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Triangle read by rows. T(n, k) = (n - k + 1)! * H(k, n - k), where H are the hyperharmonic numbers. For 0 <= k <= n.
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%I #33 Jul 03 2022 10:25:47

%S 1,1,1,2,3,1,6,11,5,1,24,50,26,7,1,120,274,154,47,9,1,720,1764,1044,

%T 342,74,11,1,5040,13068,8028,2754,638,107,13,1,40320,109584,69264,

%U 24552,5944,1066,146,15,1,362880,1026576,663696,241128,60216,11274,1650,191,17,1

%N Triangle read by rows. T(n, k) = (n - k + 1)! * H(k, n - k), where H are the hyperharmonic numbers. For 0 <= k <= n.

%C Previous name: Extended triangle related to the asymptotic expansions of the E(x, m = 2, n).

%C For the definition of the hyperharmonic numbers see the formula section.

%C This triangle is the same as triangle A165674 except for the extra left-hand column T(n, 0) = n!. The T(n) formulas for the right-hand columns generate the coefficients of this extra left-hand column.

%C _Leroy Quet_ discovered triangle A105954 which is the reversal of our triangle.

%C In square format, row k gives the (n-1)-st elementary symmetric function of {k, k+1, k+2,..., k+n}, as in the Mathematica section. - _Clark Kimberling_, Dec 29 2011

%F The hyperharmonic numbers are H(n, k) = Sum_{j=0..k} H(n - 1, j), with base condition H(0, k) = 1/(k + 1).

%F T(n, k) = (n - k + 1)*T(n - 1, k) + T(n - 1, k - 1), 1 <= k <= n-1, with T(n, 0) = n! and T(n, n) = 1.

%F From _Peter Luschny_, Jul 03 2022: (Start)

%F The rectangular array is given by:

%F A(n, k) = (k + 1)!*H(n, k).

%F A(n, k) = (k + 1)*((n + k)! / n!)*hypergeom([-k, 1, 1], [2, n + 1], 1). (End)

%e Triangle T(n, k) begins:

%e [0] 1;

%e [1] 1, 1;

%e [2] 2, 3, 1;

%e [3] 6, 11, 5, 1;

%e [4] 24, 50, 26, 7, 1;

%e [5] 120, 274, 154, 47, 9, 1;

%e [6] 720, 1764, 1044, 342, 74, 11, 1;

%e [7] 5040, 13068, 8028, 2754, 638, 107, 13, 1;

%e .

%e Seen as an array (the triangle arises when read by descending antidiagonals):

%e [0] 1, 1, 2, 6, 24, 120, 720, 5040, ...

%e [1] 1, 3, 11, 50, 274, 1764, 13068, 109584, ...

%e [2] 1, 5, 26, 154, 1044, 8028, 69264, 663696, ...

%e [3] 1, 7, 47, 342, 2754, 24552, 241128, 2592720, ...

%e [4] 1, 9, 74, 638, 5944, 60216, 662640, 7893840, ...

%e [5] 1, 11, 107, 1066, 11274, 127860, 1557660, 20355120, ...

%e [6] 1, 13, 146, 1650, 19524, 245004, 3272688, 46536624, ...

%e [7] 1, 15, 191, 2414, 31594, 434568, 6314664, 97053936, ...

%p nmax := 8; for n from 0 to nmax do a(n, 0) := n! od: for n from 0 to nmax do a(n, n) := 1 od: for n from 2 to nmax do for m from 1 to n-1 do a(n, m) := (n-m+1)*a(n-1, m) + a(n-1, m-1) od: od: seq(seq(a(n, m), m=0..n), n=0..nmax);

%p # _Johannes W. Meijer_, revised Nov 27 2012

%p # Shows the array format, using hyperharmonic numbers.

%p H := proc(n, k) option remember; if n = 0 then 1/(k+1)

%p else add(H(n - 1, j), j = 0..k) fi end:

%p seq(lprint(seq((k + 1)!*H(n, k), k = 0..7)), n = 0..7);

%p # Shows the array format, using the hypergeometric formula.

%p A := (n, k) -> (k+1)*((n + k)! / n!)*hypergeom([-k, 1, 1], [2, n + 1], 1):

%p seq(lprint(seq(simplify(A(n, k)), k = 0..7)), n = 0..7);

%p # _Peter Luschny_, Jul 03 2022

%t a[n_] := SymmetricPolynomial[n - 1, t[n]]; z = 10;

%t t[n_] := Table[k - 1, {k, 1, n}]; t1 = Table[a[n], {n, 1, z}] (* A000142 *)

%t t[n_] := Table[k, {k, 1, n}]; t2 = Table[a[n], {n, 1, z}] (* A000254 *)

%t t[n_] := Table[k + 1, {k, 1, n}]; t3 = Table[a[n], {n, 1, z}] (* A001705 *)

%t t[n_] := Table[k + 2, {k, 1, n}]; t4 = Table[a[n], {n, 1, z}] (* A001711 *)

%t t[n_] := Table[k + 3, {k, 1, n}]; t5 = Table[a[n], {n, 1, z}] (* A001716 *)

%t t[n_] := Table[k + 4, {k, 1, n}]; t6 = Table[a[n], {n, 1, z}] (* A001721 *)

%t t[n_] := Table[k + 5, {k, 1, n}]; t7 = Table[a[n], {n, 1, z}] (* A051524 *)

%t t[n_] := Table[k + 6, {k, 1, n}]; t8 = Table[a[n], {n, 1, z}] (* A051545 *)

%t t[n_] := Table[k + 7, {k, 1, n}]; t9 = Table[a[n], {n, 1, z}] (* A051560 *)

%t t[n_] := Table[k + 8, {k, 1, n}]; t10 = Table[a[n], {n, 1, z}] (* A051562 *)

%t t[n_] := Table[k + 9, {k, 1, n}]; t11 = Table[a[n], {n, 1, z}] (* A051564 *)

%t t[n_] := Table[k + 10, {k, 1, n}];t12 = Table[a[n], {n, 1, z}] (* A203147 *)

%t t = {t1, t2, t3, t4, t5, t6, t7, t8, t9, t10};

%t TableForm[t] (* A165675 in square format *)

%t m[i_, j_] := t[[i]][[j]];

%t (* A165675 as a sequence *)

%t Flatten[Table[m[i, n + 1 - i], {n, 1, 10}, {i, 1, n}]]

%t (* _Clark Kimberling_, Dec 29 2011 *)

%t A[n_, k_] := (k + 1)*((n + k)! / n!)*HypergeometricPFQ[{-k, 1, 1}, {2, n + 1}, 1];

%t Table[A[n, k], {n, 0, 7}, {k, 0, 7}] // TableForm (* _Peter Luschny_, Jul 03 2022 *)

%Y A105954 is the reversal of this triangle.

%Y A165674, A138771 and A165680 are related triangles.

%Y A080663 equals the third right hand column.

%Y A000142 equals the first left hand column.

%Y A093345 are the row sums.

%Y Columns include A165676, A165677, A165678 and A165679.

%K easy,nonn,tabl

%O 0,4

%A _Johannes W. Meijer_, Oct 05 2009

%E New name from _Peter Luschny_, Jul 03 2022