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A249733 Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.

%I A249733 #34 Jul 24 2025 20:48:37
%S A249733 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 A249733 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 A249733 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 A249733 Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.
%C A249733 Number of zeros on row n of A095143 (Pascal's triangle reduced modulo 9).
%C A249733 This should have a formula. See for example A062296, A006047 and A048967.
%H A249733 Antti Karttunen, <a href="/A249733/b249733.txt">Table of n, a(n) for n = 0..6561</a>
%H A249733 James G. Huard, Blair K. Spearman and Kenneth S. Williams, <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa78/aa7843.pdf">Pascal's triangle (mod 9)</a>, Acta Arithmetica (1997), Volume: 78, Issue: 4, page 331-349.
%F A249733 For all n >= 0, the following holds:
%F A249733 a(n) <= A048277(n).
%F A249733 a(n) <= A062296(n).
%F A249733 a(2*A249719(n)) > 0 and a((2*A249719(n))-1) > 0.
%F A249733 a(n) is odd if and only if n is one of the terms of A249720.
%e A249733 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 A249733 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 A249733 Total/@Table[If[Mod[Binomial[n,k],9]==0,1,0],{n,0,80},{k,0,n}] (* _Harvey P. Dale_, Feb 12 2020 *)
%o A249733 (PARI)
%o A249733 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 A249733 for(n=0, 6561, write("b249733.txt", n, " ", A249733(n)));
%o A249733 (Python)
%o A249733 import re
%o A249733 from gmpy2 import digits
%o A249733 def A249733(n):
%o A249733     s = digits(n,3)
%o A249733     n1 = s.count('1')
%o A249733     n2 = s.count('2')
%o A249733     n01 = s.count('10')
%o A249733     n02 = s.count('20')
%o A249733     n11 = len(re.findall('(?=11)',s))
%o A249733     n12 = s.count('21')
%o A249733     return n+1-(((3*(n01+1)+(n02<<2)+n12<<2)+3*n11)*(3**n2<<n1)//12) # _Chai Wah Wu_, Jul 24 2025
%Y A249733 Cf. A007318, A048277, A048967, A062296, A095143, A249343, A249723, A249731, A249732, A051382, A249719, A249720, A006047.
%K A249733 nonn
%O A249733 0,10
%A A249733 _Antti Karttunen_, Nov 04 2014