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

%I A249434 #18 Aug 29 2021 12:01:12
%S A249434 0,1,2,4,6,10,12,16,18,22,28,30,35,36,39,40,42,46,52,58,60,62,66,70,
%T A249434 72,78,79,82,83,88,89,96,100,102,104,106,107,108,112,126,130,131,136,
%U A249434 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
%N A249434 Integers m such that m! divides the product of elements on row m of Pascal's triangle.
%C A249434 Integers m such that A249151(m) >= m.
%C A249434 Equally: Integers m such that A249431(m) is nonnegative.
%C A249434 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.
%H A249434 Antti Karttunen, <a href="/A249434/b249434.txt">Table of n, a(n) for n = 1..1421</a>
%e A249434 0! = 1 divides the product of binomial coefficients on row 0 of A007318, namely {1}, thus a(1) = 0.
%e A249434 1! = 1 divides the product of row 1 (1*1), thus a(2) = 1.
%e A249434 2! = 2 divides the product of row 2 (1*2*1), thus a(3) = 2.
%e A249434 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.
%o A249434 (Scheme, with _Antti Karttunen_'s IntSeq-library)
%o A249434 (define A249434 (MATCHING-POS 1 0 (COMPOSE not negative? A249431)))
%Y A249434 Complement: A249433.
%Y A249434 Subsequences: A006093 (conjectured), A249429, A249430, A249432.
%Y A249434 Cf. A000142, A001142, A007318, A249151, A249431.
%K A249434 nonn
%O A249434 1,3
%A A249434 _Antti Karttunen_, Nov 02 2014