A067368 - OEIS

%I #52 May 31 2024 05:48:26

%S 2,6,10,14,16,18,22,26,30,34,38,42,46,48,50,54,58,62,66,70,74,78,80,

%T 82,86,90,94,98,102,106,110,112,114,118,122,126,128,130,134,138,142,

%U 144,146,150,154,158,162,166,170,174,176,178,182,186,190,194,198,202,206

%N a(n) is the smallest positive even integer that cannot be expressed as the product of two or three previous terms (not necessarily distinct).

%C a(n+1) - a(n) = 2 or 4 for all n >= 1. See A067395 for the sequence of differences.

%C From _Jianing Song_, Sep 21 2018: (Start)

%C Numbers of the form 2^(3t+1)*s where s is an odd number.

%C Also positions of 1 in A191255. (End)

%C The asymptotic density of this sequence is 2/7. - _Amiram Eldar_, May 31 2024

%H Robert Israel, <a href="/A067368/b067368.txt">Table of n, a(n) for n = 1..10000</a>

%F Conjecture: a(n) = a(n-1) + 2 if (n = 2a(k) + k + 1) or (n = 2a(k) + k) for some k, otherwise a(n) = a(n-1) + 4. This has been confirmed for several hundred terms.

%F The above conjecture is correct because there are 2*(a(k+1)-a(k)) terms that are not divisible by 4 in the k-th interval which are determined by terms that are divisible by 4. For example, there are 2*(a(2)-a(1)) = 2*(6-2) = 8 terms between a(5) = 16 and a(14) = 48 because numbers of the form 2*s are always terms where s is an odd number. So first differences of a(n) determine the corresponding intervals and the formula above always holds. - _Altug Alkan_, Sep 24 2018

%F a(n) = 2*A191257(n) = A213258(n)/2. - _Jianing Song_, Sep 21 2018

%e 8 = 2*2*2, but 10 = 2*5 cannot be expressed with factors 2 and 6, so a(3) = 10.

%p N:= 1000:

%p A:= {seq(seq(2^(3*k+1)*s,s=1..N/2^(3*k+1),2),k=0..floor(log[2](N/2)/3))}:

%p sort(convert(A,list)); # _Robert Israel_, Jul 23 2019

%t t = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0, 2}, 2 -> {0, 3}, 3 -> {0, 1}}] &, {0}, 9] (* A191255 *)

%t Flatten[Position[t, 0]] (* A005408, the odds *)

%t a = Flatten[Position[t, 1]] (* this sequence *)

%t b = Flatten[Position[t, 2]] (* A213258 *)

%t a/2  (* A191257 *)

%t b/4  (* a/2 *)

%t (* _Clark Kimberling_, May 28 2011 *)

%o (PARI) isok(n) = valuation(n, 2)%3==1; \\ _Altug Alkan_, Sep 21 2018

%Y Cf. A067395, A067396, A191255, A191257, A213258.

%K nonn,easy

%O 1,1

%A Jeremiah K. Hower (jhower(AT)vt.edu), Jan 20 2002

%E Edited by _John W. Layman_, Jan 23 2002