(n); etc. So T(n,0)*T(n,1)=3*n*(n+1)=A028896(n) (6 times triangular numbers). T(n,1)*T(n,2)/10=3*n*(n+1)/2=A045943(n) for n>0 T(n,2)*T(n,3)/10=7/2*n*(n+1)=A024966(n) for n>1 (7 times triangular numbers), etc.
%C From _Gary W. Adamson_, Apr 25 2010: (Start)
%C Consider the following array, signed as shown:
%C ...
%C 1, 3, 5, 7, 9, 11, ...
%C 2, -6, 10, -14, 18, -22, ...
%C 3, 9, 15, 21, 27, 33, ...
%C 4, -12, 20, -28, 36, -44, ...
%C 5, 15, 25, 35, 45, 55, ...
%C 6, -18, 30, -42, 54, -66, ...
%C 7, 21, 35, 49, 63, 77, ...
%C ...
%C Let each term (+, -)k = (+, -) phi^(-k).
%C Consider the inverse terms of the Lucas series (1/1, 1/3, 1/4, 1/7, ...).
%C By way of example, let q = phi = 1.6180339...; then
%C ...
%C 1/1 = q^(-1) + q^(-3) + q^(-5) + q^(-7) + q^-9) + ...
%C 1/3 = q^(^2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
%C 1/4 = q^(-3) + q^(-9) + q^(-15) + q^(-21) + q^(-27) +...
%C 1/7 = q^(-4) - q^-12) + q^(-20) - q^(-28) + q^(-36) + ...
%C 1/11 = q^(-5) + q^-15) + q^(-25) + q^(-35) + q^(-45) + ...
%C ...
%C Relating to the Pell series, the corresponding "Lucas"-like series is (2, 6, 14, 34, 82, 198, ...) such that herein, q = 2.414213... = (1 + sqrt(2)).
%C Then analogous to the previous set,
%C ...
%C 1/2 = q^(-1) + q^(-3) + q^(-5) + q^(-7) + ...
%C 1/6 = q^(-2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
%C ... (End)
%e From _Bruno Berselli_, Feb 10 2014: (Start)
%e Triangle begins:
%e 1;
%e 2, 3;
%e 3, 6, 5;
%e 4, 9, 10, 7;
%e 5, 12, 15, 14, 9;
%e 6, 15, 20, 21, 18, 11;
%e 7, 18, 25, 28, 27, 22, 13;
%e 8, 21, 30, 35, 36, 33, 26, 15;
%e 9, 24, 35, 42, 45, 44, 39, 30, 17;
%e 10, 27, 40, 49, 54, 55, 52, 45, 34, 19;
%e 11, 30, 45, 56, 63, 66, 65, 60, 51, 38, 21;
%e etc.
%e (End)
%t t[n_, k_] := If[n < k, 0, (2*k + 1)*(n - k + 1)]; Flatten[ Table[ t[n, k], {n, 0, 11}, {k, 0, n}]] (* _Robert G. Wilson v_, Jan 20 2005 *)
%o (PARI) T(n,k)=if(n<k,0,(2*k+1)*(n-k+1)) for(i=0,15, for(j=0,i,print1(T(i,j),","));print())
%Y Cf. A094728 (triangle generated by B*A), A000330.
%K nonn,tabl
%O 0,2
%A Lambert Klasen (lambert.klasen(AT)gmx.de) and _Gary W. Adamson_, Jan 19 2005
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