A260488 Numbers of the form 2^m * (6k + 1) for m, k >= 0, and 0.

0, 1, 2, 4, 7, 8, 13, 14, 16, 19, 25, 26, 28, 31, 32, 37, 38, 43, 49, 50, 52, 55, 56, 61, 62, 64, 67, 73, 74, 76, 79, 85, 86, 91, 97, 98, 100, 103, 104, 109, 110, 112, 115, 121, 122, 124, 127, 128, 133, 134, 139, 145, 146, 148, 151, 152, 157

Offset

0,3

Comments

Alternate definition: starting with a(0) = 0, include 2n if n is in the sequence, and include 2n+1 if no two previous terms sum to 2n+1.

It suffices to prove this for odd n. If n == 3(6), n-2 == 1 (mod 6); if n == 5 (mod 6), n-4 == 1 (mod 6). However, if n == 1 (mod 6), any even k in the sequence, 0 < k < n, will have k !== 0 (mod 3), and so n-k != 1 (mod 3), so it is not in the sequence; thus n must be.

Every nonnegative integer is the sum of two members of this sequence; every positive integer is the sum of two distinct members of this sequence. For odd n, this is by the construction in the alternate definition; and for even n, by induction n/2 = i + j, and so n = 2i + 2j.

It follows that:

* No member of this sequence except 0 is a multiple of 3.

* The sequence has a density of 1/3.

* The difference between consecutive terms is always one of {1, 2, 3, 5, 6}, and each of these occurs infinitely often, with 1 having density 1/3 and the others having density 1/6.

* The sequence is closed under multiplication.

* The primes in the sequence are A045375.

Links

Franklin T. Adams-Watters, Table of n, a(n) for n = 0..10000

Formula

n is in the sequence if and only if n = 0 or A000265(n) == 1 (mod 6). [Clarified by Peter Munn, Jun 11 2021]

n is in the sequence if n = 0 or b(n) is nonzero where b = A113448, A115235, or A123863. - Michael Somos, Jul 29 2015

Example

Using the alternate definition:
1 is in the sequence because it is not the sum of 2 elements from {0}.
2 is in the sequence because 2 = 2*1, and 1 is in the sequence.
3 is not in the sequence because 3 = 1 + 2, and 1 and 2 are in the sequence.
6 is not in the sequence because 6 = 2*3, and 3 is not in the sequence.

Maple

N:= 1000: # to get all terms <= N
sort([0, seq(seq(2^m*(6*k+1), k = 0 .. floor((N/2^m - 1)/6)), m = 0 .. ilog2(N))]);  # Robert Israel, Aug 25 2015

Mathematica

mx=160; Join[{0}, Sort@Flatten@Table[2^m*(6k+1), {m, 0, Log2[mx]}, {k, 0, mx/(6*2^m)}]] (* Robert G. Wilson v, Aug 16 2015 *)

Prog

(PARI) alist(n) = my(r=vector(n), j, k); r[1]=0; j=1; while(j<n, k++; if(k\2^valuation(k, 2)%6==1, r[j++]=k)); r
(PARI) alim(n)={my(p=1, p2=p, r, j);
  for(k=1, n,
    if(if(k%2==0, polcoeff(p, k\2), polcoeff(p2, k)==0), p+=x^k; p2+=x^k*p));
  r=vector(subst(p, x, 1)); for(k=0, n, if(polcoeff(p, k), r[j++]=k)); r}

Crossrefs

Cf. A126684, A260489, primes A045375, A000265.

Cf. A113448, A115235, A123863.

Sequence in context: A353579 A236217 A320628 * A190808 A359727 A018614

Adjacent sequences: A260485 A260486 A260487 * A260489 A260490 A260491

Keyword

nonn,nice

Author

Franklin T. Adams-Watters, Jul 27 2015

Status

approved