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.
Original entry on oeis.org
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
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.
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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 *)
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{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),", "));)
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
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.
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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 *)
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{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),", "))
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
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 + ...
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{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),", "))
Showing 1-3 of 3 results.
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