A379168
Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where column k is the expansion of e.g.f. B(x)^k, where B(x) is the e.g.f. of A140049.
Original entry on oeis.org
1, 1, 0, 1, 1, 0, 1, 2, 5, 0, 1, 3, 12, 55, 0, 1, 4, 21, 140, 1005, 0, 1, 5, 32, 261, 2600, 26601, 0, 1, 6, 45, 424, 4965, 68752, 941863, 0, 1, 7, 60, 635, 8304, 132003, 2414188, 42372177, 0, 1, 8, 77, 900, 12845, 223104, 4617675, 107385896, 2336926665, 0
Offset: 0
Square array begins:
1, 1, 1, 1, 1, 1, 1, ...
0, 1, 2, 3, 4, 5, 6, ...
0, 5, 12, 21, 32, 45, 60, ...
0, 55, 140, 261, 424, 635, 900, ...
0, 1005, 2600, 4965, 8304, 12845, 18840, ...
0, 26601, 68752, 132003, 223104, 350125, 522576, ...
0, 941863, 2414188, 4617675, 7806424, 12296935, 18477828, ...
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a(n, k) = if(k==0, 0^n, k*sum(j=0, n, (n+k)^(j-1)*binomial(n, j)*a(n-j, j)));
A162659
E.g.f. A(x) satisfies A(x) = exp(x*A(x*A(x))).
Original entry on oeis.org
1, 1, 3, 22, 281, 5396, 142297, 4865806, 207407489, 10710044776, 655655874641, 46789973764634, 3840103504940881, 358443042637767868, 37700333788138306937, 4432826052558222878206, 578707468284010393533953, 83384676375176176768112720, 13190759232920144271864441505
Offset: 0
E.g.f.: A(x) = 1 + x + 3*x^2/2! + 22*x^3/3! + 281*x^4/4! + 5396*x^5/5! +...
A(x*A(x)) = 1 + x + 5*x^2/2! + 49*x^3/3! + 777*x^4/4! + 17581*x^5/5! +...
log(A(x)) = x + 2*x^2/2! + 15*x^3/3! + 196*x^4/4! + 3885*x^5/5! + 105486*x^6/6! +...
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terms = 19; A[] = 0; Do[A[x] = Exp[x*A[x*A[x]]] + O[x]^terms // Normal, terms]; CoefficientList[A[x], x]Range[0,terms-1]! (* Stefano Spezia, Mar 24 2025 *)
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{a(n,m=1)=if(n==0,1,if(m==0,0^n,sum(k=0,n,binomial(n,k)*m*(n-k+m)^(k-1)*a(n-k,k))))}
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/* Log(A(x)) = x*A(x*A(x)) = Sum_{n>=1} L(n)*x^n/n! where: */
{L(n)=if(n<1,0,sum(k=1,n,binomial(n,k)*(n-k)^(k-1)*a(n-k,k)))}
A144681
E.g.f. satisfies: A(x/A(x)) = exp(x).
Original entry on oeis.org
1, 1, 3, 22, 305, 6656, 204337, 8226436, 414585425, 25315924960, 1828704716801, 153433983789164, 14739472821255481, 1602471473448455104, 195300935112810494801, 26470100501608768436716
Offset: 0
E.g.f. A(x) = 1 + x + 3*x^2/2! + 22*x^3/3! + 305*x^4/4! +...
A(x/A(x)) = 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! +...
1/A(x) = 1 + x - x^2/2! + 10*x^3/3! - 159*x^4/4! + 3816*x^5/5! -+...
A(log(A(x))) = 1 + x + 5*x^2/2! + 55*x^3/3! + 1005*x^4/4! + 26601*x^5/5! +...
ILLUSTRATE FORMULA a(n+1) = [x^n/n!] exp(x)*A(x)^(n+1) as follows.
Form a table of coefficients of x^k/k! in exp(x)*A(x)^n for n>=1, k>=0:
exp(x)*A(x)^1: [(1), 2, 6, 35, 416, 8437, 249340, ...];
exp(x)*A(x)^2: [1,(3), 13, 93, 1145, 22593, 645741, ...];
exp(x)*A(x)^3: [1, 4,(22), 181, 2320, 45199, 1257364, ...];
exp(x)*A(x)^4: [1, 5, 33,(305), 4097, 79825, 2177329, ...];
exp(x)*A(x)^5: [1, 6, 46, 471,(6656), 131001, 3529836, ...];
exp(x)*A(x)^6: [1, 7, 61, 685, 10201,(204337), 5477005, ...];
exp(x)*A(x)^7: [1, 8, 78, 953, 14960, 306643,(8226436), ...]; ...
then the terms along the main diagonal form this sequence shift left.
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{a(n)=local(A=1+x+x*O(x^n));for(n=0,n,A=exp(serreverse(x/A)));n!*polcoeff(A,n)}
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{a(n)=local(A=1+x+sum(k=2,n-1,a(k)*x^k/k!)+x*O(x^n));if(n==0,1,(n-1)!*polcoeff(exp(x+x*O(x^n))*A^n,n-1))}
A384689
E.g.f. A(x) satisfies A(x) = exp( x*A(x)^2 * A(x*A(x)) ).
Original entry on oeis.org
1, 1, 7, 106, 2593, 89796, 4085029, 232694806, 16053415249, 1308960150472, 123811136509861, 13387049625793746, 1635128238889494793, 223420020463904387020, 33872693045213102767093, 5658826351169923606739206, 1035543935182601250745181089, 206506472947550295487980305424
Offset: 0
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terms = 18; A[] = 0; Do[A[x] = Exp[x*A[x]^2*A[x*A[x]]] + O[x]^terms // Normal, terms]; Range[0,terms-1]!CoefficientList[A[x], x] (* Stefano Spezia, Jun 07 2025 *)
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a(n, k=1) = if(k==0, 0^n, k*sum(j=0, n, (n+j+k)^(j-1)*binomial(n, j)*a(n-j, j)));
A384787
E.g.f. A(x) satisfies A(x) = exp( x*A(x) * A(2*x*A(x)) ).
Original entry on oeis.org
1, 1, 7, 148, 7381, 801536, 186678019, 93865986880, 102755888482153, 245872091164966912, 1285664593514044479391, 14650473009515606022910976, 362327823926201727094352145661, 19359048028300511200690402408529920, 2224311455921555052696103713299884826395
Offset: 0
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a(n, k=1) = if(k==0, 0^n, k*sum(j=0, n, 2^(n-j)*(n+k)^(j-1)*binomial(n, j)*a(n-j, j)));
Showing 1-5 of 5 results.