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A232696 E.g.f. A(x) satisfies: A'(x) = A(x/A'(x)) with A(0)=1. 7

%I #11 Dec 03 2013 20:51:28

%S 1,1,1,-1,2,6,-264,5370,-93750,1315706,-3543736,-880688376,

%T 56549341380,-2612825765748,99009763750128,-2593891139126560,

%U -31860555469490020,12585468136754891100,-1364794494618494224128,114095029934565534862680,-7984695190944325311086112,419424013080533747232201968

%N E.g.f. A(x) satisfies: A'(x) = A(x/A'(x)) with A(0)=1.

%C CONJECTURES.

%C a(n) == 0 (mod 2) for n>=4.

%C a(n) == 0 (mod 2^2) for n>=10.

%C a(n) == 0 (mod 2^3) for n>=18.

%C a(n) == 0 (mod 2^k) for n>=(8*n-6) for k>1.

%H Paul D. Hanna, <a href="/A232696/b232696.txt">Table of n, a(n) for n = 0..160</a>

%F E.g.f. satisfies: A(x) = A'(x*A(x)).

%F E.g.f. satisfies: A(x) = 1/x * Series_Reversion( x/A'(x) ).

%F a(n) = [x^n/n!] A'(x)^(n+1)/(n+1) for n>=0.

%e 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! +...

%e such that A(x) = A'(x*A(x)) and

%e A(x/A'(x)) = A'(x) = 1 + x - x^2/2! + 2*x^3/3! + 6*x^4/4! - 264*x^5/5! +...

%e 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:

%e A'^1: [1, 1, -1, 2, 6, -264, 5370, -93750, 1315706, ...];

%e A'^2: [1, 2, 0, -2, 34, -508, 7472, -100392, 774076, ...];

%e A'^3: [1, 3, 3, -6, 48, -522, 6036, -54030, -435618, ...];

%e A'^4: [1, 4, 8, -4, 36, -336, 2832, 3672, -1469680, ...];

%e A'^5: [1, 5, 15, 10, 10, -100, -130, 44490, -1964390, ...];

%e A'^6: [1, 6, 24, 42, 6, 36, -1680, 59520, -1938564, ...];

%e A'^7: [1, 7, 35, 98, 84, 42, -1848, 54978, -1605394, ...];

%e A'^8: [1, 8, 48, 184, 328, 128, -1504, 42960, -1194368, ...];

%e A'^9: [1, 9, 63, 306, 846, 864, -1278, 32202, -843750, ...]; ...

%e then the diagonal in the above table generates this sequence:

%e [1/1, 2/2, 3/3, -4/4, 10/5, 36/6, -1848/7, 42960/8, -843750/9, ...].

%e SUMS OF TERM RESIDUES MODULO 2^n.

%e 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.

%e Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:

%e [4, 16, 40, 144, 432, 1008, 3184, 6384, 15600, 33520, 75504, 159472, 356080, 798448, 1797872, 3895024, 8089328, 16609008, 37842672, 76639984, 166817520, ...].

%o (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)}

%o for(n=0, 25, print1(a(n), ", "))

%o (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)}

%o for(n=0, 25, print1(a(n), ", "))

%Y Cf. A231619, A231866, A231899, A232694, A232695.

%K sign

%O 0,5

%A _Paul D. Hanna_, Nov 28 2013

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Last modified April 24 22:17 EDT 2024. Contains 371964 sequences. (Running on oeis4.)