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Symmetrical triangle T(n, m) = floor(Eulerian(n+1, m)/2), read by rows.
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%I #13 Sep 08 2022 08:45:51

%S 2,5,5,13,33,13,28,151,151,28,60,595,1208,595,60,123,2146,7809,7809,

%T 2146,123,251,7304,44117,78095,44117,7304,251,506,23920,227596,655177,

%U 655177,227596,23920,506,1018,76318,1101744,4869057,7862124,4869057,1101744,76318,1018

%N Symmetrical triangle T(n, m) = floor(Eulerian(n+1, m)/2), read by rows.

%C Row sums are: {2, 10, 59, 358, 2518, 20156, 181439, 1814398, 19958398, 239500796, 3113510398, 43589145596, 653837183996, ...}.

%H G. C. Greubel, <a href="/A174098/b174098.txt">Rows n = 2..100 of triangle, flattened</a>

%F T(n, m) = floor(Eulerian(n+1, m)/2), where Eulerian(n,k) = A008292(n,k).

%e Triangle begins as:

%e 2;

%e 5, 5;

%e 13, 33, 13;

%e 28, 151, 151, 28;

%e 60, 595, 1208, 595, 60;

%e 123, 2146, 7809, 7809, 2146, 123;

%e 251, 7304, 44117, 78095, 44117, 7304, 251;

%e 506, 23920, 227596, 655177, 655177, 227596, 23920, 506;

%t Eulerian[n_, k_]:= Sum[(-1)^j*Binomial[n+1, j]*(k-j+1)^n, {j, 0, k+1}];

%t Table[Floor[Eulerian[n+1, m]/2], {n, 2, 12}, {m, 1, n-1}]//Flatten (* _G. C. Greubel_, Apr 25 2019 *)

%o (PARI) {T(n,k) = (sum(j=0,k+1, (-1)^j*binomial(n+2,j)*(k-j+1)^(n+1)))\2};

%o for(n=2,12, for(k=1,n-1, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Apr 25 2019

%o (Magma) [[Floor((&+[(-1)^j*Binomial(n+2,j)*(k-j+1)^(n+1): j in [0..k+1]] )/2): k in [1..n-1]]: n in [2..12]]; // _G. C. Greubel_, Apr 25 2019

%o (Sage) [[floor(sum((-1)^j*binomial(n+2,j)*(k-j+1)^(n+1) for j in (0..k+1))/2) for k in (1..n-1)] for n in (2..12)] # _G. C. Greubel_, Apr 25 2019

%Y Cf. A008292, A166454.

%K nonn,tabl

%O 2,1

%A _Roger L. Bagula_, Mar 07 2010

%E Edited by _G. C. Greubel_, Apr 25 2019