A262439 -id:A262439 - cp's OEIS frontend

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

Zhi-Wei Sun, Sep 23 2015

nonn

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

			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).

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.

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.

Cf. A000217, A000720, A262403, A262439.

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}]

A262403 Number of ways to write pi(T(n)) = pi(T(k)) + pi(T(m)) with 1 < k < m < n, where T(x) is the triangular number x*(x+1)/2, and pi(x) is the number of primes not exceeding x.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 2, 2, 1, 2, 1, 3, 4, 4, 4, 3, 3, 3, 3, 5, 4, 3, 4, 6, 4, 5, 2, 3, 6, 4, 1, 5, 8, 3, 2, 6, 1, 4, 5, 4, 2, 7, 2, 4, 5, 5, 5, 3, 4, 9, 9, 4, 5, 4, 8, 7, 6, 9, 4, 7, 5, 6, 2, 5, 9, 3, 8, 5, 6, 8, 5, 4, 3, 8, 4, 8, 7, 8, 5, 7, 8, 7, 4, 6, 2, 7, 7, 8, 7, 4, 5, 6, 4, 6, 4, 6, 4, 6, 6

Offset: 1

Zhi-Wei Sun, Sep 21 2015

nonn

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,....
See also A262408 and A262409 for related conjectures involving powers.

			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)).

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.

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.

Cf. A000217, A000720, A111208, A262408, A262409, A262439, A262446.

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}]

A262995 Number of ordered pairs (k,m) with k > 0 and m > 0 such that n = pi(k(k+1)/2) + pi(1+m(m+1)/2), where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 1, 1, 3, 3, 3, 3, 5, 3, 6, 3, 6, 4, 6, 4, 7, 4, 6, 6, 6, 4, 8, 5, 6, 6, 7, 5, 8, 4, 9, 5, 7, 9, 5, 7, 8, 6, 9, 5, 9, 7, 7, 8, 8, 6, 8, 8, 8, 6, 7, 10, 8, 4, 12, 6, 8, 7, 9, 6, 10, 6, 8, 10, 8, 6, 12, 4, 12, 6, 11, 6, 11, 6, 9, 10, 8, 7, 11, 7, 10, 8, 9, 7, 10, 7, 13, 5, 7, 11, 9, 6, 8, 12, 8, 7, 11, 7, 12, 5

Offset: 1

Zhi-Wei Sun, Oct 07 2015

nonn

Conjecture: a(n) > 0 for all n > 0.
This has been verified for n up to 10^5. See also A262999 for a similar conjecture.
By Chebyshev's inequality, pi(n*(n+1)/2) > n-1 for all n > 1.
In A262403 and A262439, the author conjectured that the sequences pi(n*(n+1)/2) (n = 1,2,3,...) and pi(1+n*(n+1)/2) (n = 1,2,3,...) are both strictly increasing.

			a(1) = 1 since 1 = pi(1*2/2) + pi(1+1*2/2).
a(2) = 1 since 2 = pi(1*2/2) + pi(1+2*3/2).
a(3) = 1 since 3 = pi(2*3/2) + pi(1+1*2/2).
a(4) = 3 since 4 = pi(1*2/2) + pi(1+3*4/2) = pi(2*3/2) + pi(1+2*3/2) = pi(3*4/2) + pi(1+1*2/2).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000217, A000720, A111208, A262403, A262439, A262446, A262999.

s[k_]:=s[k]=PrimePi[k(k+1)/2+1]
t[n_]:=t[n]=PrimePi[n(n+1)/2]
Do[r=0;Do[If[s[k]>n,Goto[bb]];Do[If[t[j]>n-s[k],Goto[aa]];If[t[j]==n-s[k],r=r+1];Continue,{j,1,n-s[k]+1}];Label[aa];Continue,{k,1,n}];Label[bb];Print[n," ",r];Continue,{n,1,100}]

cp's OEIS Frontend

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

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A262403 Number of ways to write pi(T(n)) = pi(T(k)) + pi(T(m)) with 1 < k < m < n, where T(x) is the triangular number x*(x+1)/2, and pi(x) is the number of primes not exceeding x.

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A262995 Number of ordered pairs (k,m) with k > 0 and m > 0 such that n = pi(k(k+1)/2) + pi(1+m(m+1)/2), where pi(x) denotes the number of primes not exceeding x.

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cp's OEIS Frontend

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

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A262403 Number of ways to write pi(T(n)) = pi(T(k)) + pi(T(m)) with 1 < k < m < n, where T(x) is the triangular number x*(x+1)/2, and pi(x) is the number of primes not exceeding x.

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Programs

Mathematica

A262995 Number of ordered pairs (k,m) with k > 0 and m > 0 such that n = pi(k*(k+1)/2) + pi(1+m*(m+1)/2), where pi(x) denotes the number of primes not exceeding x.

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Comments

Examples

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Mathematica

A262995 Number of ordered pairs (k,m) with k > 0 and m > 0 such that n = pi(k(k+1)/2) + pi(1+m(m+1)/2), where pi(x) denotes the number of primes not exceeding x.