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A084019
a(n) = 9's complement of n-th palindrome (A002113).
2
9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 88, 77, 66, 55, 44, 33, 22, 11, 0, 898, 888, 878, 868, 858, 848, 838, 828, 818, 808, 797, 787, 777, 767, 757, 747, 737, 727, 717, 707, 696, 686, 676, 666, 656, 646, 636, 626, 616
OFFSET
1,1
COMMENTS
Leading zeros in the 9's complement are omitted.
For palindromes of the form 10^k-1 the corresponding entry is zero. Apart from this the entries are distinct.
LINKS
EXAMPLE
The 20th palindromic number A002113(20) is 101 having 9's complement 898 (999 - 101 = 898). So a(20) = 898. - Indranil Ghosh, Jan 30 2017
PROG
(Python)
# Program for generating the b-file
def a(n):
return 10**len(str(n))-n-1
i=0
j=1
while j<=250:
if i==int(str(i)[::-1]):
print(str(j)+" "+str(a(i)))
j+=1
i+=1 # Indranil Ghosh, Jan 30 2017
(Python)
def A084019(n):
if n == 1: return 9
y = 10*(x:=10**(len(str(n>>1))-1))
if n<x+y:
c = n-x
c = 10**len(str(c))-1-c
return c*x+int(str(c)[-2::-1]or 0)
else:
c = n-y
c = 10**len(str(c))-1-c
return c*y+int(str(c)[-1::-1]or 0) # Chai Wah Wu, Jun 13 2024
CROSSREFS
KEYWORD
base,easy,nonn
AUTHOR
Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 23 2003
STATUS
approved