A262446 Number of ways to write A262439(n) = A262439(k) + A262439(m) with 0 < k < m < n.

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, 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, 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

Offset: 1

Author

Zhi-Wei Sun, Sep 23 2015

Keywords

• nonn

Comments

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

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

I have verified the conjecture for n up to 10^5. - Zhi-Wei Sun, Sep 27 2015

Examples

			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).
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).
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).
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).
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).
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).
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).
		

References

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

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Neil Clift, Prime Count and Addition Chains, 2024.
• Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.

Crossrefs

Cf. A000217, A000720, A262403, A262439.

Programs

Mathematica

f[n_]:=PrimePi[n(n+1)/2+1]
T[n_]:=Table[f[k],{k,1,n}]
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}]