A078633 Smallest number of sticks of length 1 needed to construct n squares with sides of length 1.

4, 7, 10, 12, 15, 17, 20, 22, 24, 27, 29, 31, 34, 36, 38, 40, 43, 45, 47, 49, 52, 54, 56, 58, 60, 63, 65, 67, 69, 71, 74, 76, 78, 80, 82, 84, 87, 89, 91, 93, 95, 97, 100, 102, 104, 106, 108, 110, 112, 115, 117, 119, 121, 123, 125, 127, 130, 132, 134, 136, 138, 140, 142

Offset: 1

Author

Mambetov Timur and Takenov Nurdin (timur_teufel(AT)mail.ru), Dec 12 2002

Keywords

• nonn
• easy
• nice

Comments

A182834(a(n)) mod 2 = 0, or, where even terms occur in A182834. - Reinhard Zumkeller, Aug 05 2014

Examples

			a(2)=7 because we have following construction:
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Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Ralph H. Buchholz and Warwick De Launey, The square, the triangle and the hexagon, 1996.
• Douglas A. Torrance, Enumeration of planar Tangles, arXiv:1906.01541 [math.CO], 2019.
• Douglas A. Torrance, Enumeration of planar Tangles, Proc. Math. Sci. 130 (50) (2020).

Crossrefs

Cf. A027434, A027709, A135708, A182834.

Programs

Haskell

a078633 n = 2 * n + ceiling (2 * sqrt (fromIntegral n))
-- Reinhard Zumkeller, Aug 05 2014

Mathematica

Table[2n+Ceiling[2Sqrt[n]],{n,70}] (* Harvey P. Dale, Jun 20 2011 *)

PARI

a(n) = 2*n + ceil(2*sqrt(n)); \\ Michel Marcus, Mar 26 2018

Python

from math import isqrt
def A078633(n): return (m:=n<<1)+1+isqrt((m<<1)-1) # Chai Wah Wu, Jul 28 2022

Formula

a(n) = 2*n + ceiling(2*sqrt(n)) = 2*n + A027434(n).

a(n) = (4*n + A027709(n))/2. - Tanya Khovanova, Mar 07 2008