cp's OEIS Frontend

T(n,k) + T(n,n-k) = A014105(n);

row sums give A059270; Sum_{k=0..n-1} T(n,k) = A000578(n);

central terms give A007742; T(2*n+1,n) = A016754(n);

T(n,0) = A000217(n);

T(n,1) = A000096(n) for n > 0;

T(n,2) = A055998(n) for n > 1;

T(n,3) = A055999(n) for n > 2;

T(n,4) = A056000(n) for n > 3;

T(n,5) = A056115(n) for n > 4;

T(n,6) = A056119(n) for n > 5;

T(n,7) = A056121(n) for n > 6;

T(n,8) = A056126(n) for n > 7;

T(n,10) = A101859(n-1) for n > 9;

T(n,n-3) = A095794(n-1) for n > 2;

T(n,n-2) = A045943(n-1) for n > 1;

T(n,n-1) = A000326(n) for n > 0;

T(n,n) = A005449(n).

Complementary with A130481 regarding triangular numbers, in that A130481(n) + 3*a(n) = n(n+1)/2 = A000217(n).

Apart from offset, the same as A062781. - R. J. Mathar, Jun 13 2008

Apart from offset, the same as A001840. - Michael Somos, Sep 18 2010

The sum of any three consecutive terms is a triangular number. - J. M. Bergot, Nov 27 2014

Maximum dimension of Euclidean spaces which suffice for every smooth compact Riemannian n-manifold to be realizable as a sub-manifold. - comment edited by Gene Ward Smith, Jan 15 2017

This is a subsequence of A045943. - Michel Marcus, Apr 26 2014

As a rectangle, the accumulation array of A051340.

From Clark Kimberling, Feb 05 2011: (Start)

Here all the weights are divided by two where they aren't in Cahn.

As a rectangle, A141419 is in the accumulation chain

... < A051340 < A141419 < A185874 < A185875 < A185876 < ...

(See A144112 for the definition of accumulation array.)

row 1: A000027

col 1: A000217

diag (1,5,...): A000326 (pentagonal numbers)

diag (2,7,...): A005449 (second pentagonal numbers)

diag (3,9,...): A045943 (triangular matchstick numbers)

diag (4,11,...): A115067

diag (5,13,...): A140090

diag (6,15,...): A140091

diag (7,17,...): A059845

diag (8,19,...): A140672

(End)

Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011

Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and 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 T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column 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 t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: 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]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012

Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013

n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014

T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014

A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016

Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018

From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)

A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).

The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.

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 A270877; this is the only one that contains its own row numbers only once.

(End)

If Y is a 5-subset of an n-set X then, for n >= 5, a(n-4) is equal to the number of 5-subsets of X having exactly three elements in common with Y. Y is a 5-subset of an n-set X then, for n >= 6, a(n-6) is the number of (n-5)-subsets of X having exactly two elements in common with Y. - Milan Janjic, Dec 28 2007

Also sequence found by reading the line from 0, in the direction 0, 10, ... and the same line from 0, in the direction 0, 30, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. Axis perpendicular to A195148 in the same spiral. - Omar E. Pol, Sep 18 2011

Also the number of 6-cycles in the (n+5)-triangular honeycomb acute knight graph. - Eric W. Weisstein, Jun 25 2017

Triangle read by rows, T(n,k) = A000217(n) - A000217(k), 0 <= k < n. - Philippe Deléham, Mar 07 2013

Subtriangle of triangle in A049780. - Philippe Deléham, Mar 07 2013

No primes and all composite numbers (except 2^x) are generated after the first two columns of the square array for this sequence. In other words, no primes and all composites except 2^x are generated when m-n >= 2. - Bob Selcoe, Jun 18 2013

Diagonal sums in the square array equal partial sums of squares (A000330). - Bob Selcoe, Feb 14 2014

From Bob Selcoe, Oct 27 2014: (Start)

The following apply to the triangle as a square array read by rows unless otherwise specified (see Table link);

Conjecture: There is at least one prime in interval [T(n,k), T(n,k+1)]. Since T(n,k+1)/T(n,k) decreases to (k+1)/k as n increases, this is true for k=1 ("Bertrand's Postulate", first proved by P. Chebyshev), k=2 (proved by El Bachraoui) and k=3 (proved by Loo).

Starting with T(1,1), The falling diagonal of the first 2 numbers in each column (read by column) are the generalized pentagonal numbers (A001318). That is, the coefficients of T(1,1), T(2,1), T(2,2), T(3,2), T(3,3), T(4,3), T(4,4) etc. are the generalized pentagonal numbers. These are A000326 and A005449 (Pentagonal and Second pentagonal numbers: n*(3*n+1)/2, respectively), interweaved.

Let D(n,k) denote falling diagonals starting with T(n,k):

Treating n as constant: pentagonal numbers of the form n*k + 3*k*(k-1)/2 are D(n,1); sequences A000326, 005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542 are formed by n = 1 through 12, respectively.

Treating k as constant: D(1,k) are (3*n^2 + (4k-5)*n + (k-1)*(k-2))/2. When k = 2(mod3), D(1,k), is same as D(k+1,1) omitting the first (k-2)/3 numbers in the sequences. So D(1,2) is same as D(3,1); D(1,5) is same as D(6,1) omitting the 6; D(1,8) is same as D(9,1) omitting the 9 and 21; etc.

D(1,3) and D(1,4) are sequences A095794 and A140229, respectively.

(End)