

A222532


a(1)=2; for n >= 1, a(n+1) is the least prime p_m such that a(n)=p_mp_{m1}+...+(1)^{mk}p_k for some 0<k<m.


7



2, 5, 7, 13, 17, 23, 31, 37, 43, 53, 59, 67, 73, 83, 89, 101, 109, 113, 131, 149, 157, 163, 173, 179, 197, 223, 257, 263, 269, 277, 283, 311, 347, 389, 401, 421, 431, 487, 503, 523, 557, 569, 577, 601, 613, 641, 661, 709, 733, 739, 773, 823, 827, 857, 883, 929, 947, 953, 977, 983, 997, 1009, 1019, 1031, 1039, 1051, 1097, 1117, 1129, 1151, 1181, 1223, 1229, 1237, 1249, 1279, 1327, 1361, 1373, 1423, 1459, 1481, 1499, 1543, 1559, 1571, 1601, 1621, 1627, 1669, 1693, 1699, 1721, 1733, 1759, 1783, 1823, 1873, 1973, 2011
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OFFSET

1,1


COMMENTS

Conjecture: For any given prime p, if we define b(1)=p and let b(n+1) be the least prime p_m such that b(n)=p_mp_{m1}+...+(1)^{mk}p_k for some 0<k<m, then a(n)=b(n') for some positive integers n and n'. In other words, if we take all the primes as vertices of a simple graph T and let two vertices p and q>p adjacent if and only if q is the least prime p_m such that p=p_mp_{m1}+...+(1)^{mk}p_k for some 0<k<m, then the graph T is a tree!
Clearly the graph T contains no cycle. The vertices on the unique path connecting 2 and 71 are listed (in order) below: 2, 5, 7, 13, 17, 23, 31, 37, 43, 53, 59, 67, 73, 83, 89, 101, 109, 113, 131, 149, 139, 107, 97, 79, 71.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 27942812.


EXAMPLE

a(2)=5 and a(3)=7 since 2=53 and 5=75+3.


MATHEMATICA

k=1
n=1
s[0_]:=0
s[n_]:=s[n]=Prime[n]s[n1]
Do[If[m==1, Print[n, " ", 2]]; If[m==k, n=n+1; Do[If[s[j](1)^(ji)*s[i]==Prime[m], k=j; Print[n, " ", Prime[j]]; Goto[aa]], {j, m+1, PrimePi[3Prime[m]]}, {i, 0, j2}]];
Label[aa]; Continue, {m, 1, 1000}]


CROSSREFS

Cf. A000040, A008347, A222566, A163846, A163847.
Sequence in context: A045352 A278494 A107426 * A144256 A082555 A160794
Adjacent sequences: A222529 A222530 A222531 * A222533 A222534 A222535


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Feb 24 2013


STATUS

approved



