A249990 Start with the natural numbers, reverse the order in each pair, skip one pair, reverse the order in each triple, skip one triple, and so on.
2, 1, 6, 3, 4, 12, 5, 8, 7, 16, 9, 10, 13, 14, 22, 15, 18, 11, 24, 17, 32, 25, 26, 19, 20, 23, 30, 48, 31, 38, 21, 28, 27, 34, 33, 52, 35, 42, 29, 36, 39, 40, 49, 50, 58, 51, 54, 41, 44, 37, 60, 43, 66, 53, 84, 67, 68, 61, 62, 45, 46, 55, 56, 59, 76, 94, 77, 90
Offset: 1
Keywords
Examples
Start with: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ... After the first step: 2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14, 13, 16, 15, 18, ... After the 2nd step: 2, 1, 6, 3, 4, 7, 8, 5, 12, 9, 10, 13, 14, 11, 18, 15, 16, ... After the 3rd step: 2, 1, 6, 3, 4, 12, 5, 8, 7, 14, 13, 10, 9, 16, 15, 18, 11, ...
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- Popular Computing (Calabasas, CA), Problems 195 and 196, Vol. 5 (No. 55, 1977), annotated and scanned copy of page PC55-4. See Problem 195.
- Index entries for sequences that are permutations of the natural numbers
Programs
-
Haskell
a249990 n = a249990_list !! (n-1) a249990_list = f 2 [1..] where f k xs = reverse ys ++ f (k + 1) (g zs) where g us = reverse vs ++ g ws where (vs, ws) = splitAt k us (ys, zs) = splitAt k xs -- Reinhard Zumkeller, Dec 17 2014 -
Python
TOP = 100 a = list(range(TOP)) for step in range(2,TOP): numBlocks = (len(a)-1) // step if numBlocks==0: break a = a[:(1+numBlocks*step)] for pos in range(1,len(a),step): a[pos:pos+step] = a[pos+step-1:pos-1:-1] for i in range(1, step+1): print(str(a[i]), end=',') a[1:] = a[step+1:]
Comments