OFFSET
1,5
COMMENTS
Conjecture 1: a(n) > 0 for all n > 2. In other words, each n = 3,4,... can be written as 6^i + 3^j + prime(k) - prime(k-1) + ... + (-1)^(k-1)*prime(1), where i, j and k > 0 are nonnegative integers.
Conjecture 2: If {a,b} is among {2,m} (m = 3..14), {3,4}, {3,5}, then any integer n > 2 can be written as a^i + b^j + A008347(k) with i, j and k > 0 nonnegative integers.
Using Qing-Hu Hou's program, we have verified Conjectures 1 and 2 for n up to 10^9 and 10^7 respectively. - Zhi-Wei Sun, May 28 2019
Conjecture 1 verified up to 10^10. Conjecture 2 holds up to 10^10 for all cases except {2, 12} since 4551086841 cannot be written as 2^i + 12^j + A008347(k). - Giovanni Resta, May 28 2019
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.
EXAMPLE
a(3) = 1 with 3 - (6^0 + 3^0) = 1 = A008347(2).
a(4) = 1 with 4 - (6^0 + 3^0) = 2 = A008347(1).
a(24) = 1 with 24 - (6^0 + 3^0) = 22 = A008347(13).
a(234) = 1 with 234 - (6^1 + 3^3) = 201 = A008347(90).
a(1134) = 1 with 1134 - (6^2 + 3^0) = 1097 = A008347(322).
a(4330) = 1 with 4330 - (6^3 + 3^0) = 4113 = A008347(1016).
a(5619) = 1 with 5619 - (6^1 + 3^3) = 5586 = A008347(1379).
a(6128) = 1 with 6128 - (6^0 + 3^0) = 6126 = A008347(1499).
a(16161) = 1 with 16161 - (6^3 + 3^0) = 15944 = A008347(3445).
a(133544) = 1 with 133544 - (6^0 + 3^8) = 126982 = A008347(22579).
MATHEMATICA
Pow[n_]:=Pow[n]=n>0&&IntegerQ[Log[3, n]];
s[0]=0; s[n_]:=s[n]=Prime[n]-s[n-1];
tab={}; Do[r=0; Do[If[s[k]>=n, Goto[bb]]; Do[If[Pow[n-s[k]-6^m], r=r+1], {m, 0, Log[6, n-s[k]]}]; Label[bb], {k, 1, 2n-1}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 25 2019
STATUS
approved