cp's OEIS Frontend

a(5) = 3*(10^140-3) is 141 digits long and is too large to include.

If m is in this sequence then phi(m)=r(m), so this sequence is a subsequence of A069215. a(n)=3*(10^A089675(n)-3), so a(4)=3*(10^17-3), a(5)=3*(10^140-3), a(6)=3*(10^990-3), a(7)=3*(10^1887-3), a(8)=3*(10^3530-3), a(9)=3*(10^5996-3), a(10)=3*(10^13820-3), a(11)=3*(10^21873-3) & a(12)=3*(10 ^26045-3).

Numbers of the form 3*(10^n-3) are in the sequence, so A086947 is an infinite subsequence of this sequence. Also A101700 is a subsequence of this sequence.

Let f(r,s,t,z) = 2.(9)(r+s).(34.(0)(t).65)(z).(9)(s).1 where the dot between numbers means concatenation and "(m)(n)" means number of m's is n, for example f(0,2,1,3)= 299340653406534065991, it is interesting that all numbers of the form f(r,s,t,z) where r, s, t & z are nonnegative integers and r*z=0 are in this sequence.

Except for 885 & 8221845 all known terms of this sequence are of the form f(r,s,t,z).

For all r, s & t we have f(r,s,t,0)=f(r,s,0,0)=f(r+2s,0,0,0)=A086947(r+2s+1)= 3*(10^(r+2s+1)-3).

a(1) = 21 = f(0,0,0,0), a(2) = 291 = f(1,0,0,0), a(4) = 2991 = f(2,0,0,0) = f(0,1,0,0), a(5) = 29991 = f(3,0,0,0) = f(1,1,0,0), a(6) = 234651 = f(0,0,0,1), a(7) = 299991 = f(4,0,0,0) = f(0,2,0,0), a(8) = 2340651 = f(0,0,1,1), etc. Next term is greater than 11*10^8.

From David Wasserman, Mar 27 2008: (Start)

234653406534651 is a term that doesn't fit the f(r,s,t,z) format.

We may redefine f so that t is a vector of length z, which must be symmetrical to produce a member. For example f(0,0,[0,1,0],3) = 234653406534651 is a member, but f(0,0,[1,0,0],3) = 234065346534651 is not a member.

23465934651 is another member that doesn't fit the pattern. In general there may be any number of 9's between a 5 and a 3, provided that the 9's are symmetrical. So 2346593465934651 is a member, but 23465993465934651 is not. (End)