cp's OEIS Frontend

Note that when starting from a(n)^2, equality holds between series of first n+1 and next n consecutive squares: a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2*n)^2; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Henry Bottomley, Jan 22 2001; with typos fixed by Zak Seidov, Sep 10 2015

a(n) = sum of second set of n consecutive even numbers - sum of the first set of n consecutive odd numbers: a(1) = 4-1, a(3) = (8+10+12) - (1+3+5) = 21. - Amarnath Murthy, Nov 07 2002

Partial sums of odd numbers 3 mod 4, that is, 3, 3+7, 3+7+11, ... See A001107. - Jon Perry, Dec 18 2004

If Y is a fixed 3-subset of a (2n+1)-set X then a(n) is the number of (2n-1)-subsets of X intersecting Y. - Milan Janjic, Oct 28 2007

More generally (see the first comment), for n > 0, let b(n,k) = a(n) + k*(4*n + 1). Then b(n,k)^2 + (b(n,k) + 1)^2 + ... + (b(n,k) + n)^2 = (b(n,k) + n + 1 + 2*k)^2 + ... + (b(n,k) + 2*n + 2*k)^2 + k^2; e.g., if n = 3 and k = 2, then b(n,k) = 47 and 47^2 + ... + 50^2 = 55^2 + ... + 57^2 + 2^2. - Charlie Marion, Jan 01 2011

Sequence found by reading the line from 0, in the direction 0, 10, ..., and the line from 3, in the direction 3, 21, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Nov 09 2011

a(n) is the number of positions of a domino in a pyramidal board with base 2n+1. - César Eliud Lozada, Sep 26 2012

Differences of row sums of two consecutive rows of triangle A120070, i.e., first differences of A016061. - J. M. Bergot, Jun 14 2013 [In other words, the partial sums of this sequence give A016061. - Leo Tavares, Nov 23 2021]

a(n)*Pi is the total length of half circle spiral after n rotations. See illustration in links. - Kival Ngaokrajang, Nov 05 2013

For corresponding sums in first comment by Henry Bottomley, see A059255. - Zak Seidov, Sep 10 2015

a(n) also gives the dimension of the simple Lie algebras B_n (n >= 2) and C_n (n >= 3). - Wolfdieter Lang, Oct 21 2015

With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for unsigned A130757, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 13 2015

Partial sums of squares with alternating signs, ending in an even term: a(n) = 0^2 - 1^2 +- ... + (2*n)^2, cf. Example & Formula from Berselli, 2013. - M. F. Hasler, Jul 03 2018

Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central peak and the largest Dyck path has a central valley, n > 0. (Cf. A237593.) - Omar E. Pol, Aug 28 2018

a(n) is the area of a triangle with vertices at (0,0), (2*n+1, 2*n), and ((2*n+1)^2, 4*n^2). - Art Baker, Dec 12 2018

This sequence is the largest subsequence of A000217 such that gcd(a(n), 2*n) = a(n) mod (2*n) = n, n > 0 up to a given value of n. It is the interleave of A033585 (a(n) is even) and A033567 (a(n) is odd). - Torlach Rush, Sep 09 2019

A generalization of Hasler's Comment (Jul 03 2018) follows. Let P(k,n) be the n-th k-gonal number. Then for k > 1, partial sums of {P(k,n)} with alternating signs, ending in an even term, = n*((k-2)*n + 1). - Charlie Marion, Mar 02 2021

Let U_n(H) = {A in M_n(H): A*A^H = I_n} be the group of n X n unitary matrices over the quaternions (A^H is the conjugate transpose of A. Note that over the quaternions we still have A*A^H = I_n <=> A^H*A = I_n by mapping A and A^H to (2n) X (2n) complex matrices), then a(n) is the dimension of its Lie algebra u_n(H) = {A in M_n(H): A + A^H = 0} as a real vector space. A basis is given by {(E_{st}-E_{ts}), i*(E_{st}+E_{ts}), j*(E_{st}+E_{ts}), k*(E_{st}+E_{ts}): 1 <= s < t <= n} U {i*E_{tt}, j*E_{tt}, k*E_{tt}: t = 1..n}, where E_{st} is the matrix with all entries zero except that its (st)-entry is 1. - Jianing Song, Apr 05 2021

Group the non-multiples of n as follows, e.g., for n = 4: (1,2,3), (5,6,7), (9,10,11), (13,14,15), ... Then a(n) is the sum of the members of the n-th group. Or, the sum of (n-1)successive numbers preceding n^2. - Amarnath Murthy, Jan 19 2004

Convolution of odds (A005408) and multiples of three (A008585). G.f. is the product of the g.f. of A005408 by the g.f. of A008585. - Graeme McRae, Jun 06 2006

Sums of rows of the triangle in A126890. - Reinhard Zumkeller, Dec 30 2006

Corresponds to the Wiener indices of C_{2n+1} i.e., the cycle on 2n+1 vertices (n > 0). - K.V.Iyer, Mar 16 2009

Also the product of the three numbers from A005843(n) up to A163300(n), divided by 8. - Juri-Stepan Gerasimov, Jul 26 2009

Partial sums of A033428. - Charlie Marion, Dec 08 2013

For n > 0, sum of multiples of n and (n+1) from 1 to n*(n+1). - Zak Seidov, Aug 07 2016

A generalization of Ianakiev's formula, a(n) = A005408(n)*A000217(n), follows. A005408(n+k)*A000217(n) is the sum of n+1 consecutive integers and, after skipping k integers, the sum of the n immediately higher consecutive integers. For example, for n = 3 and k = 2, 9*6 = 54 = 12+13+14+15 = 17+18+19. - Charlie Marion, Jan 25 2022

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]

a(n) is the smallest solution to m^n + (m+1)^n + ... + (m+k)^n = (m+k+1)^n + (m+k+2)^n + ... + (m+2*k)^n, or -n if no solution.

In 1879 Dostor gave all solutions for n = 2. In particular, a(2) = 25.

In 1906 Collignon proved that no solution exists for n = 3 and 4, so a(3) = -3 and a(4) = -4.

In 2013 Felten and Müller-Stach claimed to prove that no solution exists when n > 2, so if their proof is correct, a(n) = -n for n >= 3.

With n = 17 consecutive numbers we can start from k = 18 but also from k = 528. The sequence considers only the least number: a(17) = 18.

In general t = k + 2*(n-1) but sometimes it differs, e.g., for n = 17, 35, 39, 51, 93, 127, 382, etc.

The n-th row contains n numbers: n^2, n^2 + (n-1)^2, ..., n^2 + (n-1)^2 + ... + 1^2.

All numbers that appear in the table are listed in ascending order at A034705.

All numbers that appear twice or more are listed at A130052.

The left column is A000290 (the squares).

The top row is A000330 (the square pyramidal numbers).

The columns are A000290, A099776 (or a tail of A001844), a tail of A005918 or A120328, a tail of A027575, a tail of A027578, a tail of A027865, ...

The first two rows are A000330 and a tail of A168599, but subsequent rows are not currently in the OEIS, and are all tails of A000330 minus various constants.

The main diagonal is A050410.