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.
%I A232686 #16 Dec 07 2013 12:29:18
%S A232686 1,1,2,14,196,4352,135864,5550560,283148560,17454554624,1269351110560,
%T A232686 106925372050688,10284022937294400,1116367988620697600,
%U A232686 135485394041604223360,18237986446534462042112,2704852975724947252789504,439428075979518197809971200,77809530527904746552436748800
%N A232686 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x))^2.
%C A232686 CONJECTURES.
%C A232686 a(n) == 0 (mod 2) for n>=2.
%C A232686 a(n) == 0 (mod 2^2) for n>=4.
%C A232686 a(n) == 0 (mod 2^3) for n>=5.
%C A232686 a(n) == 0 (mod 2^k) for n>=(2*k-1), k>=3.
%C A232686 a(2*n) == 2^n (mod 2^(n+1)) for n>=0.
%H A232686 Paul D. Hanna, <a href="/A232686/b232686.txt">Table of n, a(n) for n = 0..100</a>
%F A232686 E.g.f. A(x) satisfies: A(x)^2 = A'(x/A(x)^2).
%F A232686 E.g.f. A(x) satisfies: A(x) = sqrt( x / Series_Reversion( x*A'(x) ) ).
%F A232686 a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n)/n for n>=1.
%e A232686 E.g.f.: A(x) = 1 + x + 2*x^2/2! + 14*x^3/3! + 196*x^4/4! + 4352*x^5/5! +...
%e A232686 such that
%e A232686 A(x*A'(x))^2 = A'(x) = 1 + 2*x + 14*x^2/2! + 196*x^3/3! + 4352*x^4/4! +...
%e A232686 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! +...
%e A232686 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:
%e A232686 A^2: [1, 2, 6, 40, 528, 11224, 339632, 13560864, ...];
%e A232686 A^4: [1, 4, 20, 152, 1912, 37808, 1075680, 40938272, ...];
%e A232686 A^6: [1, 6, 42, 384, 5016, 95832, 2590464, 94032000, ...];
%e A232686 A^8: [1, 8, 72, 784, 11088, 212896, 5575424, 194217024, ...];
%e A232686 A^10:[1, 10, 110, 1400, 21760, 431960, 11203920, 378660320, ...];
%e A232686 A^12:[1, 12, 156, 2280, 39048, 815184, 21350112, 709540704, ...];
%e A232686 A^14:[1, 14, 210, 3472, 65352, 1447768, 38853920, 1286816832, ...];
%e A232686 A^16:[1, 16, 272, 5024, 103456, 2441792, 67832064, 2265188480, ...]; ...
%e A232686 then the diagonal in the above table generates this sequence shift left:
%e A232686 [1/1, 4/2, 42/3, 784/4, 21760/5, 815184/6, 38853920/7, 2265188480/8, ...].
%e A232686 SUMS OF TERM RESIDUES MODULO 2^n.
%e A232686 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.
%e A232686 Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
%e A232686 [2, 6, 14, 30, 62, 158, 350, 990, 2782, 5342, 10462, 16606, 61662, 119006, 233694, 495838, 1151198, 2592990, 5476574, 14389470, 30118110, 51089630, ...].
%o A232686 (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)}
%o A232686 for(n=0, 25, print1(a(n), ", "))
%o A232686 (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)}
%o A232686 for(n=0, 25, print1(a(n), ", "))
%Y A232686 Cf. A233319, A231619, A231866, A231899, A232694, A232695, A232696.
%K A232686 nonn
%O A232686 0,3
%A A232686 _Paul D. Hanna_, Dec 03 2013