cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

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

Original entry on oeis.org

1, 1, 5, 54, 935, 22417, 685592, 25431764, 1106630687, 55174867339, 3097872254493, 193283918695494, 13260815963831108, 991928912663646012, 80325879518096889760, 7000127337189146831092, 653156403671376068448047, 64963788042207845593775999, 6861040250464949653809027311, 766815367797924824316405828466, 90417908118862070187113849296815
Offset: 0

Views

Author

Paul D. Hanna, May 11 2018

Keywords

Comments

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).

Examples

			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 + ...

Links

Crossrefs

Cf. A304320, A304312, A304321, A304323, A304324, A304325.
Cf. A006691, A107668.

Programs

  • PARI
    {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),", "))

Formula

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

Views

Author

Paul D. Hanna, May 11 2018

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

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

Programs

  • PARI
    {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),", "));)

Formula

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

A304323 O.g.f. A(x) satisfies: [x^n] exp( n^3 * x ) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 1, 25, 2317, 466241, 162016980, 85975473871, 64545532370208, 65062315637060121, 84756897268784533255, 138581022247955235150982, 277878562828788369685779910, 670574499099019193091230751539, 1917288315895234006935990419270242, 6409780596355519454337664637246378856, 24774712941456386970945752104780461007848, 109632095120643795798521114315908854415860345
Offset: 0

Views

Author

Paul D. Hanna, May 11 2018

Keywords

Comments

It is conjectured that the coefficients of o.g.f. A(x) consist entirely of integers.
Equals row 3 of table A304320.
O.g.f. A(x) = 1/(1 - x*B(x)), where B(x) is the o.g.f. of A107675.
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 3 inputs (A006692).

Examples

			O.g.f.: A(x) = 1 + x + 25*x^2 + 2317*x^3 + 466241*x^4 + 162016980*x^5 + 85975473871*x^6 + 64545532370208*x^7 + 65062315637060121*x^8 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(n^3*x) / A(x) begins:
n=0: [1, -1, -48, -13608, -11065344, -19317285000, -61649646030720, ...];
n=1: [1, 0, -49, -13754, -11120067, -19372748284, -61765715993765, ...];
n=2: [1, 7, 0, -14440, -11517184, -19768841352, -62587640670464, ...];
n=3: [1, 26, 627, 0, -12292251, -20908064898, -64905483973113, ...];
n=4: [1, 63, 3920, 227032, 0, -22551552136, -69768485886848, ...];
n=5: [1, 124, 15327, 1874642, 213958781, 0, -75806801733845, ...];
n=6: [1, 215, 46176, 9893016, 2100211968, 416846973816, 0, ...];
n=7: [1, 342, 116915, 39937660, 13616254341, 4604681316698, 1458047845980391, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n^3*x ) / A(x) = 0 for n>=0.
LOGARITHMIC DERIVATIVE.
The logarithmic derivative of A(x) yields the o.g.f. of A304313:
A'(x)/A(x) = 1 + 49*x + 6877*x^2 + 1854545*x^3 + 807478656*x^4 + 514798204147*x^5 + 451182323794896*x^6 + 519961864703259753*x^7 + ... + A304313(n)*x^n +...
INVERT TRANSFORM.
1/A(x) = 1 - x*B(x), where B(x) is the o.g.f. of A107675:
B(x) = 1 + 24*x + 2268*x^2 + 461056*x^3 + 160977375*x^4 + 85624508376*x^5 + 64363893844726*x^6 + ... + A107675(n)*x^n + ...

Links

Crossrefs

Cf. A304320, A304313, A304321, A304322, A304324, A304325.
Cf. A006691, A107675.

Programs

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

Formula

a(n) ~ sqrt(1-c) * 3^(3*n) * n^(2*n - 1/2) / (sqrt(2*Pi) * c^n * (3-c)^(2*n) * exp(2*n)), where c = -A226750 = -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

Views

Author

Paul D. Hanna, May 11 2018

Keywords

Comments

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.

Examples

			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 + ...

Links

Crossrefs

Cf. A304320, A304315, A304321, A304322, A304323, A304324.
Cf. A304395.

Programs

  • PARI
    {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),", "))

Formula

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

A304314 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^4 * x ) / F(x) = 0 for n>0.

Original entry on oeis.org

1, 225, 229000, 612243125, 3367384031526, 33056423981177346, 527146092112494861420, 12764850938355048224394925, 446065249480005516657138106375, 21615893741029073481369412949207860, 1406758471936562034421316174257309550136, 119755662436589797897149020637183902177930534
Offset: 0

Views

Author

Paul D. Hanna, May 11 2018

Keywords

Comments

Conjecture: a(n) is the number of connected n-state finite automata with 4 inputs.
Equals row 4 of table A304321.

Examples

			O.g.f.: L(x) = 1 + 225*x + 229000*x^2 + 612243125*x^3 + 3367384031526*x^4 + 33056423981177346*x^5 + 527146092112494861420*x^6 + ...
such that L(x) = F'(x)/F(x) where F(x) is the o.g.f. of A304324 :
F(x) = 1 + x + 113*x^2 + 76446*x^3 + 153143499*x^4 + 673638499100*x^5 + 5510097691767062*x^6 + 75312181798660695788*x^7 + ... + A304324(n)*x^n + ...
which satisfies [x^n] exp( n^4 * x ) / F(x) = 0 for n>0.

Links

Crossrefs

Cf. A304324, A304321, A304312, A304313, A304315.

Programs

  • Mathematica
    m = 25;
F = 1 + Sum[c[k] x^k, {k, m}];
s[n_] := Solve[SeriesCoefficient[Exp[n^4*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 *)
  • PARI
    {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),", "))

Formula

Logarithmic derivative of the o.g.f. of A304324.
For n>=1, a(n) = B_{n+1}((n+1)^4-0!*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) * 4^(4*(n+1)) * n^(3*n + 7/2) / (sqrt(2*Pi) * c^(n+1) * (4-c)^(3*(n+1)) * exp(3*n)), where c = -LambertW(-4*exp(-4)). - Vaclav Kotesovec, Aug 31 2020

A304394 O.g.f. A(x) satisfies: [x^n] exp(n^4 * x) * (1 - x*A(x)) = 0 for n > 0.

Original entry on oeis.org

1, 112, 76221, 152978176, 673315202500, 5508710472669120, 75300988091046198131, 1595530380622638283804672, 49561200934127182294698009969, 2161539625780059763174286300310000, 127884966535158110582342524738392563401, 9979510403062963314615799917574094659938048, 1003426348756281631241586585232930123009989117616
Offset: 0

Views

Author

Paul D. Hanna, May 12 2018

Keywords

Comments

INVERT transform of A304324.
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^4 * x) * (1 - x*A(x)) = 0 is equal to 0.

Examples

			O.g.f.: A(x) = 1 + 112*x + 76221*x^2 + 152978176*x^3 + 673315202500*x^4 + 5508710472669120*x^5 + 75300988091046198131*x^6 + ...

Crossrefs

Cf. A304324, A304396, A107668, A107675, A304395, A342202.

Programs

  • PARI
    /* From formula: [x^n] exp( n^4*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^4 +x^2*O(x^m)) * (1 - x*Ser(A)) )[m+1] ); A[n+1]}
for(n=0, 25, print1( a(n), ", "))

Formula

a(n) = (n+1)^(4*n+4)/(n+1)! - Sum_{k=1..n} (n+1)^(4*k)/k! * a(n-k) for n > 0 with a(0) = 1.
a(n) = A342202(4,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)^(4*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

A304316 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n*(n-1) * x ) / F(x) = 0 for n>0.

Original entry on oeis.org

4, 72, 1736, 53040, 1961728, 85062432, 4225904800, 236455369344, 14705880874944, 1005982098054912, 75048224139686912, 6062679436944758784, 527187725605767366144, 49092882744958427976192, 4874131922792403196021248, 513942386047796079510884352, 57356122407632751143615036416, 6754087907265415509369502427136, 836924235604443592471459956156416
Offset: 1

Views

Author

Paul D. Hanna, May 11 2018

Keywords

Examples

			O.g.f.: L(x) = 4*x + 72*x^2 + 1736*x^3 + 53040*x^4 + 1961728*x^5 + 85062432*x^6 + 4225904800*x^7 + 236455369344*x^8 + 14705880874944*x^9 + 1005982098054912*x^10 + ...
such that L(x) = F'(x)/F(x) where F(x) is the o.g.f. of A304324 :
F(x) = 1 + 2*x^2 + 24*x^3 + 436*x^4 + 10656*x^5 + 328112*x^6 + 12183456*x^7 + 529242224*x^8 + 26309617536*x^9 + 1472135847072*x^10 + ... + A304318(n)*x^n + ...
which satisfies [x^n] exp( n*(n-1) * x ) / F(x) = 0 for n>0.

Links

Crossrefs

Cf. A304318, A304317, A304321.

Programs

  • Mathematica
    m = 25;
F = 1 + Sum[c[k] x^k, {k, m}];
s[n_] := Solve[SeriesCoefficient[Exp[n*(n - 1)*x]/F, {x, 0, n}] == 0][[1]];
Do[F = F /. s[n], {n, m}];
CoefficientList[D[F, x]/F + O[x]^m, x] // Rest (* Jean-François Alcover, May 21 2018 *)
  • PARI
    {a(n) = my(A=[1],L); for(i=0, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)*(m-2) +x^2*O(x^m)) / Ser(A) )[m] ); L = Vec(Ser(A)'/Ser(A)); L[n]}
for(n=1,25, print1( a(n),", "))

Formula

Logarithmic derivative of the o.g.f. of A304318.
a(n) ~ sqrt(1-c) * 2^(2*n + 1) * n^(n + 3/2) / (sqrt(Pi) * c^(n + 1/2) * (2-c)^(n+1) * exp(n)), where c = -A226775 = -LambertW(-2*exp(-2)). - Vaclav Kotesovec, Aug 31 2020
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.