OFFSET
1,4
COMMENTS
Conjecture: a(n) > 0 for all n > 0.
We have verified this for n up to 3*10^5.
It seems that a(n) = 1 only for n = 1, 2, 3, 6, 19, 22, 48, 60, 396, 1076, 3033, 3889, 4741, 6804, 7919, 9604, 16938, 19169, 32533, 59903, 100407.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(1) = 1 since 1 = 1 + 0 = pi(2^2/2) + pi(3*1^2/2).
a(2) = 1 since 2 = 2 + 0 = pi(3^2/2) + pi(3*1^2/2).
a(3) = 1 since 3 = 0 + 3 = pi(1^2/2) + pi(3*2^2/2).
a(6) = 1 since 6 = 0 + 6 = pi(1^2/2) + pi(3*3^2/2).
a(19) = 1 since 19 = 7 + 12 = pi(6^2/2) + pi(3*5^2/2).
a(22) = 1 since 22 = 1 + 21 = pi(2^2/2) + pi(3*7^2/2).
a(48) = 1 since 48 = 1 + 47 = pi(2^2/2) + pi(3*12^2/2).
a(60) = 1 since 60 = 25 + 35 = pi(14^2/2) + pi(3*10^2/2).
a(396) = 1 since 396 = 334 + 62 = pi(67^2/2) + pi(3*14^2/2).
a(1076) = 1 since 1076 = 47 + 1029 = pi(21^2/2) + pi(3*74^2/2).
a(3033) = 1 since 3033 = 7 + 3026 = pi(6^2/2) + pi(3*136^2/2).
a(3889) = 1 since 3889 = 1808 + 2081 = pi(176^2/2) + pi(3*110^2/2).
a(4741) = 1 since 4741 = 4699 + 42 = pi(301^2/2) + pi(3*11^2/2).
a(6804) = 1 since 6804 = 6047 + 757 = pi(346^2/2) + pi(3*62^2/2).
a(7919) = 1 since 7919 = 4049 + 3870 = pi(277^2/2) + pi(3*156^2/2).
a(9604) = 1 since 9604 = 4754 + 4850 = pi(303^2/2) + pi(3*177^2/2).
a(16938) = 1 since 16938 = 2223 + 14715 = pi(198^2/2) + pi(3*327^2/2).
a(19169) = 1 since 19169 = 6510 + 12659 = pi(361^2/2) + pi(3*301^2/2).
a(32533) = 1 since 32533 = 1768 + 30765 = pi(174^2/2) + pi(3*490^2/2).
a(59903) = 1 since 59903 = 59210 + 693 = pi(1213^2/2) + pi(3*59^2/2).
a(100407) = 1 since 100407 = 7554 + 92853 = pi(392^2/2) + pi(3*894^2/2).
MATHEMATICA
s[n_]:=s[n]=PrimePi[3n^2/2]
t[n_]:=t[n]=PrimePi[n^2/2]
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 09 2015
STATUS
approved