A249434 Integers m such that m! divides the product of elements on row m of Pascal's triangle.

0, 1, 2, 4, 6, 10, 12, 16, 18, 22, 28, 30, 35, 36, 39, 40, 42, 46, 52, 58, 60, 62, 66, 70, 72, 78, 79, 82, 83, 88, 89, 96, 100, 102, 104, 106, 107, 108, 112, 126, 130, 131, 136, 138, 143, 148, 149, 150, 153, 156, 159, 162, 164, 166, 167, 172, 174, 175, 178, 179, 180, 181, 190, 192, 194, 196, 197, 198, 199, 207, 209, 210, 219, 222, 226, 228, 232, 238, 240, 250, 256

Offset: 1

Author

Antti Karttunen, Nov 02 2014

Keywords

• nonn

Comments

Integers m such that A249151(m) >= m.

Equally: Integers m such that A249431(m) is nonnegative.

It seems that A006093 gives all those k for which A249151(k) = k. If that is true, then this is a disjoint union of A006093 and A249429.

Examples

			0! = 1 divides the product of binomial coefficients on row 0 of A007318, namely {1}, thus a(1) = 0.
1! = 1 divides the product of row 1 (1*1), thus a(2) = 1.
2! = 2 divides the product of row 2 (1*2*1), thus a(3) = 2.
3! = 6 does not divide the product of row 3 (1*3*3*1), but 4! = 24 divides the product of row 4 (1*4*6*4*1), as 96 = 4*24, thus a(4) = 4.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..1421

Crossrefs

Complement: A249433.

Subsequences: A006093 (conjectured), A249429, A249430, A249432.

Cf. A000142, A001142, A007318, A249151, A249431.