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A259385
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Palindromic numbers in bases 2 and 9 written in base 10.
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16
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0, 1, 3, 5, 7, 127, 255, 273, 455, 6643, 17057, 19433, 19929, 42405, 1245161, 1405397, 1786971, 2122113, 3519339, 4210945, 67472641, 90352181, 133638015, 134978817, 271114881, 6080408749, 11022828069, 24523959661, 25636651261, 25726334461, 28829406059, 1030890430479, 1032991588623, 1085079274815, 1616662113341
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listen;
history;
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OFFSET
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1,3
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LINKS
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FORMULA
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EXAMPLE
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273 is in the sequence because 273_10 = 333_9 = 100010001_2.
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MATHEMATICA
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(* 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, 9]; If[palQ[pp, 2], AppendTo[lst, pp]; Print[pp]]; k++]; lst
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PROG
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(Python)
def nextpal(n, base): # m is the first palindrome successor of n in base base
m, pl = n+1, 0
while m > 0:
m, pl = m//base, pl+1
if n+1 == base**pl:
pl = pl+1
n = n//(base**(pl//2))+1
m, n = n, n//(base**(pl%2))
while n > 0:
m, n = m*base+n%base, n//base
return m
n, a2, a9 = 0, 0, 0
while n <= 30:
if a2 < a9:
a2 = nextpal(a2, 2)
elif a9 < a2:
a9 = nextpal(a9, 9)
else: # a2 == a9
print(a2, end=", ")
a2, a9, n = nextpal(a2, 2), nextpal(a9, 9), n+1 # A.H.M. Smeets, Jun 03 2019
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CROSSREFS
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Cf. A048268, A060792, A097856, A097928, A182232, A259374, A097929, A182233, A259375, A259376, A097930, A182234, A259377, A259378, A249156, A097931, A259380, A259381, A259382, A259383, A259384, A099145, A259385, A259386, A259387, A259388, A259389, A259390, A099146, A007632, A007633, A029961, A029962, A029963, A029964, A029804, A029965, A029966, A029967, A029968, A029969, A029970, A029731, A097855, A250408, A250409, A250410, A250411, A099165, A250412.
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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