A261422 Number of ordered triples (u,v,w) of palindromes such that u+v+w=n.
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 63, 72, 79, 84, 87, 88, 87, 84, 79, 72, 66, 55, 51, 45, 40, 36, 33, 31, 30, 30, 30, 33, 27, 34, 33, 33, 33, 33, 33, 33, 33, 33, 36, 27, 39, 36, 36, 36, 36, 36, 36, 36, 36, 39, 27, 45, 39, 39, 39, 39, 39, 39, 39, 39, 42, 27, 52, 42, 42, 42, 42, 42, 42, 42, 42, 45
Offset: 0
Examples
4 can be written as a sum of three palindromes in 15 ways: 4+0+0 (3 ways), 3+1+0 (6 ways), 2+2+0 (3 ways), and 2+1+1 (3 ways), so a(4)=15.
Links
- N. J. A. Sloane, Table of n, a(n) for n = 0..10000 (See also the extended file with 200000 terms below)
- William D. Banks, Every natural number is the sum of forty-nine palindromes, arXiv:1508.04721 [math.NT], 2015. [Establishes the much weaker result that the coefficients of P(x)^49 are positive (see Formula section below).]
- Javier Cilleruelo and Florian Luca, Every positive integer is a sum of three palindromes, arXiv: 1602.06208 [math.NT], 2016-2017.
- Erich Friedman, Problem of the Month (June 1999)
- N. J. A. Sloane, Table of n, a(n) for n=0..200000
- N. J. A. Sloane, Maple program to produce 200000 terms
Crossrefs
Programs
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Mathematica
(* This program is not suitable to compute a large number of terms. *) compositions[n_, k_] := Flatten[Permutations[PadLeft[#, k]]& /@ IntegerPartitions[n, k], 1]; a[n_] := Select[compositions[n, 3], AllTrue[#, PalindromeQ]&] // Length; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Aug 05 2018 *)
Formula
G.f. = P(x)^3, where P(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^11 + x^22 + x^33 + x^44 + x^55 + x^66 + x^77 + x^88 + x^99 + x^101 + x^111 + ... = Sum_{palindromes p} x^p.
Comments