A304323 -id:A304323 - cp's OEIS frontend

A304322 O.g.f. A(x) satisfies: [x^n] exp( n^2 * x ) / A(x) = 0 for n>0.

1, 1, 5, 54, 935, 22417, 685592, 25431764, 1106630687, 55174867339, 3097872254493, 193283918695494, 13260815963831108, 991928912663646012, 80325879518096889760, 7000127337189146831092, 653156403671376068448047, 64963788042207845593775999, 6861040250464949653809027311, 766815367797924824316405828466, 90417908118862070187113849296815

Offset: 0

Paul D. Hanna, May 11 2018

nonn

It is conjectured that the coefficients of o.g.f. A(x) consist entirely of integers.
Equals row 2 of table A304320.
O.g.f. A(x) = 1/(1 - x*B(x)), where B(x) is the o.g.f. of A107668.
Logarithmic derivative of o.g.f. A(x), A'(x)/A(x), equals o.g.f. of A304312.
Conjecture: given o.g.f. A(x), the coefficient of x^n in A'(x)/A(x) is the number of connected n-state finite automata with 2 inputs (A006691).

			O.g.f.: A(x) = 1 + x + 5*x^2 + 54*x^3 + 935*x^4 + 22417*x^5 + 685592*x^6 + 25431764*x^7 + 1106630687*x^8 + 55174867339*x^9 + 3097872254493*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(n^2*x) / A(x) begins:
n=0: [1, -1, -8, -270, -19584, -2427000, -455544000, -120136161600, ...];
n=1: [1, 0, -9, -296, -20715, -2527704, -470405285, -123376631664, ...];
n=2: [1, 3, 0, -350, -24672, -2867256, -518870528, -133753337280, ...];
n=3: [1, 8, 55, 0, -29547, -3559056, -614943333, -153534305160, ...];
n=4: [1, 15, 216, 2674, 0, -4291704, -783235520, -187656684864, ...];
n=5: [1, 24, 567, 12880, 251541, 0, -948897125, -243358236600, ...];
n=6: [1, 35, 1216, 41634, 1372320, 38884296, 0, -295870371264, ...];
n=7: [1, 48, 2295, 109000, 5106453, 230531544, 8944955227, 0, ...];
n=8: [1, 63, 3960, 248050, 15443328, 949131144, 56257429312, 2865412167360, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n^2*x ) / A(x) = 0 for n>=0.
LOGARITHMIC DERIVATIVE.
The logarithmic derivative of A(x) yields the o.g.f. of A304312:
A'(x)/A(x) = 1 + 9*x + 148*x^2 + 3493*x^3 + 106431*x^4 + 3950832*x^5 + 172325014*x^6 + 8617033285*x^7 + 485267003023*x^8 + 30363691715629*x^9 + ... + A304312(n)*x^n +...
INVERT TRANSFORM.
1/A(x) = 1 - x*B(x), where B(x) is the o.g.f. of A107668:
B(x) = 1 + 4*x + 45*x^2 + 816*x^3 + 20225*x^4 + 632700*x^5 + 23836540*x^6 + 1048592640*x^7 + 52696514169*x^8 + ... + A107668(n)*x^n + ...

Paul D. Hanna, Table of n, a(n) for n = 0..400

Cf. A304320, A304312, A304321, A304323, A304324, A304325.

Cf. A006691, A107668.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^2 +x*O(x^m)) / Ser(A) )[m] ); A[n+1]}
for(n=0,25, print1( a(n),", "))

a(n) ~ sqrt(1-c) * 2^(2*n - 1/2) * n^(n - 1/2) / (sqrt(Pi) * c^n * (2-c)^n * exp(n)), where c = -A226775 = -LambertW(-2*exp(-2)). - Vaclav Kotesovec, Aug 31 2020

A304320 Table of coefficients in row functions R(n,x) such that [x^k] exp( k^n * x ) / R(n,x) = 0 for k>=1 and n>=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 25, 54, 1, 1, 1, 113, 2317, 935, 1, 1, 1, 481, 76446, 466241, 22417, 1, 1, 1, 1985, 2246281, 153143499, 162016980, 685592, 1, 1, 1, 8065, 62861994, 43087884081, 673638499100, 85975473871, 25431764, 1, 1, 1, 32513, 1723380877, 11442690973075, 2331601789103231, 5510097691767062, 64545532370208, 1106630687, 1, 1, 1, 130561, 46836819846, 2972352315820441, 7570836550478960487, 287133439746933073357, 75312181798660695788, 65062315637060121, 55174867339, 1

Offset: 1

Paul D. Hanna, May 11 2018

It is striking that the coefficients in this table consist entirely of integers.

			This table begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
1, 1, 5, 54, 935, 22417, 685592, 25431764, 1106630687, 55174867339, ...;
1, 1, 25, 2317, 466241, 162016980, 85975473871, 64545532370208, ...;
1, 1, 113, 76446, 153143499, 673638499100, 5510097691767062, ...;
1, 1, 481, 2246281, 43087884081, 2331601789103231, 287133439746933073357, ...;
1, 1, 1985, 62861994, 11442690973075, 7570836550478960487, ...;
1, 1, 8065, 1723380877, 2972352315820441, 24013530904194819396970, ...;
1, 1, 32513, 46836819846, 765428206086770699, 75487364859452767380638650, ...;
1, 1, 130561, 1268169652561, 196425341268811084961, 236460748444613412476233431261, ...; ...
Let R(n,x) denote the o.g.f. of row n of this table, then the coefficient of x^k in exp(k^n*x)/R(n,x) = 0 for k>=1 and n>=1.

Paul D. Hanna, Table of n, a(n) for n = 1..1326 as a flattened table read by antidiagonals 1..51.

Cf. A304321, A304322 (row 2), A304323 (row 3), A304324 (row 4), A304325 (row 5), A337551 (diagonal).

{T(n,k) = my(A=[1],m); for(i=1, k, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^n +x*O(x^m)) / Ser(A) )[m] ); A[k+1]}
/* Print table: */
for(n=1,8, for(k=0,8, print1( T(n,k),", "));print(""))
/* Print as a flattened table: */
for(n=0,10, for(k=0,n, print1( T(n-k+1,k),", "));)

For fixed row r > 1 is a(n) ~ sqrt(1-c) * r^(r*n) * n^((r-1)*n - 1/2) / (sqrt(2*Pi) * c^n * (r-c)^((r-1)*n) * exp((r-1)*n)), where c = -LambertW(-r*exp(-r)). - Vaclav Kotesovec, Aug 31 2020

A304324 O.g.f. A(x) satisfies: [x^n] exp( n^4 * x ) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 1, 113, 76446, 153143499, 673638499100, 5510097691767062, 75312181798660695788, 1595682359653020033714019, 49564410138113345565513815041, 2161639124039437373346491749452440, 127889301139607880711208251726358504898, 9979766671875039854419652569806336108694074

Offset: 0

Paul D. Hanna, May 11 2018

nonn

It is conjectured that the coefficients of o.g.f. A(x) consist entirely of integers.
Equals row 4 of table A304320.
O.g.f. A(x) = 1/(1 - x*B(x)), where B(x) is the o.g.f. of A304394.
Logarithmic derivative of o.g.f. A(x), A'(x)/A(x), equals o.g.f. of A304314.
Conjecture: given o.g.f. A(x), the coefficient of x^n in A'(x)/A(x) is the number of connected n-state finite automata with 4 inputs.

			O.g.f.: A(x) = 1 + x + 113*x^2 + 76446*x^3 + 153143499*x^4 + 673638499100*x^5 + 5510097691767062*x^6 + 75312181798660695788*x^7 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(n^4*x) / A(x) begins:
n=0: [1, -1, -224, -457326, -3671476224, -80797824300000, ...];
n=1: [1, 0, -225, -458000, -3673306875, -80816186256624, ...];
n=2: [1, 15, 0, -464750, -3701040000, -81092721606624, ...];
n=3: [1, 80, 6175, 0, -3787546875, -82312696206624, ...];
n=4: [1, 255, 64800, 15951250, 0, -84756571206624, ...];
n=5: [1, 624, 389151, 242091424, 146271536901, 0, ...];
n=6: [1, 1295, 1676800, 2170415250, 2804103120000, 3524906587193376, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n^4*x ) / A(x) = 0 for n>=0.
LOGARITHMIC DERIVATIVE.
The logarithmic derivative of A(x) yields the o.g.f. of A304314:
A'(x)/A(x) = 1 + 225*x + 229000*x^2 + 612243125*x^3 + 3367384031526*x^4 + 33056423981177346*x^5 + 527146092112494861420*x^6 + ... + A304314(n)*x^n + ...
INVERT TRANSFORM.
1/A(x) = 1 - x*B(x), where B(x) is the o.g.f. of A304394:
B(x) = 1 + 112*x + 76221*x^2 + 152978176*x^3 + 673315202500*x^4 + 5508710472669120*x^5 + 75300988091046198131*x^6 + ... + A304394(n)*x^n + ...

Paul D. Hanna, Table of n, a(n) for n = 0..200

Cf. A304320, A304314, A304321, A304322, A304323, A304325.

Cf. A304394.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^4 +x*O(x^m)) / Ser(A) )[m] ); A[n+1]}
for(n=0,25, print1( a(n),", "))

a(n) ~ sqrt(1-c) * 4^(4*n) * n^(3*n - 1/2) / (sqrt(2*Pi) * c^n * (4-c)^(3*n) * exp(3*n)), where c = -LambertW(-4*exp(-4)). - Vaclav Kotesovec, Aug 31 2020

A304313 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^3 * x ) / F(x) = 0 for n>0.

Original entry on oeis.org

1, 49, 6877, 1854545, 807478656, 514798204147, 451182323794896, 519961864703259753, 762210147961330421167, 1384945048774500147047194, 3055115321627096660341307614, 8043516699726480852467167758419, 24915939138210507189761922944830006, 89709850983809128394441772076036629240, 371523831948166269091257380175120352465872

Offset: 0

Paul D. Hanna, May 11 2018

nonn

Is this sequence essentially the same as A006692?
Conjecture: a(n) is the number of connected n-state finite automata with 3 inputs (A006692).
Equals row 3 of table A304321.

			O.g.f.: L(x) = 1 + 49*x + 6877*x^2 + 1854545*x^3 + 807478656*x^4 + 514798204147*x^5 + 451182323794896*x^6 + 519961864703259753*x^7 + ...
such that L(x) = F'(x)/F(x) where F(x) is the o.g.f. of A304323 :
F(x) = 1 + x + 25*x^2 + 2317*x^3 + 466241*x^4 + 162016980*x^5 + 85975473871*x^6 + 64545532370208*x^7 + 65062315637060121*x^8 + ... + A304323(n)*x^n + ...
which satisfies [x^n] exp( n^3 * x ) / F(x) = 0 for n>0.

Paul D. Hanna, Table of n, a(n) for n = 0..300

Cf. A304323, A006692, A304321, A304312, A304314, A304315.

m = 25;
F = 1 + Sum[c[k] x^k, {k, m}];
s[n_] := Solve[SeriesCoefficient[Exp[n^3*x]/F, {x, 0, n}] == 0][[1]];
Do[F = F /. s[n], {n, m}];
CoefficientList[D[F, x]/F + O[x]^m, x] (* Jean-François Alcover, May 21 2018 *)

{a(n) = my(A=[1],L); for(i=0, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^3 +x^2*O(x^m)) / Ser(A) )[m] ); L = Vec(Ser(A)'/Ser(A)); L[n+1]}
for(n=0,25, print1( a(n),", "))

Logarithmic derivative of the o.g.f. of A304323.
For n>=1, a(n) = B_{n+1}((n+1)^3-0!*a(0),-1!*a(1),...,-(n-1)!*a(n-1),0) / n!, where B_{n+1}(...) is the (n+1)-st complete exponential Bell polynomial. - Max Alekseyev, Jun 18 2018
a(n) ~ sqrt(1-c) * 3^(3*(n+1)) * n^(2*n + 5/2) / (sqrt(2*Pi) * c^(n+1) * (3-c)^(2*(n+1)) * exp(2*n)), where c = -LambertW(-3*exp(-3)). - Vaclav Kotesovec, Aug 31 2020

A304325 O.g.f. A(x) satisfies: [x^n] exp( n^5 * x ) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 1, 481, 2246281, 43087884081, 2331601789103231, 287133439746933073357, 69929721774643572422651223, 30496192503451926066104677123329, 22113985380962062942048847693898939310, 25177466100486219354624677349405490885006591, 42994825404638061265611776726882581676486680632128

Offset: 0

Paul D. Hanna, May 11 2018

nonn

It is conjectured that the coefficients of o.g.f. A(x) consist entirely of integers.
Equals row 5 of table A304320.
O.g.f. A(x) = 1/(1 - x*B(x)), where B(x) is the o.g.f. of A304395.
Logarithmic derivative of o.g.f. A(x), A'(x)/A(x), equals o.g.f. of A304315.
Conjecture: given o.g.f. A(x), the coefficient of x^n in A'(x)/A(x) enumerates the connected n-state finite automata with 5 inputs.

			O.g.f.: A(x) = 1 + x + 481*x^2 + 2246281*x^3 + 43087884081*x^4 + 2331601789103231*x^5 + 287133439746933073357*x^6 + 69929721774643572422651223*x^7 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(n^5*x) / A(x) begins:
n=0: [1, -1, -960, -13471920, -1033995878400, -279781615181250000, ...];
n=1: [1, 0, -961, -13474802, -1034049771843, -279786785295370804, ...];
n=2: [1, 31, 0, -13534384, -1035725264896, -279947192760516048, ...];
n=3: [1, 242, 57603, 0, -1044001318107, -281045183102366562, ...];
n=4: [1, 1023, 1045568, 1054175056, 0, -284106842971323856, ...];
n=5: [1, 3124, 9758415, 30465809330, 93986716449725, 0, ...];
n=6: [1, 7775, 60449664, 469967719248, 3652476388472832, 28079364132086235696, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n^5*x ) / A(x) = 0 for n >= 0.
LOGARITHMIC DERIVATIVE.
The logarithmic derivative of A(x) yields the o.g.f. of A304315:
A'(x)/A(x) = 1 + 961*x + 6737401*x^2 + 172342090401*x^3 + 11657788116175751*x^4 + 1722786509653595220757*x^5 + 489506033977061086758261063*x^6 + ... + A304315(n)*x^n +...
INVERT TRANSFORM.
1/A(x) = 1 - x*B(x), where B(x) is the o.g.f. of A304395:
B(x) = 1 + 480*x + 2245320*x^2 + 43083161600*x^3 + 2331513459843750*x^4 + 287128730182879382976*x^5 + ... + A304395(n)*x^n + ...

Paul D. Hanna, Table of n, a(n) for n = 0..200

Cf. A304320, A304315, A304321, A304322, A304323, A304324.

Cf. A304395.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^5 +x*O(x^m)) / Ser(A) )[m] ); A[n+1]}
for(n=0,25, print1( a(n),", "))

a(n) ~ sqrt(1-c) * 5^(5*n) * n^(4*n - 1/2) / (sqrt(2*Pi) * c^n * (5-c)^(4*n) * exp(4*n)), where c = -LambertW(-5*exp(-5)). - Vaclav Kotesovec, Aug 31 2020

A107675 Column 0 of triangle A107674.

Original entry on oeis.org

1, 24, 2268, 461056, 160977375, 85624508376, 64363893844726, 64928246784463872, 84623205378726331245, 138408056280920732755000, 277597038523589348539241112, 670011760601512512626484887040

Offset: 0

Paul D. Hanna, Jun 07 2005

nonn

The o.g.f. A(x) = Sum_{m >= 0} a(m)*x^m is such that, for each integer n > 0, the coefficient of x^n in the expansion of exp(n^3*x)*(1 - x*A(x)) is equal to 0.

Given the o.g.f. A(x), the o.g.f. of A304323 equals 1/(1 - x*A(x)).

			O.g.f.: A(x) = 1 + 24*x + 2268*x^2 + 461056*x^3 + 160977375*x^4 + 85624508376*x^5 + 64363893844726*x^6 + 64928246784463872*x^7 + ...

Cf. A107671, A107674, A107676.

Cf. A304323, A107668, A304394, A304395, A342202.

{a(n)=local(P=matrix(n+1,n+1,r,c,if(r>=c,(r^3)^(r-c)/(r-c)!)), D=matrix(n+1,n+1,r,c,if(r==c,r)));(P^-1*D^2*P)[n+1,1]}
for(n=0,20, print1(a(n),", "))

/* From formula: [x^n] exp( n^3*x ) * (1 - x*A(x)) = 0 */
{a(n) = my(A=[1]); for(i=0, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*m^3 +x^2*O(x^m)) * (1 - x*Ser(A)) )[m+1] ); A[n+1]}
for(n=0, 25, print1( a(n), ", ")) \\ Paul D. Hanna, May 12 2018

/* From Recurrence: */
{a(n) = if(n==0,1, (n+1)^(3*n+3)/(n+1)! - sum(k=1,n, (n+1)^(3*k)/k! * a(n-k) ))}
for(n=0,25, print1( a(n),", ")) \\ Paul D. Hanna, May 12 2018

O.g.f. A(x) satisfies: [x^n] exp(n^3*x) * (1 - x*A(x)) = 0 for n > 0. - Paul D. Hanna, May 12 2018
a(n) = (n+1)^(3*n+3)/(n+1)! - Sum_{k=1..n} (n+1)^(3*k)/k! * a(n-k) for n > 0 with a(0) = 1. - Paul D. Hanna, May 12 2018
a(n) = A342202(3,n+1) = Sum_{r=1..(n+1)} (-1)^(r-1) * Sum_{s_1, ..., s_r} (1/(Product_{j=1..r} s_j!)) * Product_{j=1..r} (Sum_{i=1..j} s_i)^(3*s_j)), where the second sum is over lists (s_1, ..., s_r) of positive integers s_i such that Sum_{i=1..r} s_i = n+1. (Thus the second sum is over all compositions of n+1. See Michel Marcus's PARI program in A342202.) - Petros Hadjicostas, Mar 10 2021

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A304322 O.g.f. A(x) satisfies: [x^n] exp( n^2 * x ) / A(x) = 0 for n>0.

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A304320 Table of coefficients in row functions R(n,x) such that [x^k] exp( k^n * x ) / R(n,x) = 0 for k>=1 and n>=1.

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A304324 O.g.f. A(x) satisfies: [x^n] exp( n^4 * x ) / A(x) = 0 for n>0.

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A304313 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^3 * x ) / F(x) = 0 for n>0.

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A304325 O.g.f. A(x) satisfies: [x^n] exp( n^5 * x ) / A(x) = 0 for n>0.

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A107675 Column 0 of triangle A107674.

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