OFFSET
1,8
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 4, and a(n) = 1 only for n = 5, 6, 7, 10, 12, 32, 38, 445, 727.
(ii) All those numbers pi(T(n)) (n = 1,2,3,...) are pairwise distinct. Moreover, if sum_{i=j,...,k}1/pi(T(i)) and sum_{r=s,...,t}1/pi(T(r)) with 1 < j <= k and j <= s <= t have the same fractional part but the ordered pairs (j,k) and (s,t) are different, then j = 2, k = 5 and s = t = 4.
Clearly, part (i) is related to addition chains, and the first assertion in part (ii) is an analog of Legendre's conjecture that pi(n^2) < pi((n+1)^2) for all n = 1,2,3,....
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
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.
EXAMPLE
a(5) = 1 since pi(T(5)) = pi(15) = 6 = 2 + 4 = pi(3) + pi(10) = pi(T(2)) + pi(T(4)).
a(6) = 1 since pi(T(6)) = pi(21) = 8 = 2 + 6 = pi(3) + pi(15) = pi(T(2)) + pi(T(5)).
a(7) = 1 since pi(T(7)) = pi(28) = 9 = 3 + 6 = pi(6) + pi(15) = pi(T(3)) + pi(T(5)).
a(10) = 1 since pi(T(10)) = pi(55) = 16 = 2 + 14 = pi(3) + pi(45) = pi(T(2)) + pi(T(9)).
a(12) = 1 since pi(T(12)) = pi(78) = 21 = 3 + 18 = pi(6) + pi(66) = pi(T(3)) + pi(T(11)).
a(32) = 1 since pi(T(32)) = pi(528) = 99 = 9 + 90 = pi(28) + pi(465) = pi(T(7)) + pi(T(30)).
a(38) = 1 since pi(T(38)) = pi(741) = 131 = 32 + 99 = pi(136) + pi(528) = pi(T(16)) + pi(T(32)).
a(445) = 1 since pi(T(445)) = pi(99235) = 9526 = 2963 + 6563 = pi(27028) + pi(65703) = pi(T(232)) + pi(T(362)).
a(727) = 1 since pi(T(727)) = pi(264628) = 23197 = 10031 + 13166 = pi(105111) + pi(141778) = pi(T(458)) + pi(T(532)).
MATHEMATICA
f[n_]:=PrimePi[n(n+1)/2]
T[m_, n_]:=Table[f[k], {k, m, n}]
Do[r=0; Do[If[MemberQ[T[k+1, n-1], f[n]-f[k]], r=r+1]; Continue, {k, 2, n-2}]; Print[n, " ", r]; Continue, {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 21 2015
STATUS
approved