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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).

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.

For each j >= 5 and odd, k = (j^3 + 6*j^2 + 14*j + 11)/2 is a term in the sequence, and represents a 3-digit palindrome in each of three consecutive integer bases:

.

base 1st digit 2nd digit 3rd digit

---- --------- --------- ---------

j+1 (j+3)/2 (j+5)/2 (j+3)/2

j+2 (j+1)/2 (j+3)/2 (j+1)/2

j+3 (j-1)/2 (j+7)/2 (j-1)/2

.

(see 178 and 373 in the Example section). Nearly all of the first 95 terms of this sequence are terms of this form.

For each j >= 44 and divisible by 4, k = (3*j^5 + 30*j^4 + 125*j^3 + 270*j^2 + 307*j + 148)/4 is a term in the sequence, and represents a 5-digit palindrome in each of three consecutive integer bases:

.

base 1st digit 2nd digit 3rd digit 4th digit 5th digit

---- --------- --------- --------- --------- ---------

j+1 3*j/4 + 4 j/2 + 9 j/4 + 11 j/2 + 9 3*j/4 + 4

j+2 3*j/4 + 1 j/2 + 2 j/4 + 0 j/2 + 2 3*j/4 + 1

j+3 3*j/4 - 2 j/2 + 10 j/4 - 11 j/2 + 10 3*j/4 - 2

.

[Reformatted by Jon E. Schoenfield, Apr 01 2018]

From Matej Veselovac, Mar 31 2018: (Start)

Similarly to the one 3-digit and one 5-digit families given above, at least seven more infinite families exist, for 7-digit consecutive palindromes. Given a nonnegative integer n, we have the following representations palindromic in exactly three consecutive integer number bases j+1, j+2, j+3 :

1. For each j = 36+12n, k = (816 + 2474*j + 3114*j^2 + 2117*j^3 + 852*j^4 + 209*j^5 + 30*j^6 + 2*j^7)/12 is a term of the sequence.

2. For each j = 55+6n, k = (245 + 748 j + 980 j^2 + 718 j^3 + 320 j^4 + 88 j^5 + 14 j^6 + j^7)/6 is a term of the sequence.

3. For each j = 73+2n, k = (247 + 748 j + 980 j^2 + 718 j^3 + 320 j^4 + 88 j^5 + 14 j^6 + j^7)/2 is a term of the sequence.

4. For each j = 116+12n, k = (2440 + 7366 j + 9694 j^2 + 7171 j^3 + 3232 j^4 + 895 j^5 + 142 j^6 + 10 j^7)/12 is a term of the sequence.

5. For each j = 172+6n, k = (812 + 2446 j + 3290 j^2 + 2527 j^3 + 1190 j^4 + 343 j^5 + 56 j^6 + 4 j^7)/6 is a term of the sequence.

6. For each j = 288+12n, k = (1176 + 3566 j + 4374 j^2 + 2807 j^3 + 1032 j^4 + 227 j^5 + 30 j^6 + 2 j^7)/12 is a term of the sequence.

7. For each j = 277+6n, k = (1237 + 3740 j + 4900 j^2 + 3590 j^3 + 1600 j^4 + 440 j^5 + 70 j^6 + 5 j^7)/6 is a term of the sequence.

The smallest terms given by these families are of magnitudes ~ 10^10.3, 10^11.5, 10^12.8, 10^14.4, 10^15.5, 10^16.4 and 10^17. The smallest term of the next family, if it exists, is at least of magnitude ~ 10^18.

Almost all known terms of the sequence so far belong in one of the above defined families, either being 3-, 5-, or 7- digit palindromes in exactly 3 consecutive integer number bases.

There are 13 known terms that do not belong to any families: 300, 3360633, 19987816, 43443858, 532083314, 1778140759, 2721194733, 11325719295, 47622367425, 97638433343, 224678540182, 265282702996, 561091062285 (all but 300 so far are 7-digit cases).

Infinite families for consecutive palindromes longer than 7 digits, as well as any examples for those cases, have not yet been observed.

Smallest example for 9-digit consecutive palindromes does not exist within first 100 integer number bases, thus is at least > 10^16.

Similarly, no terms palindromic in 4 or more consecutive integer number bases have been found, so far.

[Extended by Matej Veselovac, Feb 05 2019] (End)