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A092959
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Least square of the form 'product of n successive terms of an arithmetic progression + 1', or 0 if no such square exists.
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1
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4, 4, 16, 25, 121, 5041, 5041, 0, 2504902401, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
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OFFSET
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1,1
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COMMENTS
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Conjecture: No term is zero.
All terms in the progression are required to be positive. Zero values are highly probable but unproved. I have checked for each a(n) up to 10^(3*n+8). - David Wasserman, Aug 11 2006
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LINKS
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EXAMPLE
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a(3) = 16 = 1*3*5 + 1, a(4) = 25 = 1*2*3*4 + 1.
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PROG
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(PARI) f(n, x, y) = prod(i = 0, n - 1, x + i*y) + 1;
for (n = 8, 24, LIMIT = 10^(3*n + 8); x = 1; y = 1; num = f(n, 1, 1); while (num < LIMIT, while (num < LIMIT, if (issquare(num), print([n, num])); y++; num = f(n, x, y)); x++; y = 1; num = f(n, x, y))); \\ David Wasserman, Aug 11 2006
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CROSSREFS
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KEYWORD
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less,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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