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(1,[99n+1]) Pascal Triangle.
3

%I #14 Apr 25 2024 09:12:50

%S 1,1,100,1,101,199,1,102,300,298,1,103,402,598,397,1,104,505,1000,995,

%T 496,1,105,609,1505,1995,1491,595,1,106,714,2114,3500,3486,2086,694,1,

%U 107,820,2828,5614,6986,5572,2780,793,1,108,927,3648,8442,12600,12558

%N (1,[99n+1]) Pascal Triangle.

%H G. C. Greubel, <a href="/A172179/b172179.txt">Rows n = 1..50 of the triangle, flattened</a>

%F T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(1,0) = T(2,0) = 1, T(2,1) = 100, T(n,k)=0 if k<0 or if k>=n. - _Philippe Deléham_, Dec 26 2013

%F From _G. C. Greubel_, Apr 27 2022: (Start)

%F T(n, k) = 99*binomial(n-1, k-2) + binomial(n-1, k-1).

%F T(n, n) = A172178(n-1).

%F Sum_{k=1..n} T(n, k) = 100*A000225(n-1) + 1. (End)

%e Triangle begins as:

%e 1;

%e 1, 100;

%e 1, 101, 199;

%e 1, 102, 300, 298;

%e 1, 103, 402, 598, 397;

%e 1, 104, 505, 1000, 995, 496;

%e 1, 105, 609, 1505, 1995, 1491, 595;

%e 1, 106, 714, 2114, 3500, 3486, 2086, 694;

%e 1, 107, 820, 2828, 5614, 6986, 5572, 2780, 793;

%e 1, 108, 927, 3648, 8442, 12600, 12558, 8352, 3573, 892;

%t Table[99*Binomial[n-1, k-2] + Binomial[n-1, k-1], {n,12}, {k,n}]//Flatten (* _G. C. Greubel_, Apr 27 2022 *)

%o (SageMath) flatten([[98*binomial(n-1,k-2) + binomial(n,k-1) for k in (1..n)] for n in (1..12)]) # _G. C. Greubel_, Apr 27 2022

%Y Cf. A000225, A007318, A172171, A172178.

%K nonn,tabl

%O 1,3

%A _Mark Dols_, Jan 28 2010