OFFSET
0,5
COMMENTS
a(n) = square root of the largest summand present among all representations of n as a minimal number of squares, A002828(n). See the last two examples.
LINKS
Antti Karttunen, Table of n, a(n) for n = 0..65536
EXAMPLE
For n = 9, we have A002828(9) = 1 because 9 is itself a perfect square. By the definition of this sequence, we find the largest k <= 3 for which A002828(9 - k^2) = A002828(9)-1 = 0, and it is k=3 that satisfies this condition, thus a(9) = 3.
For n = 27, by the other interpretation given in the Comments section, we see that the two minimal sums requiring the least number of squares (= 3 = A002828(27)) are (25 + 1 + 1) and (9 + 9 + 9). As 25 is larger than 9, we have a(27) = sqrt(25) = 5.
For n = 33, the two minimal solutions are (25 + 4 + 4) and (16 + 16 + 1). As 25 is larger than 16, we have a(33) = sqrt(25) = 5.
PROG
(Scheme, two versions)
;; The first version requires that we already know how to compute A002828 without resorting to this same sequence, e.g. by Lagrange's "Four Squares theorem":
(define (A262689 n) (if (= 1 (A010052 n)) (A000196 n) (let ((t (- (A002828 n) 1))) (let loop ((k (A000196 n))) (if (= t (A002828 (- n (* k k)))) k (loop (- k 1)))))))
;; The second version is based on a more general minimalizing approach. We use memoizing-macro definec for faster computation:
(definec (A262689 n) (let ((k (A000196 n))) (if (= 1 (A010052 n)) k (let loop ((k k) (m #f) (mk #f)) (cond ((zero? k) mk) (else (let* ((c (A002828 (- n (* k k))))) (if (or (not m) (< c m)) (loop (- k 1) c k) (loop (- k 1) m mk)))))))))
;; The latter version makes it possible to compute A002828 naively, in a simple mutual recursion with this sequence:
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Oct 03 2015
STATUS
approved