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A259374
Palindromic numbers in bases 3 and 5 written in base 10.
16
0, 1, 2, 4, 26, 52, 1066, 1667, 2188, 32152, 67834, 423176, 437576, 14752936, 26513692, 27711772, 33274388, 320785556, 1065805109, 9012701786, 9256436186, 12814126552, 18814619428, 201241053056, 478999841578, 670919564984, 18432110906024, 158312796835916, 278737550525722
OFFSET
1,3
COMMENTS
0 is only 0 regardless of the base,
1 is only 1 regardless of the base,
2 on the other hand is also 10 in base 2, denoted as 10_2,
3 is 3 in all bases greater than 3, but is 11_2 and 10_3.
FORMULA
Intersection of A014190 and A029952.
EXAMPLE
52 is in the sequence because 52_10 = 202_5 = 1221_3.
MATHEMATICA
(* first load nthPalindromeBase from A002113 *) palQ[n_Integer, base_Integer] := Block[{}, Reverse[ idn = IntegerDigits[n, base]] == idn]; k = 0; lst = {}; While[k < 21000000, pp = nthPalindromeBase[k, 5]; If[ palQ[pp, 3], AppendTo[lst, pp]; Print[pp]]; k++]; lst
b1=3; b2=5; lst={}; Do[d1=IntegerDigits[n, b1]; d2=IntegerDigits[n, b2]; If[d1==Reverse[d1]&&d2==Reverse[d2], AppendTo[lst, n]], {n, 0, 10000000}]; lst (* Vincenzo Librandi, Jul 15 2015 *)
PROG
(Python)
def nextpal(n, b): # returns the palindromic successor of n in base b
....m, pl = n+1, 0
....while m > 0:
........m, pl = m//b, pl+1
....if n+1 == b**pl:
........pl = pl+1
....n = (n//(b**(pl//2))+1)//(b**(pl%2))
....m = n
....while n > 0:
........m, n = m*b+n%b, n//b
....return m
n, a3, a5 = 0, 0, 0
while n <= 20000:
....if a3 < a5:
........a3 = nextpal(a3, 3)
....elif a5 < a3:
........a5 = nextpal(a5, 5)
....else: # a3 == a5
........print(n, a3)
........a3, a5, n = nextpal(a3, 3), nextpal(a5, 5), n+1
# A.H.M. Smeets, Jun 03 2019
KEYWORD
nonn,base
AUTHOR
Eric A. Schmidt and Robert G. Wilson v, Jul 14 2015
STATUS
approved