A255003 Lexicographically earliest permutation of positive integers such that a(a(n)+a(n+1)) is even for all n.

1, 2, 4, 3, 5, 6, 8, 10, 7, 9, 12, 11, 13, 14, 15, 16, 18, 20, 17, 19, 22, 21, 24, 26, 23, 25, 28, 27, 30, 29, 32, 31, 33, 34, 35, 36, 38, 40, 37, 39, 42, 41, 44, 43, 46, 45, 47, 48, 50, 52, 49, 51, 54, 53, 56, 55, 58, 57, 60, 59, 62, 61, 64, 66, 63, 65, 68, 67, 70, 69, 72, 71, 73

Offset: 1

Author

Eric Angelini and M. F. Hasler, Feb 11 2015

Keywords

• nonn

Comments

From Eric Angelini, Feb 11 2015: (Start)

Let S denote this sequence. Then:

a) take two adjacent integers x and y in S

b) let (x + y) = z

c) a(z) is even.

S is extended with the smallest integer not yet in S and not leading to a contradiction.

Additional remarks:

The sequence T where a(a(n)+a(n+1)) is always odd is simply A000027.

But if we force a(1)=2 we then get again a permutation of A000027:

U = 2, 1, 3, 5, 4, 6, 7, 9, 11, 13, 8, 10, 15, 12, ,14, 17, 16, 19, 18, 21, 23, 20, 22, 25, 27, 29, 31, 24, 26, 28, 33, ... (A256210).

The sequence V where a(a(n)+a(n+1)) is always prime is also a permutation of A000027:

V = 1, 2, 3, 4, 5, 6, 7, 8, 11, 9, 13, 10, 17, 12, 19, 14, 15, 23, 29, 31, 16, 37, 41, 18, ... (A255004).

(End)

At least for the first 73 elements, successive blocks of numbers of size 2^m for various m>=0 each form permutations of some set of consecutive positive integers. We see blocks [1], [2], [4, 3], ..., [8, 10, 7, 9], ..., [62, 61, 64, 66, 63, 65, 68, 67]. If b(0) is the first element in such a block and b(2^m-1) the last, then for 0 <= i <= 2^m-1, b(i) + b(2^m-i-1) is constant. For example, in the latter block, 62 + 67 = 61 + 68 = 64 + 65 = 66 + 63, etc. - David A. Corneth, Mar 22 2015

Examples

			Checking the definition:
n = 1  2  3  4  5  6  7   8  9  10 11  12  13  14  15  16  17  18  19  20  21  22 ...
S = 1, 2, 4, 3, 5, 6, 8, 10, 7, 9, 12, 11, 13, 14, 15, 16, 18, 20, 17, 19, 22, 21,...
for n=1 then a(1)=1 and a(2)=2 and a(sum) reads a(1+2) reads a(3) which is 4 (even);
for n=2 then a(2)=2 and a(3)=4 and a(sum) reads a(2+4) reads a(6) which is 6 (even);
for n=3 then a(3)=4 and a(4)=3 and a(sum) reads a(4+3) reads a(7) which is 8 (even);
for n=4 then a(4)=3 and a(5)=5 and a(sum) reads a(3+5) reads a(8) which is 10 (even);
... etc.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A000027, A256210, A255004.

Programs

Maple

N:= 100: # to get a(n) for n <= N
maxodd:= 1:
maxeven:= 0:
a[1]:= 1:
needeven:= {}:
for n from 2 to N do
  if member(n,needeven) or maxeven < maxodd then
     a[n]:= maxeven + 2;
     maxeven:= a[n];
  else
     a[n]:= maxodd + 2;
     maxodd:= a[n];
  fi;
  needeven:= needeven union {a[n-1]+a[n]};
od:
seq(a[n],n=1..N); # Robert Israel, Mar 26 2015

Mathematica

M = 100;
maxodd = 1;
maxeven = 0;
a[1] = 1;
needeven = {};
For[n = 2, n <= M, n++, If[ MemberQ[needeven, n] || maxeven < maxodd, a[n] = maxeven + 2; maxeven = a[n], a[n] = maxodd + 2; maxodd = a[n]]; needeven = needeven ~Union~ {a[n-1] + a[n]}];
Array[a, M] (* Jean-François Alcover, Apr 30 2019, after Robert Israel *)

PARI

{a=vector(100,i,1); u=[1]/* used numbers beyond u[1] */; for(n=2,#a, if( a[n] < 0, a[n]=u[1]-u[1]%2; while(setsearch(u,a[n]+=2),), a[n]=u[1]; while(setsearch(u,a[n]++),)); u=setunion(u,[a[n]]); while( #u>1 && u[2]==u[1]+1, u=u[2..#u]); a[n]+a[n-1]>#a || a[a[n]+a[n-1]]=-1)}