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a(n) is the smallest base b such that (b+1)^n in base b is a palindrome.
1

%I #22 Jan 19 2024 14:54:51

%S 2,2,2,7,11,21,36,71,127,253,463,925,1717,3433,6436,12871,24311,48621,

%T 92379,184757,352717,705433,1352079,2704157,5200301,10400601,20058301,

%U 40116601,77558761,155117521,300540196,601080391,1166803111

%N a(n) is the smallest base b such that (b+1)^n in base b is a palindrome.

%C Empirically the same as A001405(n)+1 apart from a(2) where 11^10=1001 in base 2 (3^2=9 in base 10) and a(3) where 11^11=11011 in base 2 (3^3=27 in base 10).

%F Conjecture: a(n) = binomial(n, floor(n/2))+1 for n>3.

%e For n=5 the minimum base required is 11, giving 11^5=15aa51 (12^5=248832 in base 10).

%t a[n_] := Module[{b = 2, d}, While[(d = IntegerDigits[(b + 1)^n, b]) != Reverse[d], b++ ]; b] ;

%t Table[a[n],{n,33}] (* _James C. McMahon_, Dec 13 2023 after PARI *)

%o (Python)

%o from itertools import count

%o from sympy.ntheory.factor_ import digits

%o def ispal(d): return d == d[::-1]

%o def a(n): return next(b for b in count(2) if ispal(digits((b+1)**n, b)[1:]))

%o print([a(n) for n in range(1, 21)]) # _Michael S. Branicky_, Dec 04 2023

%o (PARI) a(n) = my(b=2,d); while ((d=digits((b+1)^n, b)) != Vecrev(d), b++); b; \\ _Michel Marcus_, Dec 05 2023

%Y Cf. A001405.

%K nonn,base

%O 1,1

%A _James Carruthers_, Dec 03 2023

%E a(8)-a(33) from _Michael S. Branicky_, Dec 04 2023