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%I #31 Nov 29 2023 20:23:15
%S 1,35,30,18135,189,27,321300,23760,1188585957,1656083,26,244894427400,
%T 82093908624206325,1858717755529547,86478,21491811639746039592,
%U 26135932603458945934958445,353382195058506640426335,26780050,7859354769338288038121982384,274554988002
%N Least m>0 for which m*n^2 + 1 is a square and m*triangular(n) + 1 is a triangular number (A000217). Or -1 if no such m exists.
%H Don Reble, <a href="/A267077/b267077.txt">Table of n, a(n) for n = 0..300</a>
%e 26*10^2+1 = 2601 is a square, and 26*10*11/2+1 = 1431 = triangular(53), and 26 is the smallest such multiplier, therefore a(10) = 26.
%o (Python)
%o from math import sqrt
%o def A267077(n):
%o if n == 0:
%o return 1
%o u,v,t,w = max(8,2*n),max(4,n)**2-9,4*n*(n+1),n**2
%o while True:
%o m,r = divmod(v,t)
%o if not r and int(sqrt(m*w+1))**2 == m*w+1:
%o return m
%o v += u+1
%o u += 2 # _Chai Wah Wu_, Jan 15 2016
%o (Python)
%o #!/usr/bin/python3
%o # This sequence is easy if you use a Pell-equation solver such as labmath.py
%o # Solve the A267077 Pell equation:
%o # nx^2 - (4n+4)y^2 = 5n-4; but also y^2 == 1 mod n^2
%o # Let u = nx, then # u^2 - n*(4n+4)y^2 = n*(5n-4)
%o # and (y > n) and (u == 0 mod n) and (y^2 == 1 mod n^2)
%o # (y > n makes m > 0)
%o # Report m = (y^2 - 1) / n^2
%o import labmath
%o print(0, 1)
%o print(1, 35) # When n<2, the Pell equation is elliptical.
%o for nn in range(2,1001):
%o nsq = nn * nn
%o ps = labmath.pell(nn*(4*nn+4), nn*(5*nn-4))
%o uu,yy = next(ps[0])
%o while (yy <= nn) or ((uu % nn) != 0) or ((yy*yy) % nsq != 1):
%o uu,yy = next(ps[0])
%o print(nn, (yy*yy - 1) // nsq)
%o # From _Don Reble_, Apr 15 2022, added by _N. J. A. Sloane_, Apr 15 2022.
%Y Cf. A000217, A000290, A035096, A061782, A067872, A188621.
%K nonn
%O 0,2
%A _Alex Ratushnyak_, Jan 10 2016
%E a(12)-a(15) from _Chai Wah Wu_, Jan 16 2016
%E a(16) and beyond from _Don Reble_, Apr 15 2022
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