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For any integer b > 1, the base-b expansion of any number k < b will be a one-digit number, and will thus be trivially palindromic.

From Matej Veselovac, Sep 26 2019: (Start)

All terms of the sequence have 3 or more digits in at least one of the consecutive palindromic bases. The only term that has 2,3 digits exactly in the consecutive palindromic bases, is the first term a(1) = 10 = (1,0){10} = (2,2){4} = (1,0,1)_{3}, which is palindromic in bases 4,3 and has 2,3 digits in those bases, respectively.

If a term of the sequence has d digits in the smallest of the palindromic bases, then d must be odd. This is because an even length palindrome in base b, is divisible by b+1, and hence can't be palindromic in the base b+1 as it will end in 0. This implies that if a term has an equal number of digits in all bases, that number must be odd.

All terms that have exactly d = 3 digits in consecutive palindromic number bases b,b-1,... are given by the following two families (if and only if relation):

1. n = (x+1, y+4, x+1)_{b = 5+x+y} = (x+1)(5+x+y)^2+(y+4)(5+x+y)^1+(x+1)

2. n = (x+2, 5, x+2)_{b = x+6} = (x+2)(x+6)^2+5(6+x)^1+(x+2)

Where x, y = 0,1,2,3,... go over all nonnegative integers, where (a_1, a_2, a_3) represents digits in base {b} in terms of x, y; and where the RHS is the decimal expansion.

There are similar families for every subsequence of terms having exactly d digits in all bases, but they get much more complex for d >= 5. The d = 5 case is included at the link "Special linear Diophantine system - is it solvable in general?".

Specifically, every subsequence of terms with exactly d digits in all of the consecutive palindromic bases, is infinite. This is proven by finding the following subsequence of such subsequences:

We can construct a subsequence yielding infinitely many terms for every digit case d. For example, one such family is given by (b-1,0,b-1,0,...,0,b-1)_{b}, by alternating "b-1" and "0" digits in base b, and will be nontrivially palindromic in base b+1 as well, for all b > binomial(2k, k), where d=2k+1 is an odd number of digits, for every natural number k. That is, in the decimal expansion, these terms are equal to (b^(2k+2)-1)/(b+1), giving infinitely many terms for every k, that have d=2k+1 digits in palindromic bases b, b+1, for every b > binomial(2k, k).

In contrast, if the number of digits is not equal in all of the consecutive palindromic bases, then every subsequence that is bounded by a maximal number of d digits allowed in the consecutive palindromic bases, seems to be finite.

That is, we can say "almost all" terms in this sequence belong to the case of having an equal number of digits in all consecutive palindromic bases. The remaining terms, that do not have an equal number of digits in all consecutive palindromic bases, are given in A327810.

(End).

It is conjectured that a(n)=0 for all n>=4. - Matej Veselovac, Mar 25 2020

Nontrivially palindromic means having at least 2 digits in the palindromic base representation.

| n | term | consecutive palindromic bases representations |

+---+------+-----------------------------------------------+

| 1 | 3 | 11_2 |

| 2 | 10 | 101_3 = 22_4 |

| 3 | 178 | 454_6 = 343_7 = 262_8 |

It is not known if the fourth term exists. The problem can be looked at in context of Diophantine equations, which seem hard.

Numbers which are strictly non-palindromic up to a set of k=3 consecutive number bases. It is conjectured that such a sequence for k>=4 is empty. For a special case of k=0, we have the sequence A016038.

A subsequence of A279093. Notice that a(1),a(2),a(3),a(4) are all of the form (b^3 + 3 b^2 + 5 b + 2)/2 where b=2k+6, for k=71,101,7497,36575. Not all terms are necessarily of this form. (See comments in A279093, containing a total of 9 known such forms that generate numbers palindromic in three consecutive number bases.)

For every n, a(n) should be a prime number.