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Table by antidiagonals of n*k*(k+1)/2.
4

%I #16 Sep 08 2022 08:45:03

%S 0,0,0,0,1,0,0,3,2,0,0,6,6,3,0,0,10,12,9,4,0,0,15,20,18,12,5,0,0,21,

%T 30,30,24,15,6,0,0,28,42,45,40,30,18,7,0,0,36,56,63,60,50,36,21,8,0,0,

%U 45,72,84,84,75,60,42,24,9,0,0,55,90,108,112,105,90,70,48,27,10,0

%N Table by antidiagonals of n*k*(k+1)/2.

%H G. C. Greubel, <a href="/A062707/b062707.txt">Antidiagonals n = 0..100, flattened</a>

%F T(n, k) = T(n, 1)*T(1, k) = A001477(n)*A000217(k).

%F T(n, k) = A057145(n+2, k+1)-(k+1).

%e 0 0 0 0 0 0 0 0 0

%e 0 1 3 6 10 15 21 28 36

%e 0 2 6 12 20 30 42 56 72

%e 0 3 9 18 30 45 63 84 108

%e 0 4 12 24 40 60 84 112 144

%e 0 5 15 30 50 75 105 140 180

%e 0 6 18 36 60 90 126 168 216

%e 0 7 21 42 70 105 147 196 252

%e 0 8 24 48 80 120 168 224 288

%p seq(seq(k*binomial(n-k+1,2), k=0..n), n=0..12); # _G. C. Greubel_, Sep 02 2019

%t Table[k*Binomial[n-k+1, 2], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Sep 02 2019 *)

%o (PARI) T(n,k) = k*binomial(n-k+1,2);

%o for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Sep 02 2019

%o (Magma) [k*Binomial(n-k+1,2): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Sep 02 2019

%o (Sage) [[k*binomial(n-k+1,2) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, Sep 02 2019

%o (GAP) Flat(List([0..12], n-> List([0..n], k-> k*Binomial(n-k+1,2)))); # _G. C. Greubel_, Sep 02 2019

%Y Main diagonal is A002411. Sum of antidiagonals is A000332.

%K easy,nonn,tabl

%O 0,8

%A _Henry Bottomley_, Jul 11 2001