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%I #63 Sep 08 2022 08:46:20
%S 1,1,1,1,3,1,1,6,6,1,1,10,15,10,1,1,15,29,29,15,1,1,21,49,63,49,21,1,
%T 1,28,76,118,118,76,28,1,1,36,111,201,243,201,111,36,1,1,45,155,320,
%U 452,452,320,155,45,1,1,55,209,484,781,913,781,484,209,55,1
%N Modified Pascal's triangle read by rows: T(n,k) = C(n+1,k) - n, 1 <= k <= n.
%C If we define T_m(n, k) = binomial(n+m,k) - m*n where m <= k <= n, then T_0 is Pascal's triangle A007318 and T_1 is the current triangle sequence.
%C This modified Pascal's triangle is symmetric: C(n+m, k) - m*n = C(n+m, n-k+1) - m*n for any nonnegative integer m.
%F T(n, k) = T_1(n, k) = binomial(n+1, k) - n, for 1 <= k <= n.
%e The triangle T(n, k) begins:
%e n\k 1 2 3 4 5 6 7 8 9 10
%e 1 1;
%e 2 1, 1;
%e 3 1, 3, 1;
%e 4 1, 6, 6, 1;
%e 5 1, 10, 15, 10, 1;
%e 6 1, 15, 29, 29, 15, 1;
%e 7 1, 21, 49, 63, 49, 21, 1;
%e 8 1, 28, 76, 118, 118, 76, 28, 1;
%e 9 1, 36, 111, 201, 243, 201, 111, 36, 1;
%e 10 1, 45, 155, 320, 452, 452, 320, 155, 45, 1; etc.
%p seq(seq(binomial(n+1,k)-n, k=1..n), n=1..10); # _Muniru A Asiru_, Feb 05 2018
%t Table[Binomial[n + 1, k] - n, {n, 11}, {k, n}] // Flatten (* _Michael De Vlieger_, Feb 05 2018 *)
%o (Magma) [[Binomial(n+1, k)- 1*n: k in [1..n]]: n in [1..10]];
%o (GAP) Flat(List([1..100],n->List([1..n],k->Binomial(n+1,k)-n))); # _Muniru A Asiru_, Feb 05 2018
%o (PARI) T(n, k) = binomial(n+1,k) - n;
%o tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Mar 01 2018
%Y Cf. A000012, A000217, A005286, A083858, A007318, A276472.
%K nonn,tabl,easy
%O 1,5
%A _Juri-Stepan Gerasimov_, Feb 03 2018
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