reviewed

approved

proposed

reviewed

editing

proposed

A recursion triangle sequence:fTriangle T(q,n, k)=(1 - (-, q)^ = e(n, k)/(1 + , q);q=2; , where e(n, k, q) = f((1 - (-q, k + )^(n +k- 1))/(1 + q))*e(n - 1, k, q) + (-q)^(n + k - 2)*e(n - 1, k - 1, q), e(n, 0, q) = e(n, n, q) = 1, and q = 2, read by rows.

1, 1, 1, 1, -13, 1, 1, -127, 395, 1, 1, 2635, 8857, -50645, 1, 1, 113369, -1090125, -6392903, 25929899, 1, 1, -9636493, -157388911, 2738123923, 11163788345, -53104434517, 1, 1, -1647840047, 58603503067, 1708972394545, -20846248885229, -99301333604807, 435031527557803, 1

Row sums are:

{1, 2, -11, 270, -39151, 18560242, -39369547651, 316649873125334,

-10469504736950236343, 1366047251880111590518042,...}.

The recursion sequence is an effort to get the q-Lah recursion.

G. C. Greubel, <a href="/A156539/b156539.txt">Rows n = 1..50 of the triangle, flattened</a>

R. Parthasarathy, <a href="http://arxiv.org/abs/quant-ph/0403216">q-Fermionic Numbers and Their Roles in Some Physical Problems</a>, arxivarXiv:quant-ph/0403216, 2004.

fT(n, k, q) = e(n, k, q,), where e(n, k, q) = ((1 - (-q)^(n+k-1))/(1 + q);)*e(n-1, k, q) + (-q)^(n+k-2)*e(n-1, k-1, q), e(n, 0, q) = e(n, n, q) = 1, and q = 2;;.

e(n,k)= f(q, k + n - 1)*e(n - 1, k) + (-q)^(n + k - 2)e(n - 1, k - 1).

{1},

Triangle begins as:

1;

{ 1, 1},;

{ 1, -13, 1},;

{ 1, -127, 395, 1},;

{ 1, 2635, 8857, -50645, 1},;

{ 1, 113369, -1090125, -6392903, 25929899, 1},;

{ 1, -9636493, -157388911, 2738123923, 11163788345, -53104434517, 1},;

{1, -1647840047, 58603503067, 1708972394545, -20846248885229, -99301333604807, 435031527557803, 1},

{1, 561913455515, 38338804386633, -2452767292835141, -49931154777504079, 713057053683646995, 3124896325732724281, -14255113095014110549, 1},

{1, 383786890117769, -53483266744648765, -6541471733965121303, 292767105902855250491, 6970650157511849296049, -91628813770737106650605, -418026940631041685516743, 1868446183589689497791147, 1}

Clear[e, [n, _, k, _, q_]; f:= e[n, k, q_, ]= If[k<0 || k>n, 0, If[k==1 || k_] := =n, 1, ((1 - (-q)^(n+k-1))/(1 + q))*e[n-1, k, q] + (-q)^(n+k-2)*e[n-1, k-1, q] ]];

q = 2; e[n_, 0] := 0; e[n_, 1] := 1;

e[n_, n_] := 1; e[n_, k_] := 0 /; k >= n + 1;

e[n_, k_] := f[q, k + n - 1]*e[n - 1, k] + (-q)^(n + k - 2)e[n - 1, k - 1];

T[n_, k_, q_]:= e[n, k, q];

Table[Table[eT[n, k, 2], {k, 1, n, 12}], , {k, n, 1, 10}]; //Flatten (* modified by _G. C. Greubel_, Jan 03 2022 *)

Flatten[%]

(Sage)

def e(n, k, q):

if (k<0 or k>n): return 0

elif (k==1 or k==n): return 1

else: return ((1-(-q)^(n+k-1))/(1+q))*e(n-1, k, q) + (-q)^(n+k-2)*e(n-1, k-1, q)

def T(n, k, q): return e(n, k, q)

flatten([[T(n, k, 2) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Jan 03 2022

Cf. A156535, A156538.

sign,tabl,uned

Edited by G. C. Greubel, Jan 03 2022

approved

editing

editing

approved

R.Parthasarathy,q-Fermionic Numbers and Their Roles in Some Physical Problems: http://arxiv.org/abs/quant-ph/0403216

R. Parthasarathy, <a href="http://arxiv.org/abs/quant-ph/0403216">q-Fermionic Numbers and Their Roles in Some Physical Problems</a>, arxiv:quant-ph/0403216

approved

editing

_Roger L. Bagula (rlbagulatftn(AT)yahoo.com), _, Feb 09 2009

A recursion triangle sequence:f(q,k)=(1 - (-q)^k)/(1 + q);q=2; e(n,k)= f(q, k + n - 1)*e(n - 1, k) + (-q)^(n + k - 2)e(n - 1, k - 1).

1, 1, 1, 1, -13, 1, 1, -127, 395, 1, 1, 2635, 8857, -50645, 1, 1, 113369, -1090125, -6392903, 25929899, 1, 1, -9636493, -157388911, 2738123923, 11163788345, -53104434517, 1, 1, -1647840047, 58603503067, 1708972394545, -20846248885229

1,5

Row sums are:

{1, 2, -11, 270, -39151, 18560242, -39369547651, 316649873125334,

-10469504736950236343, 1366047251880111590518042,...}.

The recursion sequence is an effort to get the q-Lah recursion.

R.Parthasarathy,q-Fermionic Numbers and Their Roles in Some Physical Problems: http://arxiv.org/abs/quant-ph/0403216

f(q,k)=(1 - (-q)^k)/(1 + q);q=2;;

e(n,k)= f(q, k + n - 1)*e(n - 1, k) + (-q)^(n + k - 2)e(n - 1, k - 1).

{1},

{1, 1},

{1, -13, 1},

{1, -127, 395, 1},

{1, 2635, 8857, -50645, 1},

{1, 113369, -1090125, -6392903, 25929899, 1},

{1, -9636493, -157388911, 2738123923, 11163788345, -53104434517, 1},

{1, -1647840047, 58603503067, 1708972394545, -20846248885229, -99301333604807, 435031527557803, 1},

{1, 561913455515, 38338804386633, -2452767292835141, -49931154777504079, 713057053683646995, 3124896325732724281, -14255113095014110549, 1},

{1, 383786890117769, -53483266744648765, -6541471733965121303, 292767105902855250491, 6970650157511849296049, -91628813770737106650605, -418026940631041685516743, 1868446183589689497791147, 1}

Clear[e, n, k, q]; f[q_, k_] := (1 - (-q)^k)/(1 + q);

q = 2; e[n_, 0] := 0; e[n_, 1] := 1;

e[n_, n_] := 1; e[n_, k_] := 0 /; k >= n + 1;

e[n_, k_] := f[q, k + n - 1]*e[n - 1, k] + (-q)^(n + k - 2)e[n - 1, k - 1];

Table[Table[e[n, k], {k, 1, n}], {n, 1, 10}];

Flatten[%]

sign,tabl,uned,new

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Feb 09 2009

approved