A248509 Length of longest sequence of distinct nonzero squares summing to n, or 0 if no such sequence exists.
1, 0, 0, 1, 2, 0, 0, 0, 1, 2, 0, 0, 2, 3, 0, 1, 2, 0, 0, 2, 3, 0, 0, 0, 2, 3, 0, 0, 3, 4, 0, 0, 0, 2, 3, 1, 2, 3, 4, 2, 3, 3, 0, 0, 3, 4, 0, 0, 3, 4, 4, 2, 3, 4, 5, 3, 4, 2, 3, 0, 3, 4, 4, 1, 4, 5, 0, 2, 3, 4, 4, 0, 2, 4, 5, 0, 3, 4, 5, 2, 4, 5, 3, 4
Offset: 1
Keywords
Examples
1 = 1^2 so a(1) = 1. 2 and 3 are not sums of distinct squares, so a(2) = 0 and a(3) = 0. 4 = 2^2 so a(4) = 1. 5 = 1^2 + 2^2 so a(5) = 2.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- P. T. Bateman, A. J. Hildebrand and G. B. Purdy, Sums of distinct squares, Acta Arithmetica 67 (1994), pp. 349-380.
- Eryk Pawilan, Math Overflow question
Programs
-
Maple
N:= 100: # to get a(1) to a(N) M:= floor(sqrt(N)): A:= Array(0..N,0..M): sj:= unapply(sum(k^2,k=1..x),x): for j from 1 to M do for n from sj(j)+1 to N do A[n,j]:= -infinity od: for n from 1 to j^2-1 do A[n,j]:= A[n,j-1] od: for n from j^2 to min(sj(j),N) do A[n,j]:= max(A[n,j-1],1+A[n-j^2,j-1]) od: od: subs(-infinity=0,[seq(A[n,M],n=1..N)]); # Robert Israel, Oct 07 2014
-
Mathematica
Nt = 100 (* = number of terms *); M = Floor[Sqrt[Nt]]; Clear[A]; A[, ] = 0; s[j_] := Range[j].Range[j]; For[j = 1, j <= M, j++, For[n = s[j] + 1, n <= Nt, n++, A[n, j] = -Infinity]; For[n = 1, n <= j^2 - 1, n++, A[n, j] = A[n, j - 1]]; For[n = j^2, n <= Min[s[j], Nt], n++, A[n, j] = Max[A[n, j-1], 1+A[nj^2, j-1]]] ]; Table[A[n, M] /. DirectedInfinity[-1] -> 0, {n, 1, Nt}] (* Jean-François Alcover, Mar 04 2019, after Robert Israel *)
Comments