A259374 Palindromic numbers in bases 3 and 5 written in base 10.

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

Author

Eric A. Schmidt and Robert G. Wilson v, Jul 14 2015

Keywords

• nonn
• base

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.

Examples

			52 is in the sequence because 52_10 = 202_5 = 1221_3.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..39
• A.H.M. Smeets, Scatterplot of log_3(number is palindromic in base 3 and base b) versus b, for b in {2,4,5, 6,7,8,10}

Crossrefs

Cf. A048268, A060792, A097856, A097928, A182232, A259374, A097929, A182233, A259375, A259376, A097930, A182234, A259377, A259378, A249156, A097931, A259380-A259384, A099145, A259385-A259390, A099146, A007632, A007633, A029961-A029964, A029804, A029965-A029970, A029731, A097855, A250408-A250411, A099165, A250412.

Programs

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 *)

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

Formula

Intersection of A014190 and A029952.