A093909 -id:A093909 - cp's OEIS frontend

A093908 Let f(k, n) be the product of n consecutive numbers beginning with k. Then a(n) is the least k > 1+n(n-1)/2 such that f(k, n) is a multiple of f(1+n(n-1)/2, n).

2, 3, 8, 39, 52, 187, 204, 863, 773, 6621, 34038, 2404, 34440, 223097, 11976, 1106290, 1980047, 85119892, 15308072, 496820597, 2590416388, 1087065675, 4736428784, 1128909067, 242793786666, 2791304683100, 273924845940

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Amarnath Murthy, Apr 24 2004

nonn

f(k, n) = A008279(n+k-1, n). 1+n*(n-1)/2 = A000124(n-1). f(1+n*(n-1)/2, n) = A057003(n).

a(28) > 88*10^12.

			a(4) = 39 because 39*40*41*42 is divisible by 7*8*9*10. No
smaller set gives a product that is a multiple of 7*8*9*10.

Cf. A000124, A008279, A057003, A093909.

Edited and extended by David Wasserman, Apr 25 2007

cp's OEIS Frontend

A093908 Let f(k, n) be the product of n consecutive numbers beginning with k. Then a(n) is the least k > 1+n(n-1)/2 such that f(k, n) is a multiple of f(1+n(n-1)/2, n).

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cp's OEIS Frontend

A093908 Let f(k, n) be the product of n consecutive numbers beginning with k. Then a(n) is the least k > 1+n*(n-1)/2 such that f(k, n) is a multiple of f(1+n*(n-1)/2, n).

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Author

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Comments

Examples

Crossrefs

Extensions

A093908 Let f(k, n) be the product of n consecutive numbers beginning with k. Then a(n) is the least k > 1+n(n-1)/2 such that f(k, n) is a multiple of f(1+n(n-1)/2, n).