A304314 -id:A304314 - cp's OEIS frontend

A304321 Table of coefficients in row functions F'(n,x)/F(n,x) such that [x^k] exp( k^n * x ) / F(n,x) = 0 for k>=1 and n>=1.

1, 1, 1, 1, 9, 1, 1, 49, 148, 1, 1, 225, 6877, 3493, 1, 1, 961, 229000, 1854545, 106431, 1, 1, 3969, 6737401, 612243125, 807478656, 3950832, 1, 1, 16129, 188580028, 172342090401, 3367384031526, 514798204147, 172325014, 1, 1, 65025, 5170118437, 45770504571813, 11657788116175751, 33056423981177346, 451182323794896, 8617033285, 1

Offset: 1

Paul D. Hanna, May 11 2018

Conjecture: T(n,k) in row n and column k gives the number of connected k-state finite automata with n inputs, for k>=0, for n>=1. For example, row 2 agrees with A006691, the number of connected n-state finite automata with 2 inputs; also, row 3 agrees with A006692, the number of connected n-state finite automata with 3 inputs.

			This table begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
1, 9, 148, 3493, 106431, 3950832, 172325014, 8617033285, 485267003023, ...;
1, 49, 6877, 1854545, 807478656, 514798204147, 451182323794896, ...;
1, 225, 229000, 612243125, 3367384031526, 33056423981177346, ...;
1, 961, 6737401, 172342090401, 11657788116175751, 1722786509653595220757, ...;
1, 3969, 188580028, 45770504571813, 37854124915368647781, ...;
1, 16129, 5170118437, 11889402239702065, 120067639589726126102806, ...;
1, 65025, 140510362000, 3061712634885743125, 377436820462509018320487276, ...;
1, 261121, 3804508566001, 785701359968473902401, 1182303741240112494973150131501, ...; ...
Let F'(n,x)/F(n,x) denote the o.g.f. of row n of this table, then the coefficient of x^k in exp(k^n*x)/F(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. A304320, A304312 (row 2), A304313 (row 3), A304314 (row 4), A304315 (row 5).

m = 10(*rows*);
row[nn_] := Module[{F, s}, F = 1 + Sum[c[k] x^k, {k, m}]; s[n_] := Solve[ SeriesCoefficient[Exp[n^nn*x]/F, {x, 0, n}] == 0][[1]]; Do[F = F /. s[n], {n, m}]; CoefficientList[D[F, x]/F + O[x]^m, x]];
T = Array[row, m];
Table[T[[n-k+1, k]], {n, 1, m}, {k, 1, n}] // Flatten (* Jean-François Alcover, Aug 27 2019 *)

{T(n,k) = my(A=[1],m); for(i=0, k, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^n +x^2*O(x^m)) / Ser(A) )[m] ); L = Vec(Ser(A)'/Ser(A)); L[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),", "));)

Row n of this table equals the logarithmic derivative of row n of table A304320.

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

A304312 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^2 * x ) / F(x) = 0 for n>0.

Original entry on oeis.org

1, 9, 148, 3493, 106431, 3950832, 172325014, 8617033285, 485267003023, 30363691715629, 2088698040637242, 156612539215405732, 12709745319947141220, 1109746209390479579732, 103724343230007402591558, 10332348604630683943445797, 1092720669631704348689818959, 122274820828415241343176467043, 14433472319311799728710020346232

Offset: 0

Paul D. Hanna, May 11 2018

nonn

Is this sequence essentially the same as A006691?
Conjecture: a(n) is the number of connected n-state finite automata with 2 inputs (A006691). [I believe the name of A006691 should be changed to read "(n+1)-state". See my comments in A006691. - Petros Hadjicostas, Feb 26 2021]
Equals row 2 of table A304321.

			O.g.f.: L(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 + ...
such that L(x) = F'(x)/F(x) where F(x) is the o.g.f. of A304322 :
F(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 + ... + A304322(n)*x^n + ...
which satisfies [x^n] exp( n^2 * x ) / F(x) = 0 for n>0.

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

Cf. A304322, A006691, A304321, A304313, A304314, A304315.

m = 25;
F = 1 + Sum[c[k] x^k, {k, m}];
s[n_] := Solve[SeriesCoefficient[Exp[n^2 * 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 20 2018 *)

{a(n) = my(A=[1],L); for(i=0, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^2 +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 A304322.
For n>=1, a(n) = B_{n+1}((n+1)^2-0!*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) * 2^(2*n + 3/2) * n^(n + 3/2) / (sqrt(Pi) * c^(n+1) * (2-c)^(n+1) * exp(n)), where c = -A226775 = -LambertW(-2*exp(-2)). - 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

A304315 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^5 * x ) / F(x) = 0 for n>0.

Original entry on oeis.org

1, 961, 6737401, 172342090401, 11657788116175751, 1722786509653595220757, 489506033977061086758261063, 243968979437942649897623460813009, 199025593654123221838381793032781035510, 251774439716905627952289102887999425054599511, 472942802381336010263584088374665504251010554412128, 1273071332950625956697135571575613091625334028239417955701

Offset: 0

Paul D. Hanna, May 11 2018

nonn

Conjecture: a(n) is the number of connected n-state finite automata with 5 inputs.

Equals row 5 of table A304321.

			O.g.f.: L(x) = 1 + 961*x + 6737401*x^2 + 172342090401*x^3 + 11657788116175751*x^4 + 1722786509653595220757*x^5 + 489506033977061086758261063*x^6 + ...
such that L(x) = F'(x)/F(x) where F(x) is the o.g.f. of A304325 :
F(x) = 1 + x + 481*x^2 + 2246281*x^3 + 43087884081*x^4 + 2331601789103231*x^5 + 287133439746933073357*x^6 + 69929721774643572422651223*x^7 + ... + A304325(n)*x^n + ...
which satisfies [x^n] exp( n^5 * x ) / F(x) = 0 for n>0.

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

Cf. A304325, A304321, A304312, A304313, A304314.

m = 25;
F = 1 + Sum[c[k] x^k, {k, m}];
s[n_] := Solve[SeriesCoefficient[Exp[n^5*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)^4 +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 A304325.
For n>=1, a(n) = B_{n+1}((n+1)^5-0!*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) * 5^(5*(n+1)) * n^(4*n + 9/2) / (sqrt(2*Pi) * c^(n+1) * (5-c)^(4*(n+1)) * exp(4*n)), where c = -LambertW(-5*exp(-5)). - Vaclav Kotesovec, Aug 31 2020

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A304321 Table of coefficients in row functions F'(n,x)/F(n,x) such that [x^k] exp( k^n * x ) / F(n,x) = 0 for k>=1 and n>=1.

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A304312 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^2 * x ) / F(x) = 0 for n>0.

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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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A304315 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^5 * x ) / F(x) = 0 for n>0.

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