A262689 - OEIS

A262689

a(n) = largest number k <= A000196(n) for which A002828(n-(k^2)) = A002828(n)-1.

0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 2, 3, 3, 3, 4, 4, 3, 3, 4, 4, 3, 3, 4, 5, 5, 5, 5, 5, 5, 5, 4, 5, 5, 5, 6, 6, 6, 6, 6, 5, 5, 5, 6, 6, 6, 6, 4, 7, 7, 7, 6, 7, 7, 7, 6, 7, 7, 7, 7, 6, 7, 7, 8, 8, 8, 7, 8, 8, 6, 7, 6, 8, 7, 7, 6, 8, 7, 7, 8, 9, 9, 9, 8, 9, 9, 9, 6, 8, 9, 9, 9, 8, 9, 9, 8, 9, 7, 9, 10, 10, 10, 10, 10, 10, 9, 9, 10, 10, 10, 10, 10, 8, 8, 9, 10, 9, 10, 10, 10, 11

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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

FORMULA

Other identities. For all n >= 0:

a(n) = A000196(A262690(n)).

a(n^2) = n.

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:

(definec (A002828 n) (if (zero? n) n (+ 1 (A002828 (- n (A000290 (A262689 n)))))))

CROSSREFS

Cf. A000196, A002828, A010052, A064876, A180466, A262690.

Differs from A064876 for the first time at n=33, where a(33) = 5, while A064876(33) = 4.

Sequence in context: A281855 A137921 A064876 * A319816 A343437 A348242

Adjacent sequences: A262686 A262687 A262688 * A262690 A262691 A262692

KEYWORD

nonn

AUTHOR

Antti Karttunen, Oct 03 2015

STATUS

approved