cp's OEIS Frontend

Lower triangular version is at A117501. - Ross La Haye, Apr 12 2006

Starts the same as A049862, the product of two distinct Fibonacci numbers. This sequence has an infinite number of consecutive terms that are consecutive numbers (such as 15 and 16) because fib(k)*fib(k+3) and fib(k+1)*fib(k+2) differ by one for all k >= 0.

It follows from Carmichael's theorem that if u and v are finite sets of Fibonacci numbers such that (product of all the numbers in u) = (product of all the numbers in v), then u = v. The same holds for many other 2nd order linear recurrence sequences with constant coefficients. In the following guide to related "distinct product sequences", W = Wythoff array, A035513:

base sequence distinct-product sequence

A000045 (Fibonacci) A160009

A000032 (Lucas, without 2) A274280

A000032 (Lucas, with 2) A274281

A000285 (1,4,5,...) A274282

A022095 (1,5,6,...) A274283

A006355 (2,4,6,...) A274284

A013655 (2,5,7,...) A274285

A022086 (3,6,9,...) A274191

row 2 of W: (4,7,11,...) A274286

row 3 of W: (6,10,16,...) A274287

row 4 of W: (9,15,24,...) A274288

- Clark Kimberling, Jun 17 2016

Literally the same as A013655(n+1), since A001060(-1) = A013655(0) = 3. - Eric W. Weisstein, Jun 30 2017

Used by the Sofia Gubaidulina and other composers. - Ian Stewart, Jun 07 2012

From a(2) on, sums of five consecutive Fibonacci numbers; the subset of primes is essentially in A153892. - R. J. Mathar, Mar 24 2010

Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (is this A001175?). - R. J. Mathar, Aug 10 2012

Also the number of independent vertex sets and vertex covers in the (n+1)-pan graph. - Eric W. Weisstein, Jun 30 2017

From Wajdi Maaloul, Jun 10 2022: (Start)

For n > 0, a(n) is the number of ways to tile the figure below with squares and dominoes (a strip of length n+1 that contains a vertical strip of height 3 in its second tile). For instance, a(4) is the number of ways to tile this figure (of length 5) with squares and dominoes.

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(End)

This is the third member, q=3, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with T(0,0)=2, not 1).

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} T(n,m)*x^m is G(z,x) = g(z)/(1-x*z*f(z)). Here: g(x) = (3-2*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (3-2*z)/(1-(1+x)*z).

The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} T(n-1-k,k) = A000285(n-2), n>=2, with n=1 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Central terms: T(2*n,n) = A028329(n) = A100320(n) for n > 0, A028329 are the central terms of triangle A028326. - Reinhard Zumkeller, Apr 08 2012

Let P be Pascal's triangle, A007318 and R the Riordan array, A097805. Then Pascal triangle (1,q) = ((q-1) * R) + P. Example: Pascal triangle (1,3) = (2 * R) + P. - Gary W. Adamson, Sep 12 2015

The array F(4;n,m) gives in the columns m >= 1 the figurate numbers based on A016813, including the hexagonal numbers A000384 (see the W. Lang link).

This is the fourth member, d=4, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653 and A093560, for d=1..3.

This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). 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) = (1+3*z)/(1-(1+x)*z).

The SW-NE diagonals give A000285(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

The old name was: The Pibonacci numbers (a Fibonacci-type sequence): each term is the sum of the two previous terms.

The 6th row in the Wythoff array begins with the 6th term of the sequence (14, 23, 37, 60, 97, 157, ...). a(n) = f(n-3) + f(n+2) for the Fibonacci numbers f(n) = f(n-1) + f(n-2); f(0) = 0, f(1) = 1.

(a(2*k), a(2*k+1)) give for k >= 0 the proper positive solutions of one of two families (or classes) of solutions (x, y) of the indefinite binary quadratic form x^2 + x*y - y^2 of discriminant 5 representing 11. The other family of such solutions is given by (x2, y2) = (b(2*k), b(2*k+1)) with b = A013655. See the formula in terms of Chebyshev S polynomials S(n, 3) = A001906(n+1) below, which follows from the fundamental solution (3, 1) by applying positive powers of the automorphic matrix given in a comment in A013655. See also A089270 with the Alfred Brousseau link with D = 11. - Wolfdieter Lang, May 28 2019

a(0,k) = A000045(k-1); a(1,k) = A000032(k); a(2,k) = A000285(k+1).

a(n,1) = a(n-1,1) + a(n-1,3) for n > 0; a(n,1) = A001047(n) = 2^(2n) - A083324(n); a(n,2) = A000244(n) = 2^(2n) - A005061(n); a(n,3) = 2a(n-1,4) for n > 0; a(n,3) = A027649(n); a(n,4) = A083313(n+1); a(n,5) = A084171(n+1).

Sum[a(n-k,k), {k,0,n}] = A098703(n+1), antidiagonal sums.

Let R, S and T be binary relations on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xRy if x is a subset of y or y is a subset of x, xSy if x is a subset of y and xTy if x is a proper subset of y. Then a(n,3) = |R|, a(n,2) = |S| and a(n,1) = |T|. Note that a binary relation W on P(A) can be defined also such that for every element x, y of P(A) xWy if x is a proper subset of y and there are no z in P(A) such that x is a proper subset of z and z is a proper subset of y. A090802(n,1) = |W|. Also, a(n,0) = |P(A)|.

Binomial transform of A000285. - R. J. Mathar, Oct 26 2011

In a row of the Wythoff array, either every two consecutive terms are relatively prime or else no two consecutive terms are relatively prime. In the first case, we call the row primitive; otherwise, the row is an integer multiple of a tail of a preceding row. Conjectures: the maximal number of consecutive primitive rows is 5, and the limiting proportion of primitive rows exists and is approximately 0.608.