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 A300728 a(n) is the smallest positive number k such that k^2 + k*n + n^2 is a perfect square, or 0 if no such k exists. 1
 1, 0, 0, 5, 0, 3, 10, 8, 7, 15, 6, 24, 20, 35, 16, 9, 5, 63, 30, 80, 12, 24, 48, 120, 11, 15, 70, 45, 32, 195, 18, 224, 10, 7, 126, 13, 60, 323, 160, 16, 24, 399, 48, 440, 96, 27, 240, 528, 15, 56, 30, 40, 140, 675, 90, 33, 9, 55, 390, 840, 36, 899, 448, 17, 20, 39, 14, 1088, 252, 91, 26, 1224, 33, 1295, 646, 45 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS A positive a(n) cannot be 1 or a multiple of n for n > 0 since there is no square in A002061 except 1. Also it is easy to show that a(n) cannot be 2 or 4 since a(2) = a(4) = 0. From Robert Israel, Mar 12 2018: If n>=5 is odd, a(n) <= n^2/4-n/2-3/4, with a(n)=n^2/4-n/2-3/4 if n is a prime >= 5. If n>=10 and n == 2 (mod 4), a(n) <= n^2/8-n/2-3/2, with equality if n/2 is a prime >= 5. If n>=16 and n == 0 (mod 4), 1 < a(n) <= n^2/16 - n/2 - 3, with equality if n/4 is 4 or a prime >= 5. (End) LINKS Altug Alkan, Table of n, a(n) for n = 0..10000 Altug Alkan, Scatterplot of first differences for n <= 10^4 EXAMPLE a(2) = 0 because k^2 + 2*k + 4 = (k + 1)^2 + 3 cannot be a square for k > 0. a(4) = 0 because k^2 + 4*k + 16 = (k + 2)^2 + 12 cannot be a square for k > 0. a(5) = 3 because 3^2 + 3*5 + 5^2 = 7^2 and 3 is the least positive number with this property. MAPLE f:= proc(n) local k; for k from 1 do if issqr(k^2 + k*n + n^2) then return k fi od end proc: f(1):= 0: f(2):= 0: f(4):= 0: map(f, [\$0..200]); # Robert Israel, Mar 12 2018 CROSSREFS Cf. A000290, A003136, A055527. Sequence in context: A299621 A182567 A339457 * A321418 A225424 A202626 Adjacent sequences:  A300725 A300726 A300727 * A300729 A300730 A300731 KEYWORD nonn,look AUTHOR Altug Alkan, Mar 11 2018 STATUS approved

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Last modified April 14 22:47 EDT 2021. Contains 342971 sequences. (Running on oeis4.)