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Triangle T(n,m) read by rows: m! if m <= floor(n/2), and (n-m)! elseotherwise.

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_Roger L. Bagula _ and _Gary W. Adamson (rlbagulatftn(AT)yahoo.com), _, May 30 2008

Triangular sequence from a symmetrical factorial function: f(n,m)=If[m (nonascii character here) Floor[n/2], m!, (n - m)! ].

Triangle T(n,m) read by rows: m! if m <= floor(n/2), and (n-m)! else.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 6, 2, 1, 1, 1, 1, 2, 6, 6, 2, 1, 1, 1, 1, 2, 6, 24, 6, 2, 1, 1, 1, 1, 2, 6, 24, 24, 6, 2, 1, 1, 1, 1, 2, 6, 24, 120, 24, 6, 2, 1, 1, 1, 1, 2, 6, 24, 120, 120, 24, 6, 2, 1, 1, 1, 1, 2, 6, 24, 120, 720, 120, 24, 6, 2, 1, 1, 1, 1, 2, 6, 24

1,0,13

Row sums: 1, 2, 3, 4, 6, 8, 14, 20, 44, 68, 188,... which is

{1, 2, 3, 4, 6, 8, 14, 20, 44, 68, 188};

This sequence comes from consideration of an interesting (non-traditional)infinite sequence:

This infinite series comes from looking for a fourth type of Infinite series

2*A003422((n+1)/2) if n is odd, and owes it's life to hypergeometric type functions:A003422(n/2)+A003422(1+n/2) if n is even.

Sum[1/Limit[Binomial[n, m], n -> Infinity], {m, 0, Infinity}]

Mathematica can't do it:

I get terms like:middle is zero:

Limit[1/Binomial[2*n, n], n -> Infinity]=0

{1,1,1/2,1/6,...0,....1/6,1/2,1,1}

which would give limit 2*Exp[1],

but doing:

a = Table[N[Sum[1/Binomial[n, m], {m, 0, n}]], {n, 1, 1000}];

it looks like it goes toward 2 instead.

f[n_, m_] := If[m (nonascii character here) Floor[n/2], 1/m!, 1/(n - m)! ]

The conjecture is that :

Limit[Sum[1/Binomial[n, m], {m, 0, n}]] - Sum[f[n, m], {m, 0, n}]/Exp[1],n->Infinity]=0

It is by inverting the series function that I get this sequence.

fT(n,m) =If[ A000142(m (nonascii character here) Floorif m<=[n/2], = A000142(n-m) if m!, (>[n - /2], 0<=m)! ]<=n.

Conjecture: limit_{n->infinity} sum_{m=0..n} ( 1/binomial(n,m) - T(n,m) ) = 0.

T(n,m) = T(n,n-m).

{1},

1,

{1, 1},,

{1, 1, 1},,

{1, 1, 1, 1},,

{1, 1, 2, 1, 1},,

{1, 1, 2, 2, 1, 1},,

{1, 1, 2, 6, 2, 1, 1},,

{1, 1, 2, 6, 6, 2, 1, 1},,

{1, 1, 2, 6, 24, 6, 2, 1,1},,

{1, 1, 2, 6, 24, 24, 6, 2, 1, 1},,

{1, 1, 2, 6, 24, 120, 24, 6, 2, 1, 1}

g[n_, m_] := If[m (nonascii character here) <= Floor[n/2], m!, (n - m)! ]; w = Table[Table[g[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[w] Table[Apply[Plus, Table[g[n, m], {m, 0, n}]], {n, 0, 10}]

nonn,uned,easy,tabl,new

Non-Ascii characters corrected, offset set to 0, reported Mma experiments removed - The Assoc. Editors of the OEIS, Oct 31 2009

Triangular sequence from a symmetrical factorial function: f(n,m)=If[m �≁� (nonascii character here) Floor[n/2], m!, (n - m)! ].

f[n_, m_] := If[m �≁� (nonascii character here) Floor[n/2], 1/m!, 1/(n - m)! ]

f(n,m)=If[m �≁� (nonascii character here) Floor[n/2], m!, (n - m)! ].

g[n_, m_] := If[m �≁� (nonascii character here) Floor[n/2], m!, (n - m)! ]; w = Table[Table[g[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[w] Table[Apply[Plus, Table[g[n, m], {m, 0, n}]], {n, 0, 10}]

nonn,uned,tabl,new

Triangular sequence from a symmetrical factorial function: f(n,m)=If[m �≁� Floor[n/2], m!, (n - m)! ].

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 6, 2, 1, 1, 1, 1, 2, 6, 6, 2, 1, 1, 1, 1, 2, 6, 24, 6, 2, 1, 1, 1, 1, 2, 6, 24, 24, 6, 2, 1, 1, 1, 1, 2, 6, 24, 120, 24, 6, 2, 1, 1

1,13

Row sums:

{1, 2, 3, 4, 6, 8, 14, 20, 44, 68, 188};

This sequence comes from consideration of an interesting (non-traditional)infinite sequence:

This infinite series comes from looking for a fourth type of Infinite series

and owes it's life to hypergeometric type functions:

Sum[1/Limit[Binomial[n, m], n -> Infinity], {m, 0, Infinity}]

Mathematica can't do it:

I get terms like:middle is zero:

Limit[1/Binomial[2*n, n], n -> Infinity]=0

{1,1,1/2,1/6,...0,....1/6,1/2,1,1}

which would give limit 2*Exp[1],

but doing:

a = Table[N[Sum[1/Binomial[n, m], {m, 0, n}]], {n, 1, 1000}];

it looks like it goes toward 2 instead.

f[n_, m_] := If[m �≁� Floor[n/2], 1/m!, 1/(n - m)! ]

The conjecture is that :

Limit[Sum[1/Binomial[n, m], {m, 0, n}]] - Sum[f[n, m], {m, 0, n}]/Exp[1],n->Infinity]=0

It is by inverting the series function that I get this sequence.

f(n,m)=If[m �≁� Floor[n/2], m!, (n - m)! ].

{1},

{1, 1},

{1, 1, 1},

{1, 1, 1, 1},

{1, 1, 2, 1, 1},

{1, 1, 2, 2, 1, 1},

{1, 1, 2, 6, 2, 1, 1},

{1, 1, 2, 6, 6, 2, 1, 1},

{1, 1, 2, 6, 24, 6, 2, 1,1},

{1, 1, 2, 6, 24, 24, 6, 2, 1, 1},

{1, 1, 2, 6, 24, 120, 24, 6, 2, 1, 1}

g[n_, m_] := If[m �≁� Floor[n/2], m!, (n - m)! ]; w = Table[Table[g[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[w] Table[Apply[Plus, Table[g[n, m], {m, 0, n}]], {n, 0, 10}]

nonn,uned,tabl

Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), May 30 2008

approved