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%I #40 Feb 14 2023 06:30:19
%S 1,3,1,9,4,1,27,13,5,1,81,40,18,6,1,243,121,58,24,7,1,729,364,179,82,
%T 31,8,1,2187,1093,543,261,113,39,9,1,6561,3280,1636,804,374,152,48,10,
%U 1,19683,9841,4916,2440,1178,526,200,58,11,1,59049,29524,14757,7356,3618,1704,726,258,69,12,1
%N A Pascal-like triangle based on 3^n.
%C Row sums are A027649. Antidiagonal sums are A106517.
%C From _Wolfdieter Lang_, Jan 09 2015: (Start)
%C Alternating row sums give A025192. The A-sequence of this Riordan lower triangular matrix is [1, 1, repeat(0, )] (leading to the Pascal recurrence for T(n,k) for n >= k >= 1. The Z-sequence is [3, repeat(0, )] (leading to the recurrence T(n,0) = 3*T(n-1,0), n >= 1. For A- and Z-sequences see the W. Lang link under A006232.
%C The inverse of this Riordan matrix is Tinv = ((1 - 2*x)/(1 + x), x/(1 + x)) given as a signed version of A093560: Tinv(n,m) = (-1)^(n-m)*A093560(n,m). (End)
%F Riordan array (1/(1-3x), x/(1-x)); Number triangle T(n, 0)=A000244(n), T(n, k)=T(n-1, k-1)+T(n-1, k); T(n, k)=sum{j=0..n, binomial(n, k+j)2^j}.
%F From _Peter Bala_, Jul 16 2013: (Start)
%F T(n,k) = binomial(n,k) + 2*sum {i = 1..n} 3^(i-1)*binomial(n-i,k).
%F O.g.f.: (1 - t)/( (1 - 3*t)*(1 - (1 + x)*t) ) = 1 + (3 + x)*t + (9 + 4*x + x^2)*t^2 + ....
%F The n-th row polynomial R(n,x) = 1/(x - 2)*( x*(x + 1)^n - 2*3^n ). (End)
%F Closed-form formula for arbitrary left and right borders of Pascal-like triangle see A228196. - _Boris Putievskiy_, Aug 19 2013
%F T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 3*T(n-2,k-1), T(0,0)=1, T(1,0)=3, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - _Philippe Deléham_, Dec 26 2013
%F From _Peter Bala_, Dec 23 2014: (Start)
%F exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(27 + 13*x + 5*x^2/2! + x^3/3!) = 27 + 40*x + 58*x^2/2! + 82*x^3/3! + 113*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
%F Let M denote the present triangle. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
%F /I_k 0\
%F \ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A143495 (but with a different offset). See the Example section. Cf. A055248. (End)
%e The triangle T(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e 0: 1
%e 1: 3 1
%e 2: 9 4 1
%e 3: 27 13 5 1
%e 4: 81 40 18 6 1
%e 5: 243 121 58 24 7 1
%e 6: 729 364 179 82 31 8 1
%e 7: 2187 1093 543 261 113 39 9 1
%e 8: 6561 3280 1636 804 374 152 48 10 1
%e 9: 19683 9841 4916 2440 1178 526 200 58 11 1
%e 10: 59049 29524 14757 7356 3618 1704 726 258 69 12 1
%e ... reformatted and extended. - _Wolfdieter Lang_, Jan 06 2015
%e ----------------------------------------------------------
%e With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
%e / 1 \/1 \/1 \ /1 \
%e | 3 1 ||0 1 ||0 1 | | 3 1 |
%e | 9 4 1 ||0 3 1 ||0 0 1 |... = | 9 7 1 |
%e |27 13 5 1 ||0 9 4 1 ||0 0 3 1 | |27 37 12 1 |
%e |... ||0 27 13 5 1 ||0 0 9 4 1| |... |
%e |... ||... ||... | |... |
%e = A143495. - _Peter Bala_, Dec 23 2014
%t a106516[n_] := Block[{a, k},
%t a[x_] := Flatten@ Last@ Reap[For[k = -1, k < x, Sow[Binomial[x, k] +
%t 2 Sum[3^(i - 1)*Binomial[x - i, k], {i, 1, x}]], k++]]; Flatten@Array[a, n, 0]]; a106516[11] (* _Michael De Vlieger_, Dec 23 2014 *)
%Y Columns 1, 2, 3, 4, 5: A003462, A000340, A052150, A097786, A097787.
%Y Cf. A055248, A143495.
%K easy,nonn,tabl
%O 0,2
%A _Paul Barry_, May 05 2005
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