OFFSET
1,2
COMMENTS
This sequence is infinite. The fundamental solution of m^2 + 1 = x^2 y^3 is (m,x,y) = (682,61,5), which means the Pellian equation m^2 - 125x^2 = -1 has the solution (m,x) = (682,61) = (m(1),x(1)). This Pellian equation admits an infinity of solutions (m(2k+1),x(2k+1)), k=1,2,..., given by the following recursive relation, starting with m(1)=682, x(1)= 61: m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1).
Squares of these terms are in A060355, since both a(n)^2 and a(n)^2 + 1 are powerful (A001694). - Charles R Greathouse IV, Nov 16 2012
It appears that y = A077426. - Robert G. Wilson v, Nov 16 2012
Also m^2 + 1 is powerful. Other solutions arise from solutions x to x^2 - k^3*y^2 = -1. - Georgi Guninski, Nov 17 2012
Although it is believed that the b-file is complete for all terms m < 10^100, the search only looked for y < 100000. - Robert G. Wilson v, Nov 17 2012
REFERENCES
Albert H. Beiler, "The Pellian" (Chap. 22), Recreations in the Theory of Numbers, 2nd ed. NY: Dover, 1966.
A. Cayley, Report of a committee appointed for the purpose of carrying on the tables connected with the Pellian equation ..., Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 430-443.
J. M. De Koninck, Ces nombres qui nous fascinent, Ellipses, 2008, p. 108.
LINKS
Robert G. Wilson v, Table of n, a(n) for n = 1..292. [This list has not been proved to be complete! - N. J. A. Sloane, Nov 17 2012]
E. E. Whitford, The Pell equation, New York, 1912.
Wikipedia, Pell's equation
FORMULA
m(1)=682, x(1) = 61 and m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1) m(2k+1) = C(2k+1,0) * m(1)^(2k+1) + C(2k+1,2)*m(1)^(2k-1)*x(1)^2 + ...
EXAMPLE
For m=682, m^2 + 1 = 465125 = 61^2 * 5^3.
MAPLE
C:=array(0..20, 0..20):C[1, 1]=1: C[2, 1]=1: n1:=682:x1:=61:for nn from 1 by 2 to 15 do:s:=0:for i from 2 to 15 do:for j from 1 to i do:C[i, j]:= C[i-1, j] + C[i-1, j-1]: od:od:for n from 1 by 2 to nn+1 do:s:=s + C[nn+1, n] * n1^(nn-n+1)*x1^(n-1)*125^((n-1)/2):od:print (s):od: # Michel Lagneau
# 2nd program R. J. Mathar, Mar 16 2016:
# print (nonsorted!) all solutions of A175155 up to search limit
with(numtheory):
# upper limit for solutions n
nsearchlim := 10^40 :
A175155y := proc(y::integer)
local disc;
disc := y^3 ;
cfrac(sqrt(disc), periodic, quotients) ;
end proc:
for y from 2 do
if issqrfree(y) then
# find continued fraction for x^2-(y^3=disc)*y^2=-1, sqrt(disc)
cf := A175155y(y) ;
nlen := nops(op(2, cf)) ;
if type(nlen, odd) then
# fundamental solution
fuso := numtheory[nthconver](cf, nlen-1) ;
fusolx := numer(fuso) ;
fusoly := denom(fuso) ;
solx := fusolx ;
soly := fusoly ;
while solx <= nsearchlim do
rhhs := solx^2-y^3*soly^2 ;
if rhhs = -1 then
# print("n=", solx, "x=", soly, "y=", y^3) ;
print(solx) ;
end if;
# solutions from fundamental solution
tempx := fusolx*solx+y^3*fusoly*soly ;
tempy := fusolx*soly+fusoly*solx ;
solx := tempx ;
soly := tempy ;
end do;
end if;
fi;
end do:
MATHEMATICA
nmax = 10^50; ymax = 100; instances = 10; fi[y_] := n /. FindInstance[0 <= n <= nmax && x > 0 && n^2 + 1 == x^2*y^3, {n, x}, Integers, instances]; yy = Select[Range[1, ymax, 2], !IntegerQ[Sqrt[#]] && OddQ[ Length[ ContinuedFraction[Sqrt[#]][[2]]]]&]; Join[{0}, fi /@ yy // Flatten // Union // Most] (* Jean-François Alcover, Jul 12 2017 *)
PROG
(PARI) is(n)=ispowerful(n^2+1) \\ Charles R Greathouse IV, Nov 16 2012
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Feb 27 2010
EXTENSIONS
Added condition that x and y must be positive. Added missing initial term 0. Added warning that b-file has not been proved to be correct - there could be missing entries. - N. J. A. Sloane, Nov 17 2012
STATUS
approved