cp's OEIS Frontend

Products of two consecutive triangular numbers (A000217).

a(n) is the number of Lyndon words of length 4 on an n-letter alphabet. A Lyndon word is a primitive word that is lexicographically earliest in its cyclic rotation class. For example, a(2)=3 counts 1112, 1122, 1222. - David Callan, Nov 29 2007

For n >= 2 this is the second rightmost column of A163932. - Johannes W. Meijer, Oct 16 2009

Partial sums of A059270. - J. M. Bergot, Jun 27 2013

Using the integers, triangular numbers, and squares plot the points (A001477(n),A001477(n+1)), (A000217(n), A000217(n+1)), and (A000290(n),A000290(n+1)) to create the vertices of a triangle. One-half the area of this triangle = a(n). - J. M. Bergot, Aug 01 2013

a(n) is the Wiener index of the triangular graph T(n+1). - Emeric Deutsch, Aug 26 2013

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).

Row sums give A110450; central terms give A110451;

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

T(n,1) = A005449(n) for n>0;

T(n,2) = A005475(n) for n>1;

T(n,3) = A022265(n) for n>2;

T(n,4) = A022267(n) for n>3;

T(n,5) = A022269(n) for n>4;

T(n,6) = A022271(n) for n>5;

T(n,7) = A022263(n) for n>6;

T(n+1,n-1) = A059270(n) for n>1;

T(n,n-1) = A081436(n) for n>1;

T(n,n) = A085786(n).

The analog for sums of integers is A059270, and the analog for sums of triangular numbers is A222716. - Jonathan Sondow, Mar 07 2013

In 1879, Dostor gave formulas for all solutions -- see the Dickson link. - Jonathan Sondow, Jun 21 2014

Also: skip 1, take 0, skip 2, take 1, skip 3, take 2, ...

Integers for which the periodic part of the continued fraction for the square root of n begins with a 1. - Robert G. Wilson v, Nov 01 2001

a(n) belongs to the sequence if and only if a(n) > floor(sqrt(a(n))) * ceiling(sqrt(a(n))), i.e. a(n) in (k*(k+1),k^2), k >= 0. - Daniel Forgues, Apr 17 2011

Any integer between (k - 1/2)^2 and k^2 exclusive, for k > 1, is in this sequence. If we take this sequence and remove each term that is one more than the previous term, we obtain the central polygonal numbers (A002061). If instead we remove each term that is one less than the next term, we obtain numbers that are one less than squares (A005563). - Alonso del Arte, Dec 28 2013

The n+1 consecutive triangular numbers start with the A028387(n-2)-th triangular number A000217(n^2-n-1), while the n-1 consecutive triangular numbers start with the A000290(n)-th triangular number A000217(n^2).

Similar sums of consecutive integers are A059270.

Similar sums of consecutive squares are A059255.

Berselli points out that a(n) = 10*A024166(n-1) = A000292(n-1)*(3*n^2 - 2). Since a(n) is a sum of triangular numbers, 10=1+2+3+4 is the 4th triangular number, A024166 is a sum of cubes, and A000292 is a tetrahedral number, is there a geometric proof of Berselli's formula? (Compare Nelsen and Unal's "Proof Without Words: Runs of Triangular Numbers.") [Jonathan Sondow, Mar 04 2013]

See illustration in links.

The corresponding triangle with column sums is found in A251630. - Wolfdieter Lang, Dec 09 2014

sqrt((Sum_{k=0..n} 2*a(k)) + 1) = A056220(n+2). - Doug Bell, Mar 09 2009

Second leg of Pythagorean triangles with hypotenuse a square: A057769(n)^2 + a(n-1)^2 = A007204(n)^2. - Martin Renner, Nov 12 2011

Numbers which are both the sum of 2*n + 4 consecutive odd integers and the sum of the 2*n + 2 immediately higher consecutive odd integers. In general, let f(k,n) = 3*k^3*A000330(n). Then f(k,n) is both the sum of k*n + k consecutive terms from the arithmetic progression with first term A000217(k) and constant difference k and the immediately higher k*n terms from the same progression. When k = 1, f(k,n) = A059270(n). - Charlie Marion, Aug 23 2021

Frank Buss gave this as a puzzle; K. L. Metlov solved it, submitting his result in the J language created by Kenneth Iverson. The program given below is only five tokens long. J defines a series of three functions to be a "fork" defined by x (f g h ) y = (x f y) g (f h y) - a generalization of the usual mathematical practice of writing (f + g) y to mean (f y) + (g y). J also has a primitive "half" and has a dummy function "cap" whose purpose is to permit more forks to be written. 3 (* * +) 5 is thus (3 * 5) * (3 + 5) or 120. cap half 3 (* * +) 5 is thus 60.

This sequence is the dimensions of the various irreducible representations of SU(3). In the language of physics, the integers m and n are one more than the numbers of quarks or antiquarks, respectively, that label the representation. - Alex Meiburg, Dec 13 2020 =

Comment on the previous one: D(n, m) = f(m+1, n+1) = (n+1)*(m+1)*(n+m+2), for 0 <= m <= n, (given as array D(n,m) as example in A212331) is the dimension of the irreducible SU(3) multiplet (n, m), denoted also by D(n, m). The multiplet (m, n) is denoted also by a bar over D(n, m). The irreducuble tensor t(n, m) is symmetric in n upper indices from {1,2,3}, symmetric in m lower indices, and traceless in every pair of an upper and a lower index. See the Coleman reference for a derivation. - Wolfdieter Lang, Dec 18 2020