Search: id:A027465 Results 1-1 of 1 results found. %I A027465 %S A027465 1,3,1,9,6,1,27,27,9,1,81,108,54,12,1,243,405,270,90,15,1,729,1458, %T A027465 1215,540,135,18,1,2187,5103,5103,2835,945,189,21,1,6561,17496,20412, %U A027465 13608,5670,1512,252,24,1,19683,59049,78732,61236,30618,10206,2268 %N A027465 Cube of lower triangular normalized binomial matrix. %C A027465 Row sums are powers of 4 (A000302), antidiagonal sums are A006190 (a(n) = 3*a(n-1) + a(n-2)). - Gerald McGarvey (gerald.mcgarvey(AT)comcast.net), May 17 2005 %C A027465 Triangle of coefficients in expansion of (3+x)^n. %C A027465 Also: Pure Galton board of scheme (3,1). Also: Multiplicity (number) of pairs of n-dimensional binary vectors with dot product (overlap) k. There are 2^n=A000079(n) binary vectors of length n and 2^(2n)=4^n=A000302(n) different pairs to form dot products k=Sum(i=1..n)v[i]*u[i] between these, 0<=k<=n. (Since dot products are symmetric, there are only 2^n(2^n-1)/2 different non-ordered pairs, actually). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 17 2006 %C A027465 Mirror image of A013610. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 25 2007 %C A027465 T(i,j) is the number of i-permutations of 4 objects a,b,c,d, with repetition allowed, containing j a's. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Dec 21 2007 %C A027465 Anti-diagonals square array give bisection Fibonacci sequence: A001906. Example: 81-(27-1)=55. Differential rule applied to rows give A000079. [From M. Dols (markdols99(AT)yahoo.com), Sep 01 2009] %D A027465 B. N. Cyvin et al., Isomer enumeration of unbranched catacondensed polygonal systems with pentagons and heptagons, Match, No. 34 (Oct 1996), pp. 109-121. %D A027465 E. Neuwirth, Recursively defined combinatorial functions: extending Galton's board, Disc. Math 239 (2001) 33-51 %F A027465 Numerators of lower triangle of (b^2)[ i, j ] where b[ i, j ] = binomial(i-1, j-1)/2^(i-1) if j<=i, 0 if j>i. %F A027465 Triangle whose (i, j)-th entry is binomial(i, j)*3^(i-j). %F A027465 a(n, m)= 4^(n-1)*sum(b(n, j)*b(j, m), j=m..n)= 3^(n-m)*binomial(n-1, m-1), n >= m >= 1; a(n, m) := 0, n