cp's OEIS Frontend

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Let R be the array in A211790 and let R' be the array in A211793. Then R(k,n) + R'(k,n) = 3^(n-1). Moreover, (row k of R) =(row k of A211796) for k>2, by Fermat's last theorem; likewise, (row k of R')=(row k of A211799) for k>2.

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Generalizations: Suppose that b,c,d are nonzero integers, and let U(k,n) be the number of ordered triples (w,x,y) with all terms in {1,...,n} and b*w*k c*x^k+d*y^k, where the relation is one of these: <, >=, <=, >. What additional assumptions force the limiting row sequence to be essentially one of these: A002412, A000330, A016061, A174723, A051925?

In the following guide to related arrays and sequences, U(k,n) denotes the number of (w,x,y) as described in the preceding paragraph:

first 3 rows limiting row sequence

A211790: w^k < x^k+y^k A004068, A211635, A211650 A002412

A211793: w^k >= x^k+y^k A000292, A211636, A211651 A000330

A211796: w^k <= x^k+y^k A002413, A211634, A211650 A002412

A211799: w^k > x^k+y^k A000292, A211637, A211651 A000330

A211802: 2w^k < x^k+y^k A182260, A211800, A211801 A016061

A211805: 2w^k >= x^k+y^k A055232, A211803, A211804 A000330

A211808: 2w^k <= x^k+y^k A055232, A211806, A211807 A174723

A182259: 2w^k > x^k+y^k A182260, A211810, A211811 A051925

Consider the set B(n) = {1,2,3,...n}. Let a(0) = 0. Then a(n) = Sum [ b(i)^2 - b(j)^2] for all i, j = 1 to n, b(i) belongs to B(n). E.g., a(3) = (3^2-1^2) + (3^2-2^2) + (2^2-1^2) = 16. - Amarnath Murthy, Jun 01 2001

Partial sums of A016061. - J. M. Bergot, Jun 18 2013

For n >= 3, a(n-2) is the number of permutations of n symbols that 3-commute with an n-cycle (see A233440 for definition). - Luis Manuel Rivera Martínez, Feb 24 2014

a(n) is the sum of all pairs with repetitions allowed drawn from the set of triangular numbers from A000217(0) to A000217(n). This is similar to A027480 but uses triangular numbers instead of the integers. Example for n=2: 0+1, 0+3, 1+1, 1+3, 3+3 gives sum of 16 = a(2). - J. M. Bergot, Mar 23 2016

From Mircea Dan Rus, Jul 29 2020: (Start)

a(n) is the number of lattice rectangles (squares included) inside half of an Aztec diamond of order n. This shape is obtained by stacking n rows of consecutive unit lattice squares, with the centers of rows vertically aligned and consisting successively of 2n, 2n-2,..., 4, 2 squares. See below the representation for n=6.

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(End)

a(n-1) = (n+1)*binomial(n+1, 3) is the number of certain rectangles (squares included) in an n X n square filled with 1 X 1 squares. Divide the n X n square, for n >= 2, into two complementary staircases by the boundary consisting of 2*n length 1 edges. For n = 1 there is no boundary. See a A000332 figure in the Mircea Dan Rus comment for the staircase with basis length n = 4. The complementary staircase is upside down with basis length n-1 = 3. Then a(n-1) is the number of rectangles in the n X n square which have at least one border link in their interior. This counting is based on the binomial identity given in the formula section, using A096948 (for n=m), A000332(n+3) and A000332(n+2). - Wolfdieter Lang, Sep 22 2020

In the following guide to related sequences,

c(n) = Sum_{0

t(n) = Sum_{0

s(k).................c(n)........t(n)

k....................A000217.....A000292

k^2..................A016061.....A004320

k^3..................A206808.....A206809

k^4..................A206810.....A206811

k!...................A206816.....A206817

prime(k).............A152535.....A062020

prime(k+1)...........A185382.....A206803

2^(k-1)..............A000337.....A045618

k(k+1)/2.............A007290.....A034827

k-th quarter-square..A049774.....A206806