A226775 -id:A226775 - cp's OEIS frontend

A107668 Column 0 of triangle A107667.

1, 4, 45, 816, 20225, 632700, 23836540, 1048592640, 52696514169, 2976295383100, 186548057815801, 12845016620629488, 963644465255618276, 78224633235142116240, 6830914919397129328500, 638477522900795994967040, 63599377775480137499907561, 6725771848938288950491594140

Offset: 0

Paul D. Hanna, Jun 07 2005

nonn

Shift right of column 1 of triangle A107670, which is the matrix square of triangle A107667.
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^2*x)*(1 - x*A(x)) is equal to 0.
Given the o.g.f. A(x), the o.g.f. of A304322 equals 1/(1 - x*A(x)).
Also, a(n) is the number of 2-symbol Turing Machine state graphs in which n states are reached in canonical order. A canonical TM state graph lists for each state 1..n, and each of 2 symbols 0,1 in lexicographic order, a next state that is either the halt state, an already listed state, or the least unlisted state, as in the Haskell program below. Multiplied by 4^(2*n), this gives a much smaller number of TMs to be considered for the Busy Beaver function than given by A052200. - John Tromp, Oct 15 2024

			O.g.f.: A(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 + 2976295383100*x^9 + ...
From _Petros Hadjicostas_, Mar 10 2021: (Start)
We illustrate the above formula for a(n) with the compositions of n + 1 for n = 2.
The compositions of n + 1 = 3 are 3, 1 + 2, 2 + 1, and 1 + 1 + 1.  Thus the above sum has four terms with (r = 1, s_1 = 3), (r = 2, s_1 = 1, s_2 = 2), (r = 2, s_1 = 2, s_2 = 1), and (r = 3, s_1 = s_2 = s_3 = 1).
The value of the denominator Product_{j=1..r} s_j! for these four terms is 6, 2, 2, and 1, respectively.
The value of the numerator Product_{j=1..r} (Sum_{i=1..j} s_i)^(2*s_j) for these four terms is 729, 81, 144, and 36.
Thus a(2) = 729/6 - 81/2 - 144/2 + 36/1 = 45. (End)

Vaclav Kotesovec, Table of n, a(n) for n = 0..300

Cf. A107667, A107669, A107670, A052200.

Cf. A304322, A107675, A304394, A304395, A342202.

-- using program for A107667
a107668 = map head a where a = [[sum [a!!n!!i * a!!i!!(k+1) | i<-[k+1..n]] | k <- [0..n-1]] ++ [fromIntegral n+1] | n <- [0..]] -- John Tromp, Oct 21 2024

-- low memory version
a107668 n = (foldl' (\r i->sum r`seq`listArray(0,n)(0:[if i+1<2*j then 0 else r!j*(n+2-j)+r!(j-1)|j<-[1..n]])) (listArray(0,n)(0:repeat 1)) [1..2*n])!n -- John Tromp, Oct 15 2024

{a(n)=local(A);if(n==0,n+1,A=(n+1)*x+x*O(x^n); for(k=0,n,A+=polcoeff(A,k)*x^k*(n+1-prod(i=0,k,1+(i-n-1)*x))); polcoeff(A,n))}
for(n=0,30, print1(a(n),", "))

/* From formula: [x^n] exp( n^2*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^2 +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)^(2*n+2)/(n+1)! - sum(k=1,n, (n+1)^(2*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^2*x ) * (1 - x*A(x)) = 0 for n > 0. - Paul D. Hanna, May 12 2018
a(n) = (n+1)^2 * A107669(n).
a(n) = (n+1)^(2*n+2)/(n+1)! - Sum_{k=1..n} (n+1)^(2*k)/k! * a(n-k) for n > 0 with a(0) = 1. - Paul D. Hanna, May 12 2018
a(n) = A342202(2,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)^(2*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 ordered partitions (i.e., compositions) of n+1. See Michel Marcus's PARI program in A342202.) - Petros Hadjicostas, Mar 10 2021
a(n) ~ sqrt(1-c) * 2^(2*n + 3/2) * n^(n + 1/2) / (sqrt(Pi) * exp(n) * c^(n+1) * (2-c)^(n+1)), where c = -A226775 = -LambertW(-2*exp(-2)). - Vaclav Kotesovec, Oct 18 2024

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

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

A101481 Column 0 of triangular matrix T=A101479, in which row n equals row (n-1) of T^(n-1) followed by '1'.

Original entry on oeis.org

1, 1, 1, 3, 19, 191, 2646, 46737, 1003150, 25330125, 735180292, 24103027865, 880627477269, 35469795883964, 1561107221731851, 74528004538789830, 3835467005270307663, 211648845813188932595, 12465477924609075602136

Offset: 0

Paul D. Hanna, Jan 21 2005

nonn

From Gerhard Kirchner, Apr 20 2017: (Start)
Also: Let U(n,i,k), k<=i<=n, be a triangular matrix with elements selected as "0" or "1" such that the partial sum of the first m rows is s(m)<=m for 1<=mA101481(n+1) is the number of possible selections.
U(n,i,k)  has r(n) = n*(n+1)/2 elements. There are c(n) = binomial(r(n), n) ways to select n elements, some of which, however, are forbidden, see example. This yields the estimation a(n+1) < c(n).
Derivation of the recurrence:
s(m-1) <= m-1, say s(m-1)= j with 0 <= j <= m-1. Let f(m-1, j) be the number of associated submatrices. In order to determine f(m, k), k>=j, we have to distribute "1" k-j times among the m elements of row number m. There are binomial(m, k-j) ways to do that. The distribution must be repeated for each j. The recurrence describes this procedure. (End)

			G.f. = 1 + x + x^2 + 3*x^3 + 19*x^4 + 191*x^5 + 2646*x^6 + 46737*x^7 + ...
This sequence can also be generated in the following manner.
Start a table with the all 1's sequence in row 0; from then on, row n+1 can be formed from row n by dropping the initial n-1 terms of row n and taking partial sums of the remaining terms to obtain row n+1.
The table below illustrates this method:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...;
[1], 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, ...;
[3, 9], 19, 34, 55, 83, 119, 164, 219, 285, 363, 454, ...;
[19, 53, 108], 191, 310, 474, 693, 978, 1341, 1795, 2354, ...;
[191, 501, 975, 1668], 2646, 3987, 5782, 8136, 11169, 15017, ...;
[2646, 6633, 12415, 20551, 31720], 46737, 66570, 92358, ...; ...;
In the above table, drop the initial n-1 terms in row n (enclosed in square brackets) and then take partial sums to obtain row n+1 for n>=1;
this sequence then forms the first column of the resultant table.
Note: column k of the above table equals column 0 of matrix power T^(k+1) where T=A101479, for k>=0.
From _Gerhard Kirchner_, Apr 20 2017: (Start)
n=3: 0         0        1       forbidden: 1
     0 0       1 0      0 1                1 1
     1 1 1     0 1 1    1 0 1              0 0 0
     s(2)=0    s(2)=1   s(2)=2             s(2)=3>2
c(3) = binomial(6,3) = 20. There is only one forbidden matrix.
Thus: a(n+1) = a(4) = 19
Using the recurrence:
f(2,0) = f(1,0) = 1
f(2,1) = 2*f(1,0) + f(1,1) = 3
a(3) = f(2,2) = f(1,0) + 2*f(1,1) = 3
a(4) = f(3,3) = f(2,0) + 3*f(2,1) + 3*f(2,2) = 19
(End)

Vaclav Kotesovec, Table of n, a(n) for n = 0..300

Cf. A101479, A101482, A101483, A226775.

a[ n_, k_: 1] := a[n, k] = If[ n < 2, Boole[n >= 0], Sum[a[n - 1, i], {i, n + k - 2}]]; (* Michael Somos, Nov 29 2016 *)
f[m_, k_] := f[m, k] = If[(m == 0 && k == 0) || (m == 0 && k == 1) || (m == 1 && k == 0) || (m == 1 && k == 1), 1, Sum[f[m-1, j]*Binomial[m, k-j], {j, 0, m-1}]]; Flatten[{1, Table[f[n-1, n-1], {n, 1, 20}]}] (* Vaclav Kotesovec, Apr 20 2017 after Gerhard Kirchner *)

{a(n)=local(A=Mat(1),B);for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,B[i,j]=(A^(i-2))[i-1,j]);));A=B);return(A[n+1,1])}

{a(n, k=1) = if( n<2, n>=0, sum(i=1, n+k-2, a(n-1, i)))}; /* Michael Somos, Nov 29 2016 */

From Benedict W. J. Irwin, Nov 29 2016: (Start)
Conjecture: a(n) is described by a series of nested sums,
a(2) = Sum_{i=1..1} 1,
a(3) = Sum_{i=1..1+1}Sum_{j=1..i} 1,
a(4) = Sum_{i=1..1+2}Sum_{j=1..i+1}Sum_{k=1..j} 1,
a(5) = Sum_{i=1..1+3}Sum_{j=1..i+2}Sum_{k=1..j+1}Sum_{l=1..k} 1,
(End)
From Gerhard Kirchner, Apr 20 2017: (Start)
Recurrence: f(m,k) = Sum_{j=0..m-1} f(m-1,j)*binomial(m,k-j) with f(1,0) = f(1,1)= 1. a(n+1) = f(n,n). (End)
a(n) ~ c * exp(n) * n^(n-3/2) / 2^n, where c = (2 + LambertW(-2*exp(-2))) / (exp(2) * sqrt(2*Pi)) = 0.08604131405842589281435... - Vaclav Kotesovec, Apr 20 2017, updated Dec 03 2017

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

A304319 O.g.f. A(x) satisfies: [x^n] exp( n(n+1) x ) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 2, 10, 104, 1772, 42408, 1303504, 48736000, 2139552016, 107629121888, 6094743943584, 383305860004992, 26491391713168640, 1994924925169038848, 162537118868301414912, 14243360542620058589184, 1335710880923054761115904, 133461369304858515494530560, 14154134380237986764584033792, 1587931951984022880659170662400

Offset: 0

Paul D. Hanna, May 11 2018

nonn

It is striking that the coefficients of o.g.f. A(x) consist entirely of integers.

Note that if [x^n] exp( (n+1)*(n+2)*x ) / G(x) = 0 then G(x) does not consist entirely of integer coefficients.

			O.g.f.: A(x) = 1 + 2*x + 10*x^2 + 104*x^3 + 1772*x^4 + 42408*x^5 + 1303504*x^6 + 48736000*x^7 + 2139552016*x^8 + 107629121888*x^9 + 6094743943584*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(n*(n+1)*x) / A(x) begins:
n=0: [1, -2, -12, -432, -32640, -4176000, -804504960, -216834831360, ...];
n=1: [1, 0, -16, -520, -36432, -4520768, -856647680, -228458074752, ...];
n=2: [1, 4, 0, -648, -46032, -5341824, -974612736, -254049782400, ...];
n=3: [1, 10, 84, 0, -56832, -6922368, -1194341760, -299397745152, ...];
n=4: [1, 18, 308, 4448, 0, -8528000, -1573784960, -376524725760, ...];
n=5: [1, 28, 768, 20088, 444720, 0, -1938504960, -502258872960, ...];
n=6: [1, 40, 1584, 61560, 2286768, 72032832, 0, -618983309952, ...];
n=7: [1, 54, 2900, 154352, 8074368, 404450176, 17201640064, 0, ...];
n=8: [1, 70, 4884, 339120, 23357568, 1583068032, 102886277760, 5682964174848, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n*(n+1)*x ) / A(x) = 0 for n>=0.
LOGARITHMIC DERIVATIVE.
The logarithmic derivative of A(x) yields the o.g.f. of A304317:
A'(x)/A(x) = 2 + 16*x + 260*x^2 + 6200*x^3 + 191832*x^4 + 7235152*x^5 + 320372320*x^6 + 16243028896*x^7 + 926219213216*x^8 + 58608051937536*x^9 + 4072302306624576*x^10 + ...+ A304317(n)*x^n + ...

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

Cf. A304317, A304318, A304320, A304863, A304864, A304865.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*m*(m-1) +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) * n^(n - 1/2) / (sqrt(Pi) * c^(n + 1/2) * (2-c)^n * exp(n)), where c = -A226775 = -LambertW(-2*exp(-2)). - Vaclav Kotesovec, Aug 31 2020

A230218 G.f. A(x) satisfies: [x^n] A(x)^(n^2+n+1) = 0 for n>1.

Original entry on oeis.org

1, 1, -3, 14, -85, 504, -4424, 6796, -878157, -25703710, -1270518018, -65772588300, -3848787714746, -248212765567326, -17520121174143210, -1343050785659060872, -111112550557260635229, -9867409274482580015370, -936234289413196544207234, -94522404087905722536648780

Offset: 0

Paul D. Hanna, Oct 11 2013

sign

			G.f.: A(x) = 1 + x - 3*x^2 + 14*x^3 - 85*x^4 + 504*x^5 - 4424*x^6 +...
Coefficients of x^k in the powers A(x)^(n^2+n+1) of g.f. A(x) begin:
n=0: [1, 1,   -3,    14,    -85,     504,    -4424,      6796, ...];
n=1: [1, 3,   -6,    25,   -153,     819,    -8664,    -18360, ...];
n=2: [1, 7,    0,     7,    -98,     210,   -10122,   -141525, ...];
n=3: [1,13,   39,     0,    -78,    -819,   -15483,   -380952, ...];
n=4: [1,21,  147,   364,      0,   -2457,   -35805,   -821916, ...];
n=5: [1,31,  372,  2139,   5580,       0,   -91698,  -1792947, ...];
n=6: [1,43,  774,  7525,  42097,  125517,        0,  -4097298, ...];
n=7: [1,57, 1425, 20482, 185877, 1089270,  3791298,         0, ...];
n=8: [1,73, 2409, 47450, 619697, 5619978, 35621518, 144591976, 0, ...]; ...
where the coefficients of x^n in A(x)^(n^2+n+1) all equal zero for n>1.
ODD TERMS:
For n>0, a(n) appears to be odd only when n is a power of 2:
a(1) = 1;
a(2) = -3;
a(4) = -85;
a(8) = -878157;
a(16) = -111112550557260635229;
a(32) = -886203693344229341179357569730608605545213045330679133; ...

Paul D. Hanna, Table of n, a(n) for n = 0..520

Cf. A185072, A229041, A229044, A171791, A292877.

{a(n)=local(A=[1,1]); for(m=2,n, A=concat(A, 0); A[#A]=-Vec(Ser(A)^(m^2+m+1))[m+1]/(m^2+m+1)); A[n+1]}
for(n=0,20,print1(a(n),", "))

for n>0, a(n) is odd iff n is a power of 2 (conjecture).
G.f. A(x) satisfies:
(1) A(x) = F(x/A(x)) where F(x) = A(x*F(x)) is the g.f. of A185072.
(2) A(x) = G(x/A(x)^2) where G(x) = A(x*G(x)^2) is the g.f. of A229041.
a(n) ~ -c * 2^(2*n) *n^(n-5/2) / (exp(n) * d^n * (2-d)^n), where d = -LambertW(-2*exp(-2)) = -A226775 = 0.40637573995995990767695812412483975821... and c = 0.015106126717978... - Vaclav Kotesovec, Sep 27 2017

A337458 O.g.f. A(x) satisfies: [x^n] exp( n(n+1) x ) / A(x)^(n+1) = 0 for n > 0.

Original entry on oeis.org

1, 1, 2, 11, 130, 2450, 63012, 2040779, 79377914, 3594766694, 185457776252, 10725423627006, 686721189003668, 48200778475446916, 3679104677398632520, 303348177377608050219, 26865664102518601306154, 2543352040870175109554654, 256296085507636954980717708, 27390678829206902911266889386

Offset: 0

Paul D. Hanna, Aug 28 2020

nonn

It is remarkable that this sequence consists entirely of integers.

			O.g.f.: A(x) = 1 + x + 2*x^2 + 11*x^3 + 130*x^4 + 2450*x^5 + 63012*x^6 + 2040779*x^7 + 79377914*x^8 + 3594766694*x^9 + 185457776252*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(n*(n+1)*x) / A(x)^(n+1) begins:
n=0: [1, -1, -2, -48, -2616, -262080, -41718240, -9630270720, ...];
n=1: [1, 0, -6, -112, -5592, -547968, -86345120, -19809990912, ...];
n=2: [1, 3, 0, -222, -10728, -958824, -144971712, -32519314080, ...];
n=3: [1, 8, 52, 0, -18648, -1693248, -236690784, -50727983616, ...];
n=4: [1, 15, 210, 2420, 0, -2739720, -399251600, -80125144800, ...];
n=5: [1, 24, 558, 12192, 221184, 0, -616918320, -131299591680, ...];
n=6: [1, 35, 1204, 40278, 1272768, 33597312, 0, -196436730672, ...];
n=7: [1, 48, 2280, 106688, 4869552, 210771456, 7654459648, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n*(n+1)*x ) / A(x)^(n+1) = 0 for n>0.
RELATED SERIES.
Define B(x) = A(x/B(x)), which begins
B(x) = 1 + x + x^2 + 7*x^3 + 93*x^4 + 1859*x^5 + 49357*x^6 + 1629227*x^7 + 64149805*x^8 + 2929386667*x^9 + ... + A337457(n)*x^n + ...
then the table of coefficients of x^k/k! in exp( n*(n-1)*x/B(x) ) begins:
n=0: [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...];
n=1: [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...];
n=2: [1, 2, 0, -16, -320, -21888, -2648576, -494325760, ...];
n=3: [1, 6, 24, 0, -1728, -88704, -9621504, -1715198976, ...];
n=4: [1, 12, 120, 864, 0, -281088, -26873856, -4328017920, ...];
n=5: [1, 20, 360, 5600, 65920, 0, -66944000, -10207436800, ...];
n=6: [1, 30, 840, 21600, 492480, 8784000, 0, -22098355200, ...];
n=7: [1, 42, 1680, 63504, 2237760, 71229312, 1814690304, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n*(n-1)*x/B(x) ) = 0 for n>0.
Also note that B(x) = x/Series_Reversion( x*A(x) ) and A(x) = B(x*A(x)).

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

Cf. A337457, A304319.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(m*(m-1)*x +x*O(x^m)) / Ser(A)^m )[m]/m);A[n+1]}
for(n=0, 20, print1(a(n), ", "))

Given o.g.f. A(x), define B(x) = A(x/B(x)), then B(x) is the o.g.f. of A337457 and satisfies [x^n] exp( n*(n-1)*x/B(x) ) = 0 for n>0.

a(n) ~ sqrt(1-c) * 2^(2*n - 3/2) * n^(n - 3/2) / (sqrt(Pi) * c^n * (2-c)^(n-1) * exp(n)), where c = -A226775 = -LambertW(-2*exp(-2)). - Vaclav Kotesovec, Aug 31 2020

A292877 G.f. A(x) satisfies: [x^(n-1)] 1/A(x)^(n^2) = 0 for n>2.

Original entry on oeis.org

1, 1, 5, 34, 273, 2331, 22110, 190450, 2540975, -1071509, 1200284739, -50263360280, 3102388877800, -199436004737160, 14155468007742978, -1088800915851203694, 90359645776680747647, -8046100605226675723225, 765244962799789283768523, -77422876485545489461403294, 8303247917673506082303329493, -940940782152450052071048090369, 112352003582903383388702940258120

Offset: 0

Paul D. Hanna, Sep 25 2017

sign

Conjectures:
(1) a(2^n) is odd for n>=0.
(2) a(n) is odd iff n is a Fibbinary number: a(A003714(k)) is odd for k>=0.
(3) The number of odd terms between a(2^n) and a(2^(n+1)-1), inclusively, is Fibonacci(n+1), for n>=0.

			G.f.: A(x) = 1 + x + 5*x^2 + 34*x^3 + 273*x^4 + 2331*x^5 + 22110*x^6 + 190450*x^7 + 2540975*x^8 - 1071509*x^9 + 1200284739*x^10 - 50263360280*x^11 + 3102388877800*x^12 - 199436004737160*x^13 + 14155468007742978*x^14 - 1088800915851203694*x^15 +...
such that the coefficient of x^n in 1/A(x)^(n^2) equals zero for n>1.
Notice that a(n) seems to be odd only when n is a Fibbinary number (A003714):
[0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37, 40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85, 128, 29, 130, 132, 133, ...].
RELATED TABLES.
(1) The table of coefficients in 1/A(x)^(n^2) begins:
n=1: [1, -1, -4, -25, -194, -1603, -15264, -122316, -1897710, ...];
n=2: [1, -4, -10, -56, -427, -3360, -33546, -218880, -5179834, ...];
n=3: [1, -9, 0, -21, -252, -1701, -25992, 2970, -7903413, ...];
n=4: [1, -16, 56, 0, -84, 784, -18656, 384896, -13426530, ...];
n=5: [1, -25, 200, -525, 0, 2695, -38600, 878150, -26292375, ...];
n=6: [1, -36, 486, -3000, 7821, 0, -101322, 1916352, -52357590, ...];
n=7: [1, -49, 980, -10241, 58898, -170079, 0, 4515000, -108626140, ...];
n=8: [1, -64, 1760, -27136, 256048, -1500352, 4979712, 0, -234893352, ...];
n=9: [1, -81, 2916, -61425, 838026, -7720839, 48097152, -184870512, 0, ...]; ...
such that the main diagonal is all zeros after the initial terms.
(2) The table of coefficients in (1/x)*Series_Reversion( x*A(x)^n ) begins:
n=1: [1, -1, -3, -14, -85, -504, -4424, -6796, -878157, ...];
n=2: [1, -2, -3, -8, -40, -60, -2604, 48112, -1747260, ...];
n=3: [1, -3, 0, 2, -15, 189, -3850, 101700, -3340845, ...];
n=4: [1, -4, 6, 0, -35, 396, -7182, 181824, -5817510, ...];
n=5: [1, -5, 15, -30, 0, 714, -13335, 315060, -9679455, ...];
n=6: [1, -6, 27, -104, 315, 0, -19957, 532848, -15864336, ...];
n=7: [1, -7, 42, -238, 1260, -5481, 0, 713796, -25010433, ...];
n=8: [1, -8, 60, -448, 3310, -23352, 136696, 0, -31112163, ...];
n=9: [1, -9, 81, -750, 7065, -66420, 598626, -4474764, 0, ...]; ...
in which the main diagonal is all zeroes after the initial terms.
(3) The table of coefficients in ((1/x)*Series_Reversion( x*A(x)^n ))^(1/n) begins:
n=1: [1, -1, -3, -14, -85, -504, -4424, -6796, -878157, ...];
n=2: [1, -1, -2, -6, -28, -70, -1446, 22302, -855032, ...];
n=3: [1, -1, -1, -1, -7, 49, -1191, 31569, -1051695, ...];
n=4: [1, -1, 0, 1, -6, 78, -1544, 40605, -1328178, ...];
n=5: [1, -1, 1, 0, -9, 117, -2118, 53232, -1699905, ...];
n=6: [1, -1, 2, -4, 0, 141, -2958, 71900, -2216860, ...];
n=7: [1, -1, 3, -11, 37, 0, -3245, 95286, -2941059, ...];
n=8: [1, -1, 4, -21, 118, -581, 0, 99086, -3760182, ...];
n=9: [1, -1, 5, -34, 259, -2002, 13212, 0, -3775221, ...];
n=10: [1, -1, 6, -50, 476, -4788, 47578, -397090, 0, ...]; ...
in which the secondary diagonal is all zeroes after the initial terms.
(4) The table of coefficients in 1/A(x)^n begins:
n=1: [1, -1, -4, -25, -194, -1603, -15264, -122316, ...];
n=2: [1, -2, -7, -42, -322, -2618, -25145, -191580, ...];
n=3: [1, -3, -9, -52, -396, -3168, -30889, -220332, ...];
n=4: [1, -4, -10, -56, -427, -3360, -33546, -218880, ...];
n=5: [1, -5, -10, -55, -425, -3286, -33990, -195585, ...];
n=6: [1, -6, -9, -50, -399, -3024, -32938, -157122, ...];
n=7: [1, -7, -7, -42, -357, -2639, -30968, -108718, ...];
n=8: [1, -8, -4, -32, -306, -2184, -28536, -54368, ...];
n=9: [1, -9, 0, -21, -252, -1701, -25992, 2970,  -7903413, ...]; ...
where the main diagonal divided by n begins:
D = [1, -2/2, -9/3, -56/4, -425/5, -3024/6, -30968/7, -54368/8, -7903413/9, ...],
D = [1, -1, -3, -14, -85, -504, -4424, -6796, -878157, 25703710, -1270518018, ...].
Compare D to:
A230218 = [1, 1, -3, 14, -85, 504, -4424, 6796, -878157, -25703710, -1270518018, ...];
the g.f. G(x) of A230218 obeys: [x^n] G(x)^(n^2+n+1) = 0 for n>1.

Paul D. Hanna, Table of n, a(n) for n = 0..520

Cf. A303562, A003714, A230218, A171791.

{a(n) = my(A=[1, 1]); for(i=2, n+1, A=concat(A, 0); A[#A]=Vec(1/Ser(A)^((#A)^2))[#A]/(#A)^2 ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

G.f. A(x) satisfies:
(1) [x^(n-1)] 1/A(x)^(n^2) = 0 for n>2.
(2) [x^(n-1)] (1/x)*Series_Reversion( x*A(x)^n ) = 0 for n>2.
(3) [x^(n-2)] ( (1/x)*Series_Reversion( x*A(x)^n ) )^(1/n) = 0 for n>3.
a(n) ~ (-1)^n * c * 2^(2*n) * n^(n-5/2) / (exp(n) * d^n * (2-d)^n), where d = -LambertW(-2*exp(-2)) = -A226775 = 0.40637573995995990767695812412483975821... and c = 0.01284812446900190... - Vaclav Kotesovec, Sep 27 2017

A304318 O.g.f. A(x) satisfies: [x^n] exp( n(n-1) x ) / A(x) = 0.

Original entry on oeis.org

1, 0, 2, 24, 436, 10656, 328112, 12183456, 529242224, 26309617536, 1472135847072, 91526938123008, 6258004268952064, 466599240364076544, 37672137946943244288, 3274012281487011586560, 304724394621209905647360, 30239686358027369113804800, 3187164738879981461171955200, 355548230503664593634743375872

Offset: 0

Paul D. Hanna, May 11 2018

nonn

It is striking that the coefficients of o.g.f. A(x) consist entirely of integers.

			O.g.f.: A(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 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(n*(n-1)*x) / A(x) begins:
n=0: [1, 0, -4, -144, -10368, -1267200, -234576000, -61085767680, ...];
n=1: [1, 0, -4, -144, -10368, -1267200, -234576000, -61085767680, ...];
n=2: [1, 2, 0, -160, -11600, -1376928, -250428416, -64479262720, ...];
n=3: [1, 6, 32, 0, -13392, -1630944, -286447104, -71981250048, ...];
n=4: [1, 12, 140, 1440, 0, -1916928, -351444096, -85338800640, ...];
n=5: [1, 20, 396, 7616, 128512, 0, -417488000, -107269127680, ...];
n=6: [1, 30, 896, 26496, 760752, 19101600, 0, -128348167680, ...];
n=7: [1, 42, 1760, 73440, 3034800, 121743072, 4260708864, 0, ...];
n=8: [1, 56, 3132, 174800, 9716608, 535021056, 28597069696, 1331047703552, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n*(n-1)*x ) / A(x) = 0 for n>=0.
LOGARITHMIC DERIVATIVE.
The logarithmic derivative of A(x) yields the o.g.f. of A304316:
A'(x)/A(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 + ... + A304316(n)*x^n + ...

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

Cf. A304316, A304319, A304320.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)*(m-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) * n^(n - 1/2) / (sqrt(Pi) * c^(n - 1/2) * (2-c)^n * exp(n)), where c = -A226775 = -LambertW(-2*exp(-2)). - Vaclav Kotesovec, Aug 31 2020

A384471 a(n) = Sum_{k=0..n} binomial(n,k)^2 * Stirling2(2k,k) Stirling2(2n-2k,n-k).

Original entry on oeis.org

1, 2, 18, 306, 8046, 296100, 14307254, 865996306, 63308257198, 5432272670376, 535074966419260, 59461066810476232, 7354069129792197762, 1001371912804041913056, 148806933109572134044158, 23958722845801073318076450, 4154065510530807075869275150, 771608888261061026185781127184

Offset: 0

Vaclav Kotesovec, May 30 2025

nonn

Cf. A187655, A187657, A384470, A384472.

Cf. A226775.

Table[Sum[StirlingS2[2*k, k]*StirlingS2[2*n-2*k, n-k]*Binomial[n, k]^2, {k, 0, n}], {n, 0, 20}]

a(n) ~ 2^(3*n + 1/2) * n^(n - 3/2) / (Pi^(3/2) * (1-w) * exp(n) * (2-w)^n * w^n), where w = -LambertW(-2*exp(-2)) = -A226775 = 0.4063757399599599...

cp's OEIS Frontend

A107668 Column 0 of triangle A107667.

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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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A101481 Column 0 of triangular matrix T=A101479, in which row n equals row (n-1) of T^(n-1) followed by '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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A304319 O.g.f. A(x) satisfies: [x^n] exp( n*(n+1) * x ) / A(x) = 0 for n>0.

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A230218 G.f. A(x) satisfies: [x^n] A(x)^(n^2+n+1) = 0 for n>1.

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A337458 O.g.f. A(x) satisfies: [x^n] exp( n*(n+1) * x ) / A(x)^(n+1) = 0 for n > 0.

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A304319 O.g.f. A(x) satisfies: [x^n] exp( n(n+1) x ) / A(x) = 0 for n>0.

A337458 O.g.f. A(x) satisfies: [x^n] exp( n(n+1) x ) / A(x)^(n+1) = 0 for n > 0.

A304318 O.g.f. A(x) satisfies: [x^n] exp( n(n-1) x ) / A(x) = 0.

A384471 a(n) = Sum_{k=0..n} binomial(n,k)^2 * Stirling2(2k,k) Stirling2(2n-2k,n-k).