A232686 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x))^2.
1, 1, 2, 14, 196, 4352, 135864, 5550560, 283148560, 17454554624, 1269351110560, 106925372050688, 10284022937294400, 1116367988620697600, 135485394041604223360, 18237986446534462042112, 2704852975724947252789504, 439428075979518197809971200, 77809530527904746552436748800
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + x + 2*x^2/2! + 14*x^3/3! + 196*x^4/4! + 4352*x^5/5! +...
such that
A(x*A'(x))^2 = A'(x) = 1 + 2*x + 14*x^2/2! + 196*x^3/3! + 4352*x^4/4! +...
A(x*A'(x)) = sqrt(A'(x)) = 1 + x + 6*x^2/2! + 80*x^3/3! + 1748*x^4/4! + 54392*x^5/5! + 2227608*x^6/6! + 114233232*x^7/7! + 7088554512*x^8/8! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n)/n, create a table of coefficients of x^k/k!, k>=0, in A(x)^(2*n) like so:
A^2: [1, 2, 6, 40, 528, 11224, 339632, 13560864, ...];
A^4: [1, 4, 20, 152, 1912, 37808, 1075680, 40938272, ...];
A^6: [1, 6, 42, 384, 5016, 95832, 2590464, 94032000, ...];
A^8: [1, 8, 72, 784, 11088, 212896, 5575424, 194217024, ...];
A^10:[1, 10, 110, 1400, 21760, 431960, 11203920, 378660320, ...];
A^12:[1, 12, 156, 2280, 39048, 815184, 21350112, 709540704, ...];
A^14:[1, 14, 210, 3472, 65352, 1447768, 38853920, 1286816832, ...];
A^16:[1, 16, 272, 5024, 103456, 2441792, 67832064, 2265188480, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 4/2, 42/3, 784/4, 21760/5, 815184/6, 38853920/7, 2265188480/8, ...].
SUMS OF TERM RESIDUES MODULO 2^n.
Given a(k) == 0 (mod 2^n) for k>=(2*n-1) for n>=3, 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:
[2, 6, 14, 30, 62, 158, 350, 990, 2782, 5342, 10462, 16606, 61662, 119006, 233694, 495838, 1151198, 2592990, 5476574, 14389470, 30118110, 51089630, ...].
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..100
Programs
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PARI
{a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A^2, 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(1/x*serreverse(x/A^2 +x*O(x^n)))); n!*polcoeff(A, n)} for(n=0, 25, print1(a(n), ", "))
Formula
E.g.f. A(x) satisfies: A(x)^2 = A'(x/A(x)^2).
E.g.f. A(x) satisfies: A(x) = sqrt( x / Series_Reversion( x*A'(x) ) ).
a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n)/n for n>=1.
Comments