A004738 Concatenation of sequences (1,2,...,n-1,n,n-1,...,2) for n >= 2.
1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 10, 9
Offset: 1
Examples
From _Boris Putievskiy_, Jan 24 2013: (Start) The start of the sequence as table: 1, 2, 3, 4, 5, 6, 7, ... 2, 1, 2, 3, 4, 5, 6, ... 3, 2, 1, 2, 3, 4, 5, ... 4, 3, 2, 1, 2, 3, 4, ... 5, 4, 3, 2, 1, 2, 3, ... 6, 5, 4, 3, 2, 1, 2, ... 7, 6, 5, 4, 3, 2, 1, ... ... The start of the sequence as triangle array read by rows: 1; 2, 1, 2; 3, 2, 1, 2, 3; 4, 3, 2, 1, 2, 3, 4; 5, 4, 3, 2, 1, 2, 3, 4, 5; 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6; 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7; ... Row number r contains 2*r - 1 numbers: r, r-1, ..., 1, 2, ..., r. (End)
References
- F. Smarandache, "Numerical Sequences", University of Craiova, 1975; [ See Arizona State University, Special Collection, Tempe, AZ, USA ].
Links
- Boris Putievskiy, Transformations [Of] Integer Sequences And Pairing Functions, arXiv preprint arXiv:1212.2732 [math.CO], 2012.
- F. Smarandache, Collected Papers, Vol. II
- F. Smarandache, Sequences of Numbers Involved in Unsolved Problems.
- Eric Weisstein's World of Mathematics, Smarandache Sequences
Programs
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Maple
A004738 := proc(n) local tri ; tri := floor(sqrt(n)+1/2) ; tri+1-abs(n-1-tri^2) ; end proc: seq(A004738(n),n=1..30) ; #R. J. Mathar, Feb 14 2019
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Mathematica
row[n_] := Range[n, 1, -1] ~Join~ Range[2, n]; Array[row, 10] // Flatten (* Jean-François Alcover, Apr 19 2020 *)
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PARI
a(n)= floor(sqrt(n)+1/2)+1-abs(n-1-(floor(sqrt(n)+1/2)-1/2)^2)
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Python
from math import isqrt def A004738(n): return abs((t:=isqrt(n-1))*(t+1)-n+1)+1 # Chai Wah Wu, Mar 01 2025
Formula
a(n) = floor(sqrt(n) + 1/2) + 1 - abs(n - 1 - (floor(sqrt(n) + 1/2))^2). - Benoit Cloitre, Feb 08 2003
From Boris Putievskiy, Jan 24 2013: (Start)
For the general case, a(n) = m*v + (2*v-1)*(t*t-n) + t, where t = floor(sqrt(n) - 1/2) + 1 and v = floor((n-1)/t) - t + 1.
For m=2, a(n) = 2*v + (2*v-1)*(t*t-n) + t, where t = floor(sqrt(n) - 1/2) + 1 and v = floor((n-1)/t) - t + 1. (End)
Extensions
More terms from Patrick De Geest, Jun 15 1998
Comments