A232696 E.g.f. A(x) satisfies: A'(x) = A(x/A'(x)) with A(0)=1.
1, 1, 1, -1, 2, 6, -264, 5370, -93750, 1315706, -3543736, -880688376, 56549341380, -2612825765748, 99009763750128, -2593891139126560, -31860555469490020, 12585468136754891100, -1364794494618494224128, 114095029934565534862680, -7984695190944325311086112, 419424013080533747232201968
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + x + x^2/2! - x^3/3! + 2*x^4/4! + 6*x^5/5! - 264*x^6/6! + 5370*x^7/7! +...
such that A(x) = A'(x*A(x)) and
A(x/A'(x)) = A'(x) = 1 + x - x^2/2! + 2*x^3/3! + 6*x^4/4! - 264*x^5/5! +...
To illustrate a(n) = [x^n/n!] A'(x)^(n+1)/(n+1), create a table of coefficients of x^k/k!, k>=0, in A'(x)^n like so:
A'^1: [1, 1, -1, 2, 6, -264, 5370, -93750, 1315706, ...];
A'^2: [1, 2, 0, -2, 34, -508, 7472, -100392, 774076, ...];
A'^3: [1, 3, 3, -6, 48, -522, 6036, -54030, -435618, ...];
A'^4: [1, 4, 8, -4, 36, -336, 2832, 3672, -1469680, ...];
A'^5: [1, 5, 15, 10, 10, -100, -130, 44490, -1964390, ...];
A'^6: [1, 6, 24, 42, 6, 36, -1680, 59520, -1938564, ...];
A'^7: [1, 7, 35, 98, 84, 42, -1848, 54978, -1605394, ...];
A'^8: [1, 8, 48, 184, 328, 128, -1504, 42960, -1194368, ...];
A'^9: [1, 9, 63, 306, 846, 864, -1278, 32202, -843750, ...]; ...
then the diagonal in the above table generates this sequence:
[1/1, 2/2, 3/3, -4/4, 10/5, 36/6, -1848/7, 42960/8, -843750/9, ...].
SUMS OF TERM RESIDUES MODULO 2^n.
Given a(k) == 0 (mod 2^n) for k>=(8*n-6) for n>1, then it is interesting to consider the sums of the residues of all terms modulo 2^n for n>=1.
Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
[4, 16, 40, 144, 432, 1008, 3184, 6384, 15600, 33520, 75504, 159472, 356080, 798448, 1797872, 3895024, 8089328, 16609008, 37842672, 76639984, 166817520, ...].
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..160
Programs
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PARI
{a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A, x, x/A' +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(x/serreverse(x*A +x*O(x^n)))); n!*polcoeff(A, n)} for(n=0, 25, print1(a(n), ", "))
Formula
E.g.f. satisfies: A(x) = A'(x*A(x)).
E.g.f. satisfies: A(x) = 1/x * Series_Reversion( x/A'(x) ).
a(n) = [x^n/n!] A'(x)^(n+1)/(n+1) for n>=0.
Comments