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A198455
Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = 1, ordered by a and then b; sequence gives b values.
3
2, 5, 9, 6, 14, 9, 20, 27, 10, 35, 13, 21, 44, 26, 54, 14, 20, 65, 17, 24, 77, 44, 90, 14, 18, 33, 51, 104, 21, 38, 119, 135, 22, 49, 75, 152, 25, 55, 84, 170, 35, 45, 189, 26, 39, 50, 68, 209, 29, 35, 75, 114, 230, 125
OFFSET
1,1
COMMENTS
See A198453.
The definition amounts to saying that T_a+T_b=T_c where T_i denotes a triangular number (A000217). - N. J. A. Sloane, Apr 01 2020
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
LINKS
J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000 [math.NT], April 2020.
EXAMPLE
2*3 + 2*3 = 3*4
3*4 + 5*6 = 6*7
4*5 + 9*10 = 10*11
5*6 + 6*7 = 8*9
5*6 + 14*15 = 15*16
6*7 + 9*10 = 11*12
PROG
(True BASIC)
input k
for a = (abs(k)-k+4)/2 to 40
for b = a to (a^2+abs(k)*a+2)/2
let t = a*(a+k)+b*(b+k)
let c =int((-k+ (k^2+4*t)^.5)/2)
if c*(c+k)=t then print a; b; c,
next b
print
next a
end
KEYWORD
nonn
AUTHOR
Charlie Marion, Oct 26 2011
STATUS
approved