cp's OEIS Frontend

This is the fourth member, q=4, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A096940 (q=5), A096956 (q=6).

This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) := Sum_{m=0..n} a(n,m)*x^m is G(z,x) = g(z)/(1 - x*z*f(z)). Here: g(x) = (4-3*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (4-3*z)/(1-(1+x)*z).

The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022095(n-2), n >= 2, with n=1 value 4. [Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.]

T(2*n,n) = A029609(n) for n > 0, A029609 are the central terms of the Pascal (2,3) triangle A029600. - Reinhard Zumkeller, Apr 08 2012

This sequence provides the first differences of A254407 and the partial sums of A017473.

Also:

a(n) - n = A022269(n);

a(n) + n = n*(11*n+5)/2: 0, 8, 27, 57, 98, 150, 213, 287, ...;

a(n) - 2*n = A022268(n);

a(n) + 2*n = n*(11*n+7)/2: 0, 9, 29, 60, 102, 155, 219, 294, ...;

a(n) - 3*n = n*(11*n-3)/2: 0, 4, 19, 45, 82, 130, 189, 259, ...;

a(n) + 3*n = A211013(n);

a(n) - 4*n = A226492(n);

a(n) + 4*n = A152740(n);

a(n) - 5*n = A180223(n);

a(n) + 5*n = n*(11*n+13)/2: 0, 12, 35, 69, 114, 170, 237, 315, ...;

a(n) - 6*n = A051865(n);

a(n) + 6*n = n*(11*n+15)/2: 0, 13, 37, 72, 118, 175, 243, 322, ...;

a(n) - 7*n = A152740(n-1) with A152740(-1) = 0;

a(n) + 7*n = n*(11*n+17)/2: 0, 14, 39, 75, 122, 180, 249, 329, ...;

a(n) - n*(n-1)/2 = A168668(n);

a(n) + n*(n-1)/2 = A049453(n);

a(n) - n*(n+1)/2 = A202803(n);

a(n) + n*(n+1)/2 = A033580(n).

Table starts

.....0.....0.....0.....0.....0.....0.....0.....0....0....0...0...0..0..0.0

.....1.....2.....3.....4.....5.....6.....7.....8....9...10..11..12.13.14

.....3.....5.....7.....9....11....13....15....17...19...21..23..25.27

.....9....15....22....30....39....49....60....72...85...99.114.130

....36....57....82...111...144...181...222...267..316..369.426

...117...181...260...355...467...597...746...915.1105.1317

...411...632...912..1257..1673..2166..2742..3407.4167

..1452..2199..3158..4357..5825..7592..9689.12148

..5040..7593.10920.15146.20404.26835.34588

.17829.26706.38385.53379.72246.95590

From Paul Curtz, Mar 26 2011: (Start)

Successive A026741(n) * A026741(n+p):

p=0: 0, 1, 1, 9, 4, 25, 9, a(n),

p=1: 0, 1, 3, 6, 10, 15, 21, A000217,

p=2: 0, 3, 2, 15, 6, 35, 12, A142705,

p=3: 0, 2, 5, 9, 14, 20, 27, A000096,

p=4: 0, 5, 3, 21, 8, 45, 15, A171621,

p=5: 0, 3, 7, 12, 18, 25, 33, A055998,

p=6: 0, 7, 4, 27, 10, 55, 18,

p=7: 0, 4, 9, 15, 22, 30, 39, A055999,

p=8: 0, 9, 5, 33, 12, 65, 21, (see A061041),

p=9: 0, 5, 11, 18, 26, 35, 45, A056000. (End)

The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016

Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

The rows of this table and that in A098737 are related. Given a function f = n/( 1 + (1+n) mod(2) ), row n of A098737 can be derived from row n of T by multiplying the latter by f(n); row n of T can be derived from row n of A098737 by dividing the latter by f(n).

For the m-th level generalization of Catalan triangle T(n,k) = C(n+mk,k)*(n-k+1)/(n+(m-1)k+1); for n >= k+m: T(n,k) = T(n-m+1,k+1) - T(n-m,k+1); and T(n,n) = T(n+m-1,n-1) = C((m+1)n,n)/(mn+1).

Antidiagonals of convolution matrix of Table 1.5, p. 397, of Hoggatt and Bicknell. - Tom Copeland, Dec 25 2019

This number triangle results from the array A(n, m) = T(n+m+1) - T(n-1), with T = A000217, for n, m >= 1. For this array see the example by Bob Selcoe, in A111774 (but with rows continued). The present triangle is obtained by reading the array by upwards antidiagonals: T(n, k) = A(n+1-k, k). See also the Jul 09 2019 comment by Ralf Steiner with the formula c_k(n) (rows k >= 1, columns n >= 3), rewritten for A(n, m) = (m+2)*(2*n+m+1)/2, leading to T(n, k) = (k+2)*(2*n-k+3)/2.

Therefore this triangle is related to the problem of giving the numbers which are sums of at least three consecutive positive integers given as sequence A111774. It allows us to find the multiplicities for the numbers of A111774. They are given in A338428(n).

To obtain the multiplicity for number N (>= 6) from A111774 one has to consider only the triangle rows n = 1, 2, ..., floor((N-3)/3).

The row reversed triangle, considered by Bob Selcoe in A111774, is T(n, n-k+1) = T(n+2) - T(k-1), for n >= 1, and k=1, 2, ..., n.

This triangle contains no odd prime numbers and no exact powers 2^m, for m >= 0. This can be seen by considering the diagonal sequences D(d, k), for d >= 1, k >= 1 or the row sequences of the array A(n, m), for n >= 1 and m >= 1. The result is A(r+1, s-2) = s*(s + 2*r + 1)/2, for r >= 0 and s >= 3 (from the g.f. of the diagonals of T given below). This is also given in the Jul 09 2019 comment by Ralf Steiner in A111774. Therefore A(r+1, s-2) is a product of two numbers >= 2, hence not a prime. And in both cases (i) s/2 integer or (ii) (s + 2*r + 1)/2 integer not both numbers can be powers of 2 by simple parity arguments.

The previous comment means that each T(n, k) has at least one odd prime as a proper divisor.

A number N appears in this triangle, or in A111774, if and only if floor(N/2) - delta(N) >= 1, where delta(N) = A055034(N). For the sequence b(n) := floor(n/2) - delta(n), for n >= 2, see A219839(n), b(1) = -1. See a W. Lang comment in A111774 for the proof.

Pisot sequences E(4,5), P(4,5), T(4,5).

In spite of the definition, this is simply 1, 3, then numbers of the form n*(n+7)/2 (A055999). In other words, a(n) = (n-3)*(n+4)/2 for n >= 3. The proof is by induction: For n > 3, a(n-1) = (n-4)*(n+3)/2 is composite, so a(n) = a(n-1) + n = (n-3)*(n+4)/2. - Dean Hickerson, T. D. Noe, Paul C. Leopardi, Labos Elemer and others, Oct 04 2004

If a 2-set Y and a 3-set Z, having one element in common, are subsets of an n-set X then a(n) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 03 2007