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.

A232694 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^4) with A(0)=1.

Original entry on oeis.org

1, 1, 1, 9, 177, 5601, 249681, 14569545, 1062623265, 93853717761, 9810385567329, 1192614883442889, 166310354311947345, 26308546859152889697, 4677436610087462937393, 927353710845763536487305, 203648424149429271943770945, 49245501579619466882211194625, 13045520297945193508654786790337
Offset: 0

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Author

Paul D. Hanna, Nov 27 2013

Keywords

Comments

CONJECTURES.
a(n) == 0 (mod 3) for n>=3.
a(n) == 1 (mod 8) for n>=0.

Examples

			E.g.f.: A(x) = 1 + x + x^2/2! + 9*x^3/3! + 177*x^4/4! + 5601*x^5/5! +...
such that
A(x*A'(x)^4) = A'(x) = 1 + x + 9*x^2/2! + 177*x^3/3! + 5601*x^4/4! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(4*n-3)/(4*n-3), create a table of coefficients of x^k/k!, k>=0, in A(x)^(4*n-3), n>=1, like so:
A^1 : [1,  1,   1,     9,    177,    5601,     249681,    14569545, ...];
A^5 : [1,  5,  25,   165,   2145,   55125,    2211225,   120873045, ...];
A^9 : [1,  9,  81,   801,  10449,  218889,    7501761,   373998465, ...];
A^13: [1, 13, 169,  2301,  35841,  731133,   21950409,   974182989, ...];
A^17: [1, 17, 289,  5049,  95217,  2102577,  60325809,  2417773881, ...];
A^21: [1, 21, 441,  9429, 211617,  5243301, 154446201,  5861076165, ...];
A^25: [1, 25, 625, 15825, 414225, 11585625, 364238625, 13752570225, ...];
A^29: [1, 29, 841, 24621, 738369, 23206989, 791747241, 30816074685, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 5/5, 81/9, 2301/13, 95217/17, 5243301/21, 364238625/25, 30816074685/29, ...].
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A, x, x*A'^4 +x*O(x^n)))); n!*polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal((1/x*serreverse(x/A^4 +x*O(x^n)))^(1/4))); n!*polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", "))

Formula

E.g.f. satisfies: A(x) = A'(x/A(x)^4).
E.g.f. satisfies: A(x) = ( x / Series_Reversion( x*A'(x)^4 ) )^(1/4).
a(n) = [x^(n-1)/(n-1)!] A(x)^(4*n-3)/(4*n-3) for n>=1.