%I #47 Jul 27 2023 11:57:10
%S 1,2,4,3,6,9,4,8,12,16,5,10,15,20,25,6,12,18,24,30,36,7,14,21,28,35,
%T 42,49,8,16,24,32,40,48,56,64,9,18,27,36,45,54,63,72,81,10,20,30,40,
%U 50,60,70,80,90,100,11,22,33,44,55,66,77,88,99,110,121,12,24,36,48,60,72,84
%N Triangle read by rows with the n-th row containing the first n multiples of n.
%C (Conjecture) Let N=2*n and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (see [Jeffery]) associated with N. Define the Chebyshev polynomials of the second kind by the recurrence U_0(x)=1, U_1(x)=2*x and U_r(x)=2*x*U_(r-1)(x)-U_(r-2)(x) (r>1). Define the column vectors V_(k-1)=(U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where B^T denotes the transpose of matrix B. Let S_N=[V_0,V_1,...,V_(n-1)] be the n X n matrix formed by taking the components of vector V_(k-1) as the entries in column k-1 (V_(k-1) gives the ordered spectrum of A_{N,k-1}). Let X_N=[S_N]^T*S_N, and let [X_N]_(i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then also T(n,k)=[X_N]_(k-1,k-1); that is, row n of the triangle is given by the main diagonal entries of X_N. Hence T(n,k) is the sum of squares T(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]=[V_(k-1)]^T*V_(k-1). - _L. Edson Jeffery_, Jan 20 2012
%C Conjecture that antidiagonal sums are A023855. - _L. Edson Jeffery_, Jan 20 2012
%C Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A066680 U {1}; this is the only one that contains its own row numbers only once. - _Peter Munn_, Dec 04 2019
%H Reinhard Zumkeller, <a href="/A075362/b075362.txt">Rows n = 1..150 of triangle, flattened</a>
%H L. E. Jeffery, <a href="/wiki/User:L._Edson_Jeffery/Unit-Primitive_Matrices">Unit-primitive matrices</a>
%F T(n,k) = n*k, 1 <= k <= n. - _Reinhard Zumkeller_, Mar 07 2010
%F T(n,k) = A050873(n,k)*A051173(n,k), 1 <= k <= n. - _Reinhard Zumkeller_, Apr 25 2011
%F T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+1-k,n-i), 1 <= k <= n. - _Mircea Merca_, Apr 11 2012
%F T(n,k) = A002260(n,k)*A002024(n,k) = (A215630(n,k)-A215631(n,k))/2, 1 <= k <= n. - _Reinhard Zumkeller_, Nov 11 2012
%F a(n) = A223544(n) - 1; a(n) = i*(t+1), where i = n - t*(t+1)/2, t = floor((-1 + sqrt(8*n-7))/2). - _Boris Putievskiy_, Jul 24 2013
%e Triangle begins:
%e 1;
%e 2, 4;
%e 3, 6, 9;
%e 4, 8, 12, 16;
%e 5, 10, 15, 20, 25;
%e 6, 12, 18, 24, 30, 36;
%p T(n,k):=piecewise(k<=n,sum(i*binomial(k,i)*binomial(n+1-k,n-i),i=1..k),k>n,0) # _Mircea Merca_, Apr 11 2012
%t Table[NestList[n+#&,n,n-1],{n,15}]//Flatten (* _Harvey P. Dale_, Jun 14 2022 *)
%o (Haskell)
%o a075362 n k = a075362_tabl !! (n-1) !! (k-1)
%o a075362_row n = a075362_tabl !! (n-1)
%o a075362_tabl = zipWith (zipWith (*)) a002260_tabl a002024_tabl
%o -- _Reinhard Zumkeller_, Nov 11 2012, Oct 04 2012
%Y A002411 gives the sum of the n-th row. A141419 is similarly derived.
%Y Cf. A066680, A223544.
%Y Cf. A003991 (square multiplication table).
%Y Main diagonal gives A000290.
%K nonn,tabl,easy
%O 1,2
%A _Amarnath Murthy_, Sep 20 2002
%E More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 20 2003