A262038 Least palindrome >= n.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 11, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 66, 66, 66, 66, 66, 66, 66, 66, 66, 66, 66, 77, 77, 77, 77, 77, 77, 77, 77, 77, 77

Offset: 0

Author

M. F. Hasler, Sep 08 2015

Keywords

• nonn
• base

Comments

Could be called nextpalindrome() in analogy to the nextprime() function A007918. As for the latter (A151800), there is the variant "next strictly larger palindrome" which equals a(n+1), and thus differs from a(n) iff n is a palindrome; see PARI code.

Might also be called palindromic ceiling function in analogy to the name "palindromic floor" proposed for A261423.

Links

• M. F. Hasler, Table of n, a(n) for n = 0..10000
• Eric Weisstein's World of Mathematics, Palindromic Number
• Wikipedia, Palindromic number
• Index entries for sequences related to palindromes

Crossrefs

Cf. A002113, A261423, A262037.

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

Programs

Haskell

a262038 n = a262038_list !! n
a262038_list = f 0 a002113_list where
   f n ps'@(p:ps) = p : f (n + 1) (if p > n then ps' else ps)
-- Reinhard Zumkeller, Sep 16 2015

Mathematica

palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; Table[k = n; While[! palQ@ k, k++]; k, {n, 0, 80}] (* Michael De Vlieger, Sep 09 2015 *)

PARI

{A262038(n, d=digits(n), p(d)=sum(i=1, #d\2, (10^(i-1)+10^(#d-i))*d[i],if(bittest(#d,0),10^(#d\2)*d[#d\2+1])))= for(i=(#d+3)\2,#d,d[i]>d[#d+1-i]&&break;(d[i]9||return(p(d));d[i]=0);10^#d+1} \\ For a function "next strictly larger palindrome", delete the i==#d and n<10... part. - M. F. Hasler, Sep 09 2015

Python

def A262038(n):
    sl = len(str(n))
    l = sl>>1
    if sl&1:
        w = 10**l
        n2 = w*10
        for y in range(n//(10**l),n2):
            k, m = y//10, 0
            while k >= 10:
                k, r = divmod(k,10)
                m = 10*m + r
            z = y*w + 10*m + k
            if z >= n:
                return z
    else:
        w = 10**(l-1)
        n2 = w*10
        for y in range(n//(10**l),n2):
            k, m = y, 0
            while k >= 10:
                k, r = divmod(k,10)
                m = 10*m + r
            z = y*n2 + 10*m + k
            if z >= n:
                return z # Chai Wah Wu, Sep 14 2022