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A092959
Least square of the form 'product of n successive terms of an arithmetic progression + 1', or 0 if no such square exists.
1
4, 4, 16, 25, 121, 5041, 5041, 0, 2504902401, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
OFFSET
1,1
COMMENTS
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
EXAMPLE
a(3) = 16 = 1*3*5 + 1, a(4) = 25 = 1*2*3*4 + 1.
PROG
(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
CROSSREFS
Sequence in context: A156232 A053441 A065732 * A330054 A183433 A322039
KEYWORD
less,nonn
AUTHOR
Amarnath Murthy, Mar 25 2004
EXTENSIONS
More terms from David Wasserman, Aug 11 2006
STATUS
approved