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A249298
Smallest positive integer k, such that s-k*n is a square where s is the smallest square >= k*n.
1
1, 2, 1, 1, 1, 2, 3, 1, 1, 4, 5, 1, 5, 6, 1, 1, 9, 2, 9, 2, 1, 12, 13, 1, 1, 14, 1, 3, 17, 2, 19, 1, 3, 20, 1, 1, 23, 24, 3, 1, 25, 2, 27, 6, 1, 30, 31, 1, 1, 2, 3, 7, 35, 4, 1, 2, 3, 40, 41, 1, 41, 42, 1, 1, 1, 4, 47, 10, 5, 2, 51, 1, 51, 52, 3, 12, 1, 6, 57, 1, 1, 60, 61, 1, 3, 62, 7, 3, 65, 2
OFFSET
1,2
COMMENTS
For any n>=3, there exists at least one positive integer k, 1 <= k <= n-1 such that the difference between the smallest square >= k*n and k*n is a square. To prove this, consider the multiplier k = n-2. Then (n-2)*n = (n-1)^2-1, thus the difference from the next square is 1, which is a square. If n = 1, k = 1 and if n = 2, k = 2.
Smallest positive integer k such that ceiling(sqrt(k*n))^2-k*n is a square.
EXAMPLE
a(10) = 4, for ceiling(sqrt(10))^2-10 = 6, ceiling(sqrt(2*10))^2-2*10 = 5, ceiling(sqrt(3*10))^2-3*10 = 6 and ceiling(sqrt(4*10))^2-4*10 = 9 = 3^2.
MATHEMATICA
dif[n_] := Ceiling[Sqrt[n]]^2 - n; a[k_] := Module[{n = 1}, While[dif[dif[n*k]] != 0, n++]; Return[n]]; Table[a[k], {k, 1, 90}]
PROG
(PARI) a(n) = {k=1; while(!issquare(ceil(sqrt(k*n))^2-k*n), k++); k; } \\ Michel Marcus, Oct 24 2014
CROSSREFS
Cf. A000290, A145236 (equals a(A000040)), A068527 (difference for k=1).
Sequence in context: A319193 A097886 A308293 * A088863 A053283 A035669
KEYWORD
nonn
AUTHOR
Valtteri Raiko, Oct 24 2014
STATUS
approved