

A075362


Triangle read by rows with the nth row containing the first n multiples of n.


13



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, 42, 49, 8, 16, 24, 32, 40, 48, 56, 64, 9, 18, 27, 36, 45, 54, 63, 72, 81, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 12, 24, 36, 48, 60, 72, 84
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OFFSET

1,2


COMMENTS

(Conjecture) Let N=2*n and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n1} be the n X n unitprimitive 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_(r1)(x)U_(r2)(x) (r>1). Define the column vectors V_(k1)=(U_(k1)(cos(Pi/N)), U_(k1)(cos(3*Pi/N)), ..., U_(k1)(cos((2*n1)*Pi/N)))^T, where B^T denotes the transpose of matrix B. Let S_N=[V_0,V_1,...,V_(n1)] be the n X n matrix formed by taking the components of vector V_(k1) as the entries in column k1 (V_(k1) gives the ordered spectrum of A_{N,k1}). 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,...,n1}. Then also T(n,k)=[X_N]_(k1,k1); 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_(k1)(cos((2*m1)*Pi/N)))^2]=[V_(k1)]^T*V_(k1).  L. Edson Jeffery, Jan 20 2012
Conjecture that antidiagonal sums are A023855.  L. Edson Jeffery, Jan 20 2012
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


LINKS

Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened
L. E. Jeffery, Unitprimitive matrices


FORMULA

T(n,k) = n*k, 1 <= k <= n.  Reinhard Zumkeller, Mar 07 2010
T(n,k) = A050873(n,k)*A051173(n,k), 1 <= k <= n.  Reinhard Zumkeller, Apr 25 2011
T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+1k,ni), 1 <= k <= n.  Mircea Merca, Apr 11 2012
T(n,k) = A002260(n,k)*A002024(n,k) = (A215630(n,k)A215631(n,k))/2, 1 <= k <= n.  Reinhard Zumkeller, Nov 11 2012
a(n) = A223544(n)  1; a(n) = i*(t+1), where i = n  t*(t+1)/2, t = floor((1 + sqrt(8*n7))/2).  Boris Putievskiy, Jul 24 2013


EXAMPLE

Triangle begins:
1;
2, 4;
3, 6, 9;
4, 8, 12, 16;
5, 10, 15, 20, 25;
6, 12, 18, 24, 30, 36;


MAPLE

T(n, k):=piecewise(k<=n, sum(i*binomial(k, i)*binomial(n+1k, ni), i=1..k), k>n, 0) # Mircea Merca, Apr 11 2012


PROG

(Haskell)
a075362 n k = a075362_tabl !! (n1) !! (k1)
a075362_row n = a075362_tabl !! (n1)
a075362_tabl = zipWith (zipWith (*)) a002260_tabl a002024_tabl
 Reinhard Zumkeller, Nov 11 2012, Oct 04 2012


CROSSREFS

A002411 gives the sum of the nth row. A141419 is similarly derived.
Cf. A066680, A223544.
Cf. A003991 (square multiplication table).
Sequence in context: A077583 A153125 A139413 * A110749 A077529 A143516
Adjacent sequences: A075359 A075360 A075361 * A075363 A075364 A075365


KEYWORD

nonn,tabl,easy


AUTHOR

Amarnath Murthy, Sep 20 2002


EXTENSIONS

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 20 2003


STATUS

approved



