Template:Sequence of the Day for November 7

Intended for: November 7, 2012

Timetable

• First draft entered by Mitch Harris on July 29, 2011 ✓
• Draft reviewed by Alonso del Arte on April 4, 2012 ✓
• Draft approved by Daniel Forgues on November 7, 2012 ✓

Yesterday's SOTD * Tomorrow's SOTD

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A002620: Quarter-squares:

n   2    n   2 , n   ≥   0

. Equivalently,

n  2 4 , n   ≥   0

.

{ 0, 0, 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, 81, 90, 100... }

This sequence is fairly simple, just the interleaving of two already simple sequences, namely the squares (A000290) and the pronic numbers (A002378), i.e. the [even index/odd index] bisections of this sequence are

     

a (2 n)  =  n 2, n ≥ 0, a (2 n + 1)  =  n  (n + 1)  =  2  tn , n ≥ 0,

where

tn

is the

n

th triangular number.

The recurrences for the [even index/odd index] bisections are

A000290 (0) = 0; A000290 (n) = A000290 (n  −  1) + 2 n  −  1, n   ≥   1, and

A002378 (0) = 0; A002378 (n) = A002378 (n  −  1) + 2 n, n   ≥   1,

which leads to the simple recurrence

A002620 (0) = 0; A002620 (n) = A002620 (n  −  1) +

n   2

, n   ≥   1,

where

n   2

is given by A004526.
Suppose you try to fill an

n  ×  n

discrete grid, node by node, keeping what is covered as compact and simple as possible. If you start with a square covering

n  ×  n

nodes, you’ll add one column (of

n

nodes) covering

n  ×  (n + 1)

nodes, then add another row (of

n + 1

nodes), to now cover

(n + 1)  ×  (n + 1)

nodes. Covering nodes in this ‘herring-bone’ pattern generates the quarter squares.