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 A249733 Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle. 8
 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, 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, 38, 28, 18, 28, 14, 0, 50, 37, 24, 36, 24, 12, 22, 11, 0, 52, 32, 12, 34, 20, 6, 16, 8, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,10 COMMENTS Number of zeros on row n of A095143 (Pascal's triangle reduced modulo 9). This should have a formula. See for example A062296, A006047 and A048967. LINKS Antti Karttunen, Table of n, a(n) for n = 0..6561 FORMULA For all n >= 0, the following holds: a(n) <= A048277(n). a(n) <= A062296(n). a(2*A249719(n)) > 0 and a((2*A249719(n))-1) > 0. a(n) is odd if and only if n is one of the terms of A249720. EXAMPLE 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. 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. MATHEMATICA Total/@Table[If[Mod[Binomial[n, k], 9]==0, 1, 0], {n, 0, 80}, {k, 0, n}] (* Harvey P. Dale, Feb 12 2020 *) PROG (PARI) 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. for(n=0, 6561, write("b249733.txt", n, " ", A249733(n))); CROSSREFS Cf. A007318, A048277, A048967, A062296, A095143, A249343, A249723, A249731, A249732, A051382, A249719, A249720, A006047. Sequence in context: A322632 A184082 A198868 * A241532 A196830 A206530 Adjacent sequences:  A249730 A249731 A249732 * A249734 A249735 A249736 KEYWORD nonn AUTHOR Antti Karttunen, Nov 04 2014 STATUS approved

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Last modified February 27 18:47 EST 2020. Contains 332308 sequences. (Running on oeis4.)