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Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.
8

%I #28 Feb 12 2020 12:50:39

%S 0,0,0,0,0,0,0,0,0,6,3,0,4,2,0,2,1,0,12,6,0,8,4,0,4,2,0,24,21,18,19,

%T 14,9,14,7,0,28,20,12,20,13,6,12,6,0,32,19,6,21,12,3,10,5,0,48,42,36,

%U 38,28,18,28,14,0,50,37,24,36,24,12,22,11,0,52,32,12,34,20,6,16,8,0

%N Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.

%C Number of zeros on row n of A095143 (Pascal's triangle reduced modulo 9).

%C This should have a formula. See for example A062296, A006047 and A048967.

%H Antti Karttunen, <a href="/A249733/b249733.txt">Table of n, a(n) for n = 0..6561</a>

%F For all n >= 0, the following holds:

%F a(n) <= A048277(n).

%F a(n) <= A062296(n).

%F a(2*A249719(n)) > 0 and a((2*A249719(n))-1) > 0.

%F a(n) is odd if and only if n is one of the terms of A249720.

%e Row 9 of Pascal's triangle is {1, 9, 36, 84, 126, 126, 84, 36, 9, 1}. The terms 9, 36, and 126 are the only multiples of nine, and each of them occurs two times on that row, thus a(9) = 2*3 = 6.

%e Row 10 of Pascal's triangle is {1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1}. The terms 45 (= 9*5) and 252 (= 9*28) are the only multiples of nine, and the former occurs twice, while the latter is alone at the center, thus a(10) = 2+1 = 3.

%t Total/@Table[If[Mod[Binomial[n,k],9]==0,1,0],{n,0,80},{k,0,n}] (* _Harvey P. Dale_, Feb 12 2020 *)

%o (PARI)

%o A249733(n) = { my(c=0); for(k=0,n\2,if(!(binomial(n,k)%9),c += (if(k<(n/2),2,1)))); return(c); } \\ Unoptimized.

%o for(n=0, 6561, write("b249733.txt", n, " ", A249733(n)));

%Y Cf. A007318, A048277, A048967, A062296, A095143, A249343, A249723, A249731, A249732, A051382, A249719, A249720, A006047.

%K nonn

%O 0,10

%A _Antti Karttunen_, Nov 04 2014