A308403 Number of ways to write n as 6^i + 3^j + A008347(k), where i, j and k > 0 are nonnegative integers.

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

Offset: 1

Author

Zhi-Wei Sun, May 25 2019

Keywords

• nonn

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

Examples

			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).
		

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.

Crossrefs

Cf. A000040, A000244, A000400, A008347, A303656, A303821, A308411.

Programs

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]