A135225 - OEIS

%I #26 Mar 27 2022 19:02:48

%S 1,1,1,1,1,1,1,1,2,1,1,1,3,3,1,1,1,4,6,4,1,1,1,5,10,10,5,1,1,1,6,15,

%T 20,15,6,1,1,1,7,21,35,35,21,7,1,1,1,8,28,56,70,56,28,8,1

%N Pascal's triangle A007318 augmented with a leftmost border column of 1's.

%C Row sums give A094373.

%C From _Peter Bala_, Sep 08 2011: (Start)

%C This augmented Pascal array, call it P, has interesting connections with the Bernoulli polynomials B(n,x). The infinitesimal generator S of P is the array such that exp(S) = P. The array S is obtained by augmenting the infinitesimal generator A132440 of the Pascal triangle with an initial column [0, 0, 1/2, 1/6, 0, -1/30, ...] on the left. The entries in this column, after the first two zeros, are the Bernoulli values B(n,1), n>=1.

%C The array P is also connected with the problem of summing powers of consecutive integers. In the array P^n, the entry in position p+1 of the first column is equal to sum {k = 1..n} k^p - see the Example section below.

%C For similar results for the square of Pascal's triangle see A062715.

%C Note: If we augment Pascal's triangle with the column [1, 1, x, x^2, x^3, ...] on the left, the resulting lower unit triangular array has the Bernoulli polynomials B(n,x) in the first column of its infinitesimal generator. The present case is when x = 1.

%C (End)

%H G. C. Greubel, <a href="/A135225/b135225.txt">Rows n = 0..100 of triangle, flattened</a>

%F A103451 * A007318 * A000012(signed), where A000012(signed) = (1; -1,1; 1,-1,1; ...); as infinite lower triangular matrices.

%F Given A007318, binomial(n,k) is shifted to T(n+1,k+1) and a leftmost border of 1's is added.

%e First few rows of the triangle:

%e   1;

%e   1, 1;

%e   1, 1, 1;

%e   1, 1, 2, 1;

%e   1, 1, 3, 3, 1;

%e   1, 1, 4, 6, 4, 1;

%e ...

%e The infinitesimal generator for P begins:

%e   /0

%e   |0.......0

%e   |1/2.....1...0

%e   |1/6.....0...2....0

%e   |0.......0...0....3....0

%e   |-1/30...0...0....0....4....0

%e   |0.......0...0....0....0....5....0

%e   |1/42....0...0....0....0....0....6....0

%e   |...

%e   \

%e The array P^n begins:

%e   /1

%e   |1+1+...+1........1

%e   |1+2+...+n........n.........1

%e   |1+2^2+...+n^2....n^2.....2*n........1

%e   |1+2^3+...+n^3....n^3.....3*n^2....3*n.......1

%e   |...

%e   \

%e More generally, the array P^t, defined as exp(t*S) for complex t, begins:

%e   /1

%e   |B(1,1+t)-B(1,1)..........1

%e   |1/2*(B(2,1+t)-B(2,1))....t.........1

%e   |1/3*(B(3,1+t)-B(3,1))....t^2.....2*t........1

%e   |1/4*(B(4,1+t)-B(4,1))....t^3.....3*t^2....3*t.......1

%e   |...

%e   \

%p T:= proc(n, k) option remember;

%p       if k=0 then 1

%p     else binomial(n-1, k-1)

%p       fi; end:

%p seq(seq(T(n, k), k=0..n), n=0..10); # _G. C. Greubel_, Nov 19 2019

%t T[n_, k_]:= T[n, k]= If[k==0, 1, Binomial[n-1, k-1]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* _G. C. Greubel_, Nov 19 2019 *)

%o (PARI) T(n,k) = if(k==0, 1, binomial(n-1, k-1)); \\ _G. C. Greubel_, Nov 19 2019

%o (Magma)

%o T:= func< n, k | k eq 0 select 1 else Binomial(n-1, k-1) >;

%o [T(n,k): k in [0..n], n in [0..10]]; // _G. C. Greubel_, Nov 19 2019

%o (Sage)

%o @CachedFunction

%o def T(n,k):

%o     if (k==0): return 1

%o     else: return binomial(n-1, k-1)

%o [[T(n,k) for k in (0..n)] for n in (0..10)] # _G. C. Greubel_, Nov 19 2019

%o (GAP)

%o T:= function(n,k)

%o     if k=0 then return 1;

%o     else return Binomial(n-1,k-1);

%o     fi; end;

%o Flat(List([0..10], n-> List([0..n], k-> T(n,k) ))); # _G. C. Greubel_, Nov 19 2019

%Y Cf. A007318, A027641, A027642, A062715, A094373, A103438, A132440.

%K nonn,easy,tabl

%O 0,9

%A _Gary W. Adamson_, Nov 23 2007

%E Corrected by _R. J. Mathar_, Apr 16 2013