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%I #19 Mar 25 2021 21:19:35
%S 0,1,1,1,2,3,3,4,6,9,5,8,12,18,27,11,16,24,36,54,81,21,32,48,72,108,
%T 162,243,43,64,96,144,216,324,486,729,85,128,192,288,432,648,972,1458,
%U 2187,171,256,384,576,864,1296,1944,2916,4374,6561,341,512,768,1152,1728,2592,3888,5832,8748,13122,19683
%N Array T(n,k) read by antidiagonals: the k-th term of the n-th iterated differences of A140429.
%C Deleting column k=0 and reading by antidiagonals yields A036561.
%C Deleting column k=0 and reading the antidiagonals downwards yields A175840.
%H Nathaniel Johnston, <a href="/A155118/b155118.txt">Table of n, a(n) for n = 0..10000</a>
%F For the square array:
%F T(n,k) = 2^n*3^(k-1), k>0.
%F T(n,k) = T(n-1,k+1) - T(n-1,k), n>0.
%F Rows:
%F T(0,k) = A140429(k) = A000244(k-1).
%F T(1,k) = A025192(k).
%F T(2,k) = A003946(k).
%F T(3,k) = A080923(k+1).
%F T(4,k) = A257970(k+3).
%F Columns:
%F T(n,0) = A001045(n) (Jacobsthal numbers J_{n}).
%F T(n,1) = A000079(n).
%F T(n,2) = A007283(n).
%F T(n,3) = A005010(n).
%F T(n,4) = A175806(n).
%F T(0,k) - T(k+1,0) = 4*A094705(k-2).
%F From _G. C. Greubel_, Mar 25 2021: (Start)
%F For the antidiagonal triangle:
%F t(n, k) = T(n-k, k).
%F t(n, k) = (2^(n-k) - (-1)^(n-k))/3 (J_{n-k}) if k = 0 else 2^(n-k)*3^(k-1).
%F Sum_{k=0..n} t(n, k) = 3^n - J_{n+1}, where J_{n} = A001045(n).
%F Sum_{k=0..n} t(n, k) = A004054(n-1) for n >= 1. (End)
%e The array starts in row n=0 with columns k>=0 as:
%e 0 1 3 9 27 81 243 729 2187 ... A140429;
%e 1 2 6 18 54 162 486 1458 4374 ... A025192;
%e 1 4 12 36 108 324 972 2916 8748 ... A003946;
%e 3 8 24 72 216 648 1944 5832 17496 ... A080923;
%e 5 16 48 144 432 1296 3888 11664 34992 ... A257970;
%e 11 32 96 288 864 2592 7776 23328 69984 ...
%e 21 64 192 576 1728 5184 15552 46656 139968 ...
%e Antidiagonal triangle begins as:
%e 0;
%e 1, 1;
%e 1, 2, 3;
%e 3, 4, 6, 9;
%e 5, 8, 12, 18, 27;
%e 11, 16, 24, 36, 54, 81;
%e 21, 32, 48, 72, 108, 162, 243;
%e 43, 64, 96, 144, 216, 324, 486, 729;
%e 85, 128, 192, 288, 432, 648, 972, 1458, 2187; - _G. C. Greubel_, Mar 25 2021
%p T:=proc(n,k)if(k>0)then return 2^n*3^(k-1):else return (2^n - (-1)^n)/3:fi:end:
%p for d from 0 to 8 do for m from 0 to d do print(T(d-m,m)):od:od: # _Nathaniel Johnston_, Apr 13 2011
%t t[n_, k_]:= If[k==0, (2^(n-k) -(-1)^(n-k))/3, 2^(n-k)*3^(k-1)];
%t Table[t[n, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Mar 25 2021 *)
%o (Magma)
%o t:= func< n,k | k eq 0 select (2^(n-k) -(-1)^(n-k))/3 else 2^(n-k)*3^(k-1) >;
%o [t(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Mar 25 2021
%o (Sage)
%o def A155118(n,k): return (2^(n-k) -(-1)^(n-k))/3 if k==0 else 2^(n-k)*3^(k-1)
%o flatten([[A155118(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Mar 25 2021
%Y Cf. A001045, A004054, A046936.
%K nonn,tabl,easy
%O 0,5
%A _Paul Curtz_, Jan 20 2009
%E a(22) - a(57) from _Nathaniel Johnston_, Apr 13 2011
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