A175498 a(1)=1. a(n) = the smallest positive integer not occurring earlier such that a(n)-a(n-1) doesn't equal a(k)-a(k-1) for any k with 2 <= k <= n-1.

1, 2, 4, 3, 6, 10, 5, 11, 7, 12, 9, 16, 8, 17, 15, 23, 13, 24, 18, 28, 14, 26, 19, 32, 20, 34, 21, 36, 25, 41, 22, 39, 30, 48, 27, 46, 29, 49, 31, 52, 37, 59, 33, 56, 40, 64, 35, 60, 38, 65, 42, 68, 43, 71, 44, 73, 45, 75, 51, 82, 47, 79, 112, 50, 84, 53, 88, 54, 90, 57, 94, 55, 93, 61, 100, 58, 98, 62, 103, 63, 105, 67

Offset: 1

Author

Leroy Quet, May 31 2010

Keywords

• nonn
• nice

Comments

This sequence is a permutation of the positive integers.

a(n+1)-a(n) = A175499(n).

Conjecture: the lexicographically earliest permutation of {1,2,...n} for which differences of adjacent numbers are all distinct (cf. A131529) has, for n-->infinity, this sequence as its prefix. - Joerg Arndt, May 27 2012

Links

• Joerg Arndt and Reinhard Zumkeller, Table of n, a(n) for n = 1..10000, first 1122 terms from Joerg Arndt
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A081145, A175499, A257465 (inverse), A257883, A131388, A131389, A257705.

Programs

Haskell

import Data.List (delete)
a175498 n = a175498_list !! (n-1)
a175498_list = 1 : f 1 [2..] [] where
   f x zs ds = g zs where
     g (y:ys) | diff `elem` ds = g ys
              | otherwise      = y : f y (delete y zs) (diff:ds)
              where diff = y - x
-- Reinhard Zumkeller, Apr 25 2015

Mathematica

a[1] = 1; d[1] = 0; k = 1; z = 10000; zz = 120;
A[k_] := Table[a[i], {i, 1, k}]; diff[k_] := Table[d[i], {i, 1, k}];
c[k_] := Complement[Range[-z, z], diff[k]];
T[k_] := -a[k] + Complement[Range[z], A[k]];
Table[{h = Min[Intersection[c[k], T[k]]], a[k + 1] = a[k] + h, d[k + 1] = h, k = k + 1}, {i, 1, zz}];
u = Table[a[k], {k, 1, zz}]  (* Clark Kimberling, May 13 2015 *)

Python

A175498_list, l, s, b1, b2 = [1,2], 2, 3, set(), set([1])
for n in range(3, 10**5):
    i = s
    while True:
        if not (i in b1 or i-l in b2):
            A175498_list.append(i)
            b1.add(i)
            b2.add(i-l)
            l = i
            while s in b1:
                b1.remove(s)
                s += 1
            break
        i += 1 # Chai Wah Wu, Dec 15 2014

Extensions

More terms from Sean A. Irvine, Jan 27 2011