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%I #25 Sep 08 2022 08:45:18
%S 1,2,1,5,3,1,12,8,4,1,29,20,12,5,1,70,49,32,17,6,1,169,119,81,49,23,7,
%T 1,408,288,200,130,72,30,8,1,985,696,488,330,202,102,38,9,1,2378,1681,
%U 1184,818,532,304,140,47,10,1,5741,4059,2865,2002,1350,836,444,187,57,11,1
%N A Pell-Pascal matrix.
%C This triangle gives the iterated partial sums of the Pell sequence A000129(n+1), n>=0. - _Wolfdieter Lang_, Oct 05 2014
%H G. C. Greubel, <a href="/A106513/b106513.txt">Rows n = 0..50 of the triangle, flattened</a>
%F Riordan array (1/(1-2*x-x^2), x/(1-x)).
%F Number triangle T(n,0) = A000129(n+1), T(n,k) = T(n-1,k-1) + T(n-1,k).
%F T(n,k) = Sum_{j=0..floor((n+1)/2)} binomial(n+1, 2*j+k+1)*2^j.
%F Sum_{k=0..n} T(n, k) = A106514(n).
%F Sum_{k=0..floor(n/2)} T(n-k, k) = A106515(n).
%F T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-3,k) - T(n-3,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - _Philippe Deléham_, Jan 14 2014
%F From _Wolfdieter Lang_, Oct 05 2014: (Start)
%F O.g.f. for row polynomials R(n,x) = Sum_{k=0..n} T(n,k)*x^k: (1 - z)/((1 - 2*z - z^2)*(1 - (1+x)*z)).
%F O.g.f. column m: (1/(1 - 2*z - z^2))*(x/(1 - x)))^m, m >= 0. (Riordan property).
%F The alternating row sums are shown in A001333.
%F A-sequence: [1, 1] (see the three term recurrence given above). Z-sequence has o.g.f. (2 + 3*x)/(1 + x), [2, 1, repeat(-1,1)] (unsigned A054977). See the W. Lang link under A006232 for Riordan A- and Z-sequences.
%F The inverse Riordan triangle is shown in A248156. (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: 2 1
%e 2: 5 3 1
%e 3: 12 8 4 1
%e 4: 29 20 12 5 1
%e 5: 70 49 32 17 6 1
%e 6: 169 119 81 49 23 7 1
%e 7: 408 288 200 130 72 30 8 1
%e 8: 985 696 488 330 202 102 38 9 1
%e 9: 2378 1681 1184 818 532 304 140 47 10 1
%e 10: 5741 4059 2865 2002 1350 836 444 187 57 11 1
%e ... Reformatted and extended. - _Wolfdieter Lang_, Oct 05 2014
%e -----------------------------------------------------
%e Recurrence from the Z-sequence (see the formula above) for T(0,n) in terms of the entries of row n-1. For example, 29 = T(4,0) = 2*12 + 1*8 + (-1)*4 + 1*1 = 29. - _Wolfdieter Lang_, Oct 05 2014
%t T[n_, k_]= Sum[Binomial[n+1, 2*j+k+1]*2^j, {j, 0, Floor[(n+1)/2]}];
%t Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Aug 05 2021 *)
%o (Magma) [ (&+[Binomial(n+1, 2*j+k+1)*2^j: j in [0..Floor((n+1)/2)]]) : k in [0..n], n in [0..12]]; // _G. C. Greubel_, Aug 05 2021
%o (Sage)
%o @CachedFunction
%o def T(n, k):
%o if (k<0 or k>n): return 0
%o elif (k==0): return lucas_number1(n+1, 2, -1)
%o else: return T(n-1,k-1) + T(n-1,k)
%o flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Aug 05 2021
%Y Cf. A000129, A001333, A106514 (row sums), A106515 (antidiagonal sums), A248156.
%K easy,nonn,tabl
%O 0,2
%A _Paul Barry_, May 05 2005
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