A calculus based on Pascal-Sierpinski row sums

[Roger Bagula]

A long time ago, now,
I connected fractional-Abel derivatives to fractals
as distributions in IFS fractals.

In studying the Modulo two iso-Pascal triangles
that give Sierpinski gaskets,
I came across a sequence of row sums.
The sequence of Pascal types are:
1,2,1-> Pascal ,k=0,A007318
<http://www.research.att.com/%7Enjas/sequences/A007318>
1,4,1->Eulerian numbers ,k=1,A008292
<http://www.research.att.com/%7Enjas/sequences/A008292>
1,6,1-> MacMahon numbers, k=2,A060187
<http://www.research.att.com/%7Enjas/sequences/A060187>
1,8,1-> new,k=3,A142458
<http://www.research.att.com/%7Enjas/sequences/A142458>
1,10,1,A142459 <http://www.research.att.com/%7Enjas/sequences/A142459>
1,12,1,A142460 <http://www.research.att.com/%7Enjas/sequences/A142460>
1,14,1,A142461 <http://www.research.att.com/%7Enjas/sequences/A142461>
etc.
There is a recursion that I found that works to give these row sums:
for the MacMahon at k=2:A000165
<http://www.research.att.com/%7Enjas/sequences/A000165>
k = 2; b[1] = 2; b[n_] := b[n] = b[n - 1] + k;
a[0] = 1; a[1] = 2; a[n_] := a[n] = a[n - 1]*b[n];
Table[a[n], {n, 0, 20}]

The generalized factorial functions that come from the row sum sequence:
factorial[n,m]={2^n,n!,2^n*n!, -Product[3*k-1,{k,0,n}],(2*n)!/n!,...}
These are the factorial[n,m] row sums.
The Taylor expansion of a function
is traditionally:derivatives at zero as f^(n)[0]=(D[f[x],{x,n}]/.x->0)
f[x]=Sum[(D[f[x],{x,n}]/.x->0)*x^n/n!,{n,0,Infinity}]
New notation:
(D[f[x],{x,n}]/.x->0)->f[0,n]
It is possible to add a variable:
f[x]=Sum[f[0,n,m]*x^n/factorial[n,m],{n,0,Infinity}]
Such that:
f[0,n,m]/factorial[n,m]=f[0,n]/n!
or
f[0,n,m]=factorial[n,m]*f[0,n]/n!
Which has a structure somewhat like Abel derivaive with it in
terms of Gamma[] instead of .
That result gives a sequence of constants: 2,E, Sqrt[E],...
Sum[1/factorial[n,m], {n, 0, Infinity}]
Mathematica:
a[n_]={2^n,n!,2^n*n!, -Product[3*k-1,{k,0,n}],(2*n)!/n!}
Table[Sum[1/a[n][[m]],{n,0,Infinity}],{m,1,Length[a[n]]}]
N[%]
{2.`, 2.718281828459045`, 1.6487212707001282`, 1.6137225566295876,
1.5922965364693267`}

This approach to a scaled Sierpinski like calculus
that involves a combinatorial Pascal symmetry
appears to be new.
As the n! can be expessed as a real valiable with an integral
involved as a Gamma function:
(z-1)!=Gamma[z]=Integrate[Exp[-t]*t^(z-1),{t,0,Infinity}]
It seems very likely some general integral form
for the generalized factorial function presented here must exist.
The forms may already exist and have never been connected
as a sequence.

[Roger Bagula]

Except for 2^n and ones in front
this is the one:
{{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600},
{2, 8, 48, 384, 3840, 46080, 645120, 10321920, 185794560, 3715891200,
81749606400},
{2, 10, 80, 880, 12320,209440, 4188800, 96342400, 2504902400,
72642169600, 2324549427200},
{2, 12,120, 1680, 30240, 665280, 17297280, 518918400, 17643225600,
670442572800, 28158588057600},
{2, 14, 168, 2856, 62832, 1696464, 54286848, 2008613376, 84361761792,
3965002804224, 206180145819648},
{2, 16, 224, 4480, 116480, 3727360, 141639680, 6232145920, 311607296000,
17450008576000, 1081900531712000},
{2, 18, 288, 6624, 198720, 7352640, 323516160, 16499324160,
956960801280, 62202452083200, 4478576549990400},
{2, 20, 360, 9360, 318240, 13366080, 668304000, 38761632000,
2558267712000, 189311810688000, 15523568476416000},
{2, 22, 440, 12760, 484880, 22789360, 1276204160, 82953270400,
6138542009600, 509498986796800, 46873906785305600},
{2, 24, 528, 16896, 709632, 36900864, 2287853568, 164725456896,
13507487465472, 1242688846823424, 126754262375989248}}

Modifying the function to:
t[n_, m_] = If[m == 0, 2^n, Product[m*k + 2, {k, 0, n}]]
covers the Sierpinski-Pascal row sums.
t[n_, m_] = If[m == 0, 2^n, m^(1+n)*Pochhammer[2/m,n+1]]


%I A153190
%S A153190 1,2,6,2,8,48,2,10,80,880,2,12,120,1680,30240,2,14,168,2856,62832,
%T A153190
1696464,2,16,224,4480,116480,3727360,141639680,2,18,288,6624,198720,
%U A153190
7352640,323516160,16499324160,2,20,360,9360,318240,13366080,668304000
%N A153190 Triangle sequence: t(n,m)=If[m == 0, 1, Product[m*k + 2, {k,
0, n}]]. %C A153190 Row sums are:
%C A153190 {1, 8, 58, 972, 32054, 1762336, 145488242, 16830398612,
2597711342062, %C A153190 515721781568424, 128010625761216554,...}.
%C A153190 The one for Pascal, Eulerian and MacMahon row sums. %F
A153190 t(n,m)=If[m == 0, 1, Product[m*k + 2, {k, 0, n}]].
t(n,m)=m^(1+n)*Pochhammer[2/m,n+1]. %e A153190 {1}, %e A153190 {2, 6},
%e A153190 {2, 8, 48}, %e A153190 {2, 10, 80, 880}, %e A153190 {2, 12,
120, 1680, 30240},
%e A153190 {2, 14, 168, 2856, 62832, 1696464}, %e A153190 {2, 16, 224,
4480, 116480, 3727360, 141639680}, %e A153190 {2, 18, 288, 6624, 198720,
7352640, 323516160, 16499324160}, %e A153190 {2, 20, 360, 9360, 318240,
13366080, 668304000, 38761632000, 2558267712000}, %e A153190 {2, 22,
440, 12760, 484880, 22789360, 1276204160, 82953270400, 6138542009600,
509498986796800}, %e A153190 {2, 24, 528, 16896, 709632, 36900864,
2287853568, 164725456896, 13507487465472, 1242688846823424,
126754262375989248} %t A153190 t[n_, m_] =If[m == 0, 1, Product[m*k + 2,
{k, 0, n}]];
%t A153190 Table[Table[t[n, m], {n, 0, m}], {m, 1, 10}];
%t A153190 Flatten[%] %Y A153190 A000142, A008544, A001813,A000165 %K
A153190 nonn
%O A153190 0,2
%A A153190 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Dec 20 2008

[Rev. 11D Meow!]

I love it when he talks geek in context.

Post by Roger Bagula
Except for 2^n and ones in front
{{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600},
{2, 8, 48, 384, 3840, 46080, 645120, 10321920, 185794560, 3715891200,
81749606400},
{2, 10, 80, 880, 12320,209440, 4188800, 96342400, 2504902400, 72642169600,
2324549427200},
{2, 12,120, 1680, 30240, 665280, 17297280, 518918400, 17643225600,
670442572800, 28158588057600},
{2, 14, 168, 2856, 62832, 1696464, 54286848, 2008613376, 84361761792,
3965002804224, 206180145819648},
{2, 16, 224, 4480, 116480, 3727360, 141639680, 6232145920, 311607296000,
17450008576000, 1081900531712000},
{2, 18, 288, 6624, 198720, 7352640, 323516160, 16499324160, 956960801280,
62202452083200, 4478576549990400},
{2, 20, 360, 9360, 318240, 13366080, 668304000, 38761632000,
2558267712000, 189311810688000, 15523568476416000},
{2, 22, 440, 12760, 484880, 22789360, 1276204160, 82953270400,
6138542009600, 509498986796800, 46873906785305600},
{2, 24, 528, 16896, 709632, 36900864, 2287853568, 164725456896,
13507487465472, 1242688846823424, 126754262375989248}}
t[n_, m_] = If[m == 0, 2^n, Product[m*k + 2, {k, 0, n}]]
covers the Sierpinski-Pascal row sums.
t[n_, m_] = If[m == 0, 2^n, m^(1+n)*Pochhammer[2/m,n+1]]
%I A153190
%S A153190
1,2,6,2,8,48,2,10,80,880,2,12,120,1680,30240,2,14,168,2856,62832,
%T A153190
1696464,2,16,224,4480,116480,3727360,141639680,2,18,288,6624,198720,
%U A153190
7352640,323516160,16499324160,2,20,360,9360,318240,13366080,668304000
%N A153190 Triangle sequence: t(n,m)=If[m == 0, 1, Product[m*k + 2, {k, 0,
%C A153190 {1, 8, 58, 972, 32054, 1762336, 145488242, 16830398612,
2597711342062, %C A153190 515721781568424, 128010625761216554,...}.
%C A153190 The one for Pascal, Eulerian and MacMahon row sums. %F A153190
t(n,m)=If[m == 0, 1, Product[m*k + 2, {k, 0, n}]].
t(n,m)=m^(1+n)*Pochhammer[2/m,n+1]. %e A153190 {1}, %e A153190 {2, 6}, %e
A153190 {2, 8, 48}, %e A153190 {2, 10, 80, 880}, %e A153190 {2, 12, 120,
1680, 30240},
%e A153190 {2, 14, 168, 2856, 62832, 1696464}, %e A153190 {2, 16, 224,
4480, 116480, 3727360, 141639680}, %e A153190 {2, 18, 288, 6624, 198720,
7352640, 323516160, 16499324160}, %e A153190 {2, 20, 360, 9360, 318240,
13366080, 668304000, 38761632000, 2558267712000}, %e A153190 {2, 22, 440,
12760, 484880, 22789360, 1276204160, 82953270400, 6138542009600,
509498986796800}, %e A153190 {2, 24, 528, 16896, 709632, 36900864,
2287853568, 164725456896, 13507487465472, 1242688846823424,
126754262375989248} %t A153190 t[n_, m_] =If[m == 0, 1, Product[m*k + 2,
{k, 0, n}]];
%t A153190 Table[Table[t[n, m], {n, 0, m}], {m, 1, 10}];
%t A153190 Flatten[%] %Y A153190 A000142, A008544, A001813,A000165 %K
A153190 nonn
%O A153190 0,2
%A A153190 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Dec 20 2008
--
Respectfully, Roger L. Bagula
11759Waterhill Road, Lakeside,Ca 92040-2905,tel: [phone]
:http://www.geocities.com/rlbagulatftn/Index.html

[Roger Bagula]

I found Narayana numbers on my own doing factoring experiments
of a factorial type.

I tried to factor the Eulerian numbers as factorial- like and failed
because they have multiple values.
This new type of recursion is the result of some thought about using
the first solution in each row as the reference factor.
Since my by hand calculations were for the Eulerian numbers,
the MacMahon numbers
were a little more difficult as they came from polynomials to start.
The result is a new method of factoring the higher Sierpinski -Pascal
type triangle sequence
as if they were factorial -like.

This method should work for other symmetrical triangular
sequences as well.
The results are closer to "combinations"
than they were before.
You can also define permutations;
like:
f[n,m]/f[m,1]
with Floor[n/2] distinct values
per n.




%I A155556
%S A155556
1,1,4,44,1144,1056,65208,53152,7824960,5450016,4677376,1932765120,
%T A155556 1119751776,786197984,970248090240,457228062720,253156757568,
%U A155556 204411475840,982861315413120,369853933363200,156721804462080
%N A155556 Multi-bifurcating recursion of a factorial type based on the
Eulerian numbers A008292 as a triangle sequence: t(n,k)=Sum[(-1)^j
Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; f(n, m) = If[m ≤
Floor[n/2], f(m, 1)*f(n - m, 1)*t(n + 1, m)];
%C A155556 Row sums are:
%C A155556 {1, 1, 4, 44, 2200, 118360, 17952352, 3838714880, 1885044386368,
%C A155556 1607186778033408, 2934910973174349312,...}.
%C A155556 The Eulerian numbers factored as factorial like to middle
Floor[n/2]:
%C A155556 t(n,m)=f(n,m)/(f[m,1]*f[n-m,1]).
%C A155556 The idea is to factor the Eulerian numbers as
%C A155556 if the coefficients were made up of equivalents to factorials.
%C A155556 The result is a multi-bifurcating recursion thast fits the
Eulerian numbers.
%e A155556 Half Eulerian numbers: Table[Table[f[n, m]/(f[m, 1]*f[n - m,
1]), {m, 0, Floor[n/2]}], {n, 0, 10}];
%e A155556 {1},
%e A155556 {1},
%e A155556 {1, 4},
%e A155556 {1, 11},
%e A155556 {1, 26, 66},
%e A155556 {1, 57, 302},
%e A155556 {1, 120, 1191, 2416},
%e A155556 {1, 247, 4293, 15619},
%e A155556 {1, 502, 14608, 88234, 156190},
%e A155556 {1, 1013, 47840, 455192, 1310354},
%e A155556 {1, 2036, 152637, 2203488, 9738114, 15724248}...
%e A155556 Factorial type triangle is:
%e A155556 {1},
%e A155556 {1},
%e A155556 {4},
%e A155556 {44},
%e A155556 {1144, 1056},
%e A155556 {65208, 53152},
%e A155556 {7824960, 5450016, 4677376},
%e A155556 {1932765120, 1119751776, 786197984},
%e A155556 {970248090240, 457228062720, 253156757568, 204411475840},
%e A155556 {982861315413120, 369853933363200, 156721804462080,
97749724795008},
%e A155556 {2001105638181112320, 592383030999851520, 187388288944496640,
87173203289103360, 66860811759785472}
%t A155556 Clear[t, n, m, f, x];
%t A155556 t[n_, k_] = Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j,
0, k + 1}];
%t A155556 f[0, 1] = 1; f[1, 1] = 1; f[2, 1] = 4;
%t A155556 f[n_, m_] := f[n, m] = If[m ≤ Floor[n/2], f[m, 1]*f[n - m,
1]*t[n + 1, m]];
%t A155556 a = Join[{{1}}, {{1}}, Table[Table[f[n, m], {m, 1,
Floor[n/2]}], {n, 2, 10}]];
%t A155556 Flatten[%]
%Y A155556 A008292
%K A155556 nonn,tabl
%O A155556 0,3
%A A155556 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 24 2009

%I A155558
%S A155558
1,1,6,138,10488,8280,2485656,1392696,1794643632,663449904,448448112,
%T A155558 3910528474128,904634615952,375908525712,25621782562486656,
%U A155558
3570752184568704,852504801992064,514242863174016,504057328351799983488
%N A155558 Multi-bifurcating recursion of a factorial type based on the
MacMahon numbers A060187 as a triangle sequence: t(n,k)=A060187[n,m]
from polynomials; f(n, m) = If[m ≤ Floor[n/2], f(m, 1)*f(n - m, 1)*t(n +
1, m)];
%C A155558 Row sums are:
%C A155558 {1, 1, 6, 138, 18768, 3878352, 2906541648, 5191071615792,
30559282412221440,
%C A155558 552452422047369288192, 31282693282240728431406336,...}.
%C A155558 The Eulerian numbers factored as factorial like to middle
Floor[n/2]:
%C A155558 t(n,m)=f(n,m)/(f[m,1]*f[n-m,1]).
%C A155558 The idea is to factor the Eulerian numbers as
%C A155558 if the coefficients were made up of equivalents to factorials.
%C A155558 The result is a multi-bifurcating recursion thast fits the
Eulerian numbers.
%F A155558 t(n,k)=A060187[n,m] from polynomials; f(n, m) = If[m ≤
Floor[n/2], f(m, 1)*f(n - m, 1)*t(n + 1, m)];
%e A155558 Half MacMahon numbers: Table[Table[f[n, m]/(f[m, 1]*f[n - m,
1]), {m, 0, Floor[n/2]}], {n, 0, 10}];
%e A155558 {1},
%e A155558 {1},
%e A155558 {1, 6},
%e A155558 {1, 23},
%e A155558 {1, 76, 230},
%e A155558 {1, 237, 1682},
%e A155558 {1, 722, 10543, 23548},
%e A155558 {1, 2179, 60657, 259723},
%e A155558 {1, 6552, 331612, 2485288, 4675014},
%e A155558 {1, 19673, 1756340, 21707972, 69413294},
%e A155558 {1, 59038, 9116141, 178300904, 906923282, 1527092468}
Factorial type triangle is:
%e A155558 {1},
%e A155558 {1},
%e A155558 {6},
%e A155558 {138},
%e A155558 {10488, 8280},
%e A155558 {2485656, 1392696},
%e A155558 {1794643632, 663449904, 448448112},
%e A155558 {3910528474128, 904634615952, 375908525712},
%e A155558 {25621782562486656, 3570752184568704, 852504801992064,
514242863174016},
%e A155558 {504057328351799983488, 41209305481499829120,
5376214172453933952, 1809574041615541632},
%e A155558 {29758536551233567425164544, 1401430695065818000286976,
96220605163557295616256, 17070311724802276269312, 9435119052983434069248}
%t A155558 Clear[t, n, m, f, x, p];
%t A155558 p[x_, n_] = (-1)^(n + 1)*(x - 1)^(n + 1)*Sum[(2* m +
1)^n*x^m, {m, 0, Infinity}];
%t A155558 t[n_, m_] :=
Table[CoefficientList[FullSimplify[ExpandAll[p[x, k]]], x], { k, 0,
10}][[n + 1, m + 1]];
%t A155558 f[0, 1] = 1; f[1, 1] = 1; f[2, 1] = 6;
%t A155558 f[n_, m_] := f[n, m] = If[m ≤ Floor[n/2], f[m, 1]*f[n - m,
1]*t[n, m]];
%t A155558 a = Join[{{1}}, {{1}}, Table[Table[f[n, m], {m, 1,
Floor[n/2]}], {n, 2, 10}]];
%t A155558 Flatten[%]
%Y A155558 A008292
%K A155558 nonn,tabl
%O A155558 0,3
%A A155558 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 24 2009



Respectfully, Roger L. Bagula
11759Waterhill Road, Lakeside,Ca 92040-2905,tel: [phone]
:http://www.geocities.com/rlbagulatftn/Index.html
alternative email: ***@sbcglobal.net

[Roger Bagula]

In my projection of Lauweier's IFS von Koch on polynomials
I found several indications that the polynomial set was Stirling like.
Here I have derived a new sort of combination that shows up like one of
my first efforts to
make a von Koch IFS: as a dust set.

%I A155742
%S A155742
1,0,0,0,1,0,0,6,6,0,0,36,121,36,0,0,240,1750,1750,240,0,0,1800,23290,
%T A155742
50625,23290,1800,0,0,15120,308700,1193640,1193640,308700,15120,0,0,
%U A155742
141120,4207896,25738720,45819361,25738720,4207896,141120,0,0,1451520
%N A155742 A symmetrical triangle sequence: t(n,m)=((-1)^(n -
m)*StirlingS1[n, m]*(-1)^(-m)*StirlingS1[n, n - m]).
%C A155742 Row sums are:
%C A155742 {1, 0, 1, 12, 193, 3980, 100805, 3034920, 105994833,
4215106728, 188097696345,...}.
%C A155742 Plotting this function gives what appears to be a von Koch
dust set:
%C A155742 a = Table[Table[t[n, m], {m, 0, n}], {n, 0, 243}];
%C A155742 b = Table[If[m ≤ n, 3 - Mod[a[[n]][[m]], 3], 0], {m, 1,
Length[a]}, {n, 1, Length[a]}];
%C A155742 ListDensityPlot[b, Mesh -> False, Frame -> False, AspectRatio
-> Automatic, ColorFunction -> (Hue[2#] &)]
%C A155742 Derivation:
%C A155742 Let:(as permutations)
%C A155742 f(n)/f(m)=(-1)^(n-m)*StrilingS1[n,m];
%C A155742 and assume:
%C A155742 f(m)=n!/((-1)^(n-m)*StrilingS1[n,m]);
%C A155742 Then:f(n-m)=n!/((-1)^(m)*StrilingS1[n,n-m]); Combinations are:
%C A155742 c(n,m)=StirlingS1[n,m]/(n!/((-1)^(m)*StrilingS1[n,n-m]));
%C A155742 Leaving out the n! you get this set.
%F A155742 t(n,m)=((-1)^(n - m)*StirlingS1[n, m]*(-1)^(-m)*StirlingS1[n,
n - m]).
%e A155742 {1},
%e A155742 {0, 0},
%e A155742 {0, 1, 0},
%e A155742 {0, 6, 6, 0},
%e A155742 {0, 36, 121, 36, 0},
%e A155742 {0, 240, 1750, 1750, 240, 0},
%e A155742 {0, 1800, 23290, 50625, 23290, 1800, 0},
%e A155742 {0, 15120, 308700, 1193640, 1193640, 308700, 15120, 0},
%e A155742 {0, 141120, 4207896, 25738720, 45819361, 25738720, 4207896,
141120, 0},
%e A155742 {0, 1451520, 59832864, 535810464, 1510458516, 1510458516,
535810464, 59832864, 1451520, 0},
%e A155742 {0, 16329600, 893121120, 11082015000, 45789404640,
72535955625, 45789404640, 11082015000, 893121120, 16329600, 0}
%t A155742 Clear[t, n, m, a];
%t A155742 t[n_, m_] = ((-1)^(n - m)*StirlingS1[n,
m]*(-1)^(-m)*StirlingS1[n, n - m]);
%t A155742 Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}];
%t A155742 Flatten[%]
%K A155742 nonn,tabl
%O A155742 0,8
%A A155742 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 26 2009