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Given any number let say $N$, how many ways this can be written as the sum of the palindrome numbers. For example $1443$ there are $20$ pairs of palindrome which have sum $1443$. $(1441, 2), (1221, 222), (999, 444), (989, 454), (979, 464), (969, 474), (959, 484), (949, 494), (898, 545), (888, 555), (878, 565), (868, 575), (858, 585), (848, 595), (797, 646), (787, 656), (777, 666), (767, 676), (757, 686) (747, 696)$

I tried this and able to find all possible pairs for small numbers $ N<=10^{14} $.

  • list all palindromic numbers in $O(\sqrt N * log N)$.
  • Iterate through the list for all 1....$\sqrt N$

        int id = lowerBound(allPanin, n / 2 + 1);
    
        for (int i = 0; i < id; i++) {
    
            int p = Collections.binarySearch(allPanin, n - allPanin.get(i));
            if (p > 0){
                System.out.println(n - allPanin.get(i) + ", " + allPanin.get(i));
                tp += 1;
            }
        }
    

But how to find all pairs for $N <= 10^{18}$

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  • $\begingroup$ This is A260254. The graph looks interesting. $\endgroup$
    – Vepir
    Aug 21, 2020 at 16:26

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