A104459 -id:A104459 - cp's OEIS frontend

A244551 Numbers k such that k +- (sum of digits of k) are both palindromes.

1, 2, 3, 4, 10, 100, 105, 181, 262, 267, 343, 348, 424, 429, 681, 762, 767, 843, 848, 924, 929, 1000, 10000, 100000, 1000000, 10000000, 63999991, 72999982, 81999973, 90999964, 100000000, 1000000000, 10000000000, 100000000000, 1000000000000, 10000000000000, 100000000000000, 1000000000000000, 10000000000000000

Offset: 1

Derek Orr, Jun 29 2014

For k = 0, 1, 2, ..., 10^k is a member of this sequence, so A011557 is a subsequence.
Also, floor(a(n)/10) must be in A244573.
Intersection of A229545 and A229621. - Michel Marcus, Jun 30 2014
The corresponding digit sums are: 1, 2, 3, 4, 1, 1, 6, 10, 10, 15, 10, 15, 10, 15, 15, 15, 20, 15, 20, 15, 20, 1, 1, 1, 1, 1, 55, 55, 55, 55, 1, 1, .... - Derek Orr, Jun 30 2014
When the digits sums are 1, the terms are a power of 10 terms, with corresponding palindromes being 2 apart (consecutive palindromes). The next instances of consecutive palindromes appear when digit sums are 55, so that the palindromes are 110 apart (see sequence A104459 that lists the possible differences between adjacent palindromes). - Michel Marcus, Jul 01 2014
There are infinitely many terms in this sequence that are not of the form 10^k for some k. Take the number 180 {59 9's} 631. This is a 65-digit number (180999...999631). This number is a member of this sequence. From this number, we can generate 34 other numbers. Keeping the 59 9's there, to preserve its properties, subtract 9 from 180 and add 90 to 631. Now we have 171 {59 9's} 721. This is also a member. The 59 9's are only to make the digit sum = 550. Thus, the two palindromes are 1100 apart, they are consecutive. If we keep doing this arithmetic (subtract 9 from the first number and add 90 to the second), we get 162 {59 9's} 811, 153 {59 9's} 901, 144 {58 9's} 991. The last number only has 58 9's in order to make sure the digit sum stays at 550. Other numbers and patterns to them have been listed in an a-file. Similarly, patterns like this will appear when considering digit sums of 5500 and larger. - Derek Orr, Jul 01 2014

			267 - (2+6+7) = 252 is a palindrome and 267 + (2+6+7) = 282 is also a palindrome. Thus 252 is a member of this sequence.

Derek Orr, More terms in the sequence

Cf. A229545, A229621, A007953.

rev(n)={r="";for(i=1,#digits(n),r=concat(Str(digits(n)[i]),r));return(eval(r))}
for(n=1,10^7,dig=digits(n);s=sum(k=1,#dig,dig[k]);sm=n-s;la=n+s;if(rev(sm)==sm&&rev(la)==la,print1(n,", ")))

def palgen(l,b=10): # generator of palindromes in base b of length <= 2*l
    if l > 0:
        yield 0
        for x in range(1,l+1):
            n = b**(x-1)
            n2 = n*b
            for y in range(n,n2):
                k, m = y//b, 0
                while k >= b:
                    k, r = divmod(k,b)
                    m = b*m + r
                yield y*n + b*m + k
            for y in range(n,n2):
                k, m = y, 0
                while k >= b:
                    k, r = divmod(k,b)
                    m = b*m + r
                yield y*n2 + b*m + k
A244551_list = []
for p in palgen(9):
    l = len(str(p))
    for i in range(1,l*9+1):
        n = p-i
        if n > 0:
            if sum((int(d) for d in str(n))) == i:
                s = str(n-i)
                if s == s[::-1]:
                    A244551_list.append(n) # Chai Wah Wu, Aug 24 2015

a(27)-a(32) from Michel Marcus, Jun 30 2014

a(33)-a(39) from Chai Wah Wu, Aug 24 2015

A109868 Numbers which can be differences of successive palindromes in order of their first occurrence.

Original entry on oeis.org

1, 2, 11, 10, 110, 100, 1100, 1000, 11000, 10000, 110000, 100000, 1100000, 1000000, 11000000, 10000000, 110000000, 100000000, 1100000000, 1000000000, 11000000000, 10000000000, 110000000000, 100000000000

Offset: 1

Amarnath Murthy, Jul 09 2005

Except a(2)=2, all other terms contain digits 1 or 0.

Almost the same as A104459. - Georg Fischer, Apr 17 2025

			11 comes before 10 as 11 = 22-11 and 10 = 111-101, 11 occurs earlier to 10 as a successive palindrome difference.

Index entries for linear recurrences with constant coefficients, signature (0,10).

Cf. A104459.

LinearRecurrence[{0,10}, {1,2,11,10}, 50] (* Georg Fischer, Apr 17 2025 *)

More terms from Erich Friedman, Aug 08 2005

Offset changed by Andrew Howroyd, Sep 17 2024

A229622 Palindromes m such that m + (sum of digits of m) is also a palindrome.

Original entry on oeis.org

0, 1, 2, 3, 4, 181, 191, 262, 343, 424, 505, 696, 767, 848, 929, 28999999999982, 37999999999973, 46999999999964, 55999999999955, 64999999999946, 73999999999937, 82999999999928, 91999999999919, 1099999999999901

Offset: 1

Derek Orr, Sep 26 2013

Palindromes in A229545.

This sequence is infinite. It is possible to generate an infinite subsequence using 1099999999999901 as a model. Look at palindromes of the form: 1, z zeros, floor(11*10^z/9) nines, z zeros, 1. The sum of the digits is 11*10^z. Adding 11*10^z to the number produces a palindrome having 4 ones. - T. D. Noe, Oct 03 2013

			262 + (2+6+2) = 272 (another palindrome). So 262 is in this sequence.

Cf. A002113, A104459, A229545.

isPal[d_List] := d[[1]] != 0 && d == Reverse[d]; check[d_List] := Module[{num = FromDigits[d]}, If[isPal[IntegerDigits[num + Total[d]]], Print[num]; AppendTo[t, num];]]; t = {0}; Do[d = IntegerDigits[n]; dig = Join[d, Reverse[d]]; check[dig]; dig = Join[d, Reverse[Most[d]]]; check[dig], {n, 0, 9999999}]; Sort[t] (* T. D. Noe, Oct 02 2013 *)

def ispal(n):
  r = ''
  for i in str(n):
    r = i + r
  return n == int(r)
def DS(n):
  s = 0
  for i in str(n):
    s += int(i)
  return s
{print(n,end=', ') for n in range(10**7) if ispal(n) and ispal(n+DS(n))}
# Simplified by Derek Orr, Apr 10 2015
# Fixed by Robert C. Lyons, Mar 16 2023

1099999999999901 from T. D. Noe, Oct 03 2013

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A244551 Numbers k such that k +- (sum of digits of k) are both palindromes.

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A109868 Numbers which can be differences of successive palindromes in order of their first occurrence.

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A229622 Palindromes m such that m + (sum of digits of m) is also a palindrome.

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