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A279092 Numbers that are nontrivially palindromic in two or more consecutive integer bases.

%I A279092 #39 Jan 03 2020 12:18:35
%S A279092 10,46,67,92,98,104,121,130,135,154,178,185,191,227,232,235,277,282,
%T A279092 292,300,326,343,373,379,410,436,446,454,455,464,483,497,543,555,562,
%U A279092 565,631,640,646,647,651,656,676,704,738,745,781,787,797,809,835,858,862
%N A279092 Numbers that are nontrivially palindromic in two or more consecutive integer bases.
%C A279092 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.
%C A279092 From _Matej Veselovac_, Sep 26 2019: (Start)
%C A279092 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.
%C A279092 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.
%C A279092 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):
%C A279092 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)
%C A279092 2. n = (x+2, 5, x+2)_{b = x+6} = (x+2)(x+6)^2+5(6+x)^1+(x+2)
%C A279092 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.
%C A279092 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?".
%C A279092 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:
%C A279092 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).
%C A279092 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.
%C A279092 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.
%C A279092 (End).
%H A279092 MathOverflow, <a href="https://mathoverflow.net/q/342464/88524">Special linear Diophantine system - is it solvable in general?</a>
%H A279092 Matej Veselovac, <a href="/A279092/a279092.txt">Table of n, a(n) for n = 1..1000</a> (first 53 terms from _Jon E. Schoenfield_)
%e A279092 10 is in the sequence because the bases in which 10 is nontrivially palindromic include 3 and 4: 10 = 101_3 = 22_4.
%e A279092 178 is in the sequence because the bases in which 178 is nontrivially palindromic include 6, 7, and 8: 178 = 454_6 = 343_7 = 262_8.
%e A279092 252 is nontrivially palindromic in 11 integer bases (2002_5 = 252_10 = ee_17 = cc_20 = 99_27 = 77_35 = 66_41 = 44_62 = 33_83 = 22_125 = 11_251), but none of these bases are consecutive integers, so 252 is not in the sequence.
%t A279092 palQ[n_Integer, base_Integer] := Boole@ Block[{}, Reverse[idn = IntegerDigits[n, base]] == idn]; fQ[n_] := Block[{b = 2}, While[b < n && {palQ[n, b], palQ[n, b + 1]} != {1, 1}, b++]; b < n]; Select[ Range@1000, fQ] (* _Robert G. Wilson v_, Jan 31 2017 *)
%t A279092 c[b1_, d_] := Pick[FromDigits[#, b1 + 1] & /@ #, PalindromeQ[#] && Length[#] > 1 & /@ #] &@ IntegerDigits[ FromDigits[#, b1] & /@ (Flatten[Outer[List, Range[1, b1 - 1], Sequence @@ ConstantArray[Range[0, b1 - 1], d + 0]], d + 0][[All, Join[Range[d + 1], Reverse[Range[1, d + 0]]]]]), b1 + 1]; a[L_] := DeleteDuplicates[Sort[Select[Flatten[Table[c[b1, d], {d, 1, Ceiling[Log[2, L]/2] + 1}, {b1, 2, Ceiling[L ^(1/(2 d))]}]], # <L &]]]; a[31220] (* _Matej Veselovac_, Sep 28 2019 *)
%Y A279092 Cf. A002113 (palindromes in base 10), A048268 (smallest palindrome greater than n in bases n and n+1).
%Y A279092 Numbers that are palindromic in bases k and k+1: A060792 (k=2), A097928 (k=3), A097929 (k=4), A097930 (k=5), A097931 (k=6), A099145 (k=7), A099146 (k=8), A029965 (k=9), A029966 (k=11).
%Y A279092 Cf. A279093 (analogous with three or more consecutive integer bases).
%Y A279092 Cf. A327810 (subsequence with different number of digits in those bases).
%K A279092 nonn,base
%O A279092 1,1
%A A279092 _Jon E. Schoenfield_, Jan 31 2017