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A267077
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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.
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2
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1, 35, 30, 18135, 189, 27, 321300, 23760, 1188585957, 1656083, 26, 244894427400, 82093908624206325, 1858717755529547, 86478, 21491811639746039592, 26135932603458945934958445, 353382195058506640426335, 26780050, 7859354769338288038121982384, 274554988002
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OFFSET
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0,2
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LINKS
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EXAMPLE
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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.
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PROG
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(Python)
from math import sqrt
if n == 0:
return 1
u, v, t, w = max(8, 2*n), max(4, n)**2-9, 4*n*(n+1), n**2
while True:
m, r = divmod(v, t)
if not r and int(sqrt(m*w+1))**2 == m*w+1:
return m
v += u+1
(Python)
#!/usr/bin/python3
# This sequence is easy if you use a Pell-equation solver such as labmath.py
# nx^2 - (4n+4)y^2 = 5n-4; but also y^2 == 1 mod n^2
# Let u = nx, then # u^2 - n*(4n+4)y^2 = n*(5n-4)
# and (y > n) and (u == 0 mod n) and (y^2 == 1 mod n^2)
# (y > n makes m > 0)
# Report m = (y^2 - 1) / n^2
import labmath
print(0, 1)
print(1, 35) # When n<2, the Pell equation is elliptical.
for nn in range(2, 1001):
nsq = nn * nn
ps = labmath.pell(nn*(4*nn+4), nn*(5*nn-4))
uu, yy = next(ps[0])
while (yy <= nn) or ((uu % nn) != 0) or ((yy*yy) % nsq != 1):
uu, yy = next(ps[0])
print(nn, (yy*yy - 1) // nsq)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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