A262446 - OEIS

%I #14 Jan 27 2024 10:33:15

%S 0,0,0,1,2,1,2,2,3,3,1,3,4,2,3,2,3,2,4,3,1,2,3,3,6,4,3,2,4,4,4,3,5,4,

%T 2,5,5,4,6,4,5,6,6,4,5,5,3,5,3,6,6,5,4,1,4,5,9,5,3,7,5,3,5,5,3,8,4,5,

%U 3,7,5,8,5,7,6,6,7,5,6,5,7,4,8,6,6,6,2,5,4,11,5,3,5,7,7,7,9,5,8,5

%N Number of ways to write A262439(n) = A262439(k) + A262439(m) with 0 < k < m < n.

%C Conjecture: a(n) > 0 for all n > 3, and a(n) = 1 only for n = 4, 6, 11, 21, 54, 253, 325.

%C This is slightly stronger than part (ii) of the conjecture in A262439.

%C I have verified the conjecture for n up to 10^5. - _Zhi-Wei Sun_, Sep 27 2015

%D R. K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004. (Cf. Section C6 on addition chains.)

%D Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

%H Zhi-Wei Sun, <a href="/A262446/b262446.txt">Table of n, a(n) for n = 1..10000</a>

%H Neil Clift, <a href="http://additionchains.com/primes.html">Prime Count and Addition Chains</a>, 2024.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641 [math.NT], 2014.

%e a(4) = 1 since pi(4*5/2+1) = pi(11) = 5 = 1 + 4 = pi(2) + pi(7) = pi(1*2/2+1) + pi(3*4/2+1).

%e a(6) = 1 since pi(6*7/2+1) = pi(22) = 8 = 2 + 6 = pi(4) + pi(16) = pi(2*3/2+1) + pi(5*6/2+1).

%e a(11) = 1 since pi(11*12/2+1) = pi(67) = 19 = 5 + 14 = pi(11) + pi(46) = pi(4*5/2+1) + pi(9*10/2+1).

%e a(21) = 1 since pi(21*22/2+1) = pi(232) = 50 = 14 + 36 = pi(46) + pi(154) = pi(9*10/2+1) + pi(17*18/2+1).

%e a(54) = 1 since pi(54*55/2+1) = pi(1486) = 235 = 30 + 205 = pi(121) + pi(1276) = pi(15*16/2+1) + pi(50*51/2+1).

%e a(253) = 1 since pi(253*254/2+1) = pi(32132) = 3447 = 747 + 2700 = pi(5672) + pi(24311) = pi(106*107/2+1) + pi(220*221/2+1).

%e a(325) = 1 since pi(325*326/2+1) = pi(52976) = 5406 = 1446 + 3960 = pi(12091) + pi(37402) = pi(155*156/2+1) + pi(37402*37403/2+1).

%t f[n_]:=PrimePi[n(n+1)/2+1]

%t T[n_]:=Table[f[k],{k,1,n}]

%t Do[r=0;Do[If[2*f[k]>=f[n],Goto[aa]];If[MemberQ[T[n],f[n]-f[k]],r=r+1];Continue,{k,1,n-1}];Label[aa];Print[n," ",r];Continue,{n,1,100}]

%Y Cf. A000217, A000720, A262403, A262439.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Sep 23 2015