OFFSET
1,2
COMMENTS
Of course Lagrange's theorem tells us that any positive integer can be written as a sum of at most four squares (cf. A004215).
Records in A053610. - Hugo van der Sanden, Jun 24 2015
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Rick L. Shepherd, Table of n, a(n) for n = 1..15
Art of Problem Solving, 2010 AMC 10A Problems/Problem 25 [From Rick L. Shepherd, Jan 28 2014]
E. Lemoine, Décomposition d'un nombre entier N en ses puissances nièmes maxima, C. R. Acad. Sci. Paris, Vol. 95, pp. 719-722, 1882 (then next pages).
E. Lemoine, Sur la décomposition d'un nombre en ses carrés maxima, Assoc. Française pour L'Avancement des Sciences (1896), 73-77.
FORMULA
For n >= 4, a(n) = a(n-1) + ((a(n-1)+1)/2)^2. - Joe K. Crump (joecr(AT)carolina.rr.com), Apr 16 2000
a(n) = n for n <= 3; for n > 3, a(n) = ((a(n-1)+3)/2)^2 - 2. - Arkadiusz Wesolowski, Mar 30 2013
a(n+2) = 2 * A053630(n) - 3. - Thomas Ordowski, Jul 14 2014
a(n+3) = A053630(n)^2 - 2. - Thomas Ordowski, Jul 19 2014
EXAMPLE
Here is why a(5) = 23: start with 23, subtract largest square <= 23, which is 16, getting 7.
Now subtract largest square <= 7, which is 4, getting 3.
Now subtract largest square <= 3, which is 1, getting 2.
Now subtract largest square <= 2, which is 1, getting 1.
Now subtract largest square <= 1, which is 1, getting 0.
Thus 23 = 16+4+1+1+1.
It took 5 steps to get to 0, and 23 is the smallest number which takes 5 steps. - N. J. A. Sloane, Jan 29 2014
PROG
(PARI) a(n) = if (n <= 3, n , ((a(n-1)+3)/2)^2 - 2) \\ Michel Marcus, May 25 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Four more terms from Rick L. Shepherd, Jan 27 2014
STATUS
approved