|
|
A367857
|
|
a(n) is the smallest base b such that (b+1)^n in base b is a palindrome.
|
|
1
|
|
|
2, 2, 2, 7, 11, 21, 36, 71, 127, 253, 463, 925, 1717, 3433, 6436, 12871, 24311, 48621, 92379, 184757, 352717, 705433, 1352079, 2704157, 5200301, 10400601, 20058301, 40116601, 77558761, 155117521, 300540196, 601080391, 1166803111
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
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).
|
|
LINKS
|
|
|
FORMULA
|
Conjecture: a(n) = binomial(n, floor(n/2))+1 for n>3.
|
|
EXAMPLE
|
For n=5 the minimum base required is 11, giving 11^5=15aa51 (12^5=248832 in base 10).
|
|
MATHEMATICA
|
a[n_] := Module[{b = 2, d}, While[(d = IntegerDigits[(b + 1)^n, b]) != Reverse[d], b++ ]; b] ;
|
|
PROG
|
(Python)
from itertools import count
from sympy.ntheory.factor_ import digits
def ispal(d): return d == d[::-1]
def a(n): return next(b for b in count(2) if ispal(digits((b+1)**n, b)[1:]))
(PARI) a(n) = my(b=2, d); while ((d=digits((b+1)^n, b)) != Vecrev(d), b++); b; \\ Michel Marcus, Dec 05 2023
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|