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A381361 E.g.f. satisfies A(x) = exp( Integral abs(1/A(x)^2) dx ), where abs(F(x)) equals the series expansion formed by the unsigned coefficients in F(x).

%I A381361 #23 May 04 2025 03:59:54
%S A381361 1,1,3,7,41,225,1435,11815,108945,1062145,12367475,154967175,
%T A381361 2052583225,30729193825,489419016075,8162349262375,149689173742625,
%U A381361 2891193923066625,58124967786683875,1262035025301370375,28655826402568055625,674057118324247590625,16913418325411880586875
%N A381361 E.g.f. satisfies A(x) = exp( Integral abs(1/A(x)^2) dx ), where abs(F(x)) equals the series expansion formed by the unsigned coefficients in F(x).
%C A381361 All terms appear to be odd.
%C A381361 Conjecture: for n > 4, a(n) == 0 (mod 5).
%H A381361 Paul D. Hanna, <a href="/A381361/b381361.txt">Table of n, a(n) for n = 0..500</a>
%F A381361 E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies the following formulas.
%F A381361 (1) A(x) = sqrt( 1 + 2*Series_Reversion( Integral 1/sqrt(1 + 8*x + 16*x^2 + 32*x^3/3) dx ) ).
%F A381361 (2) A(x) = sqrt( 1 + Integral 2*sqrt( (4*A(x)^6 - 1)/3 ) dx ).
%F A381361 (3) A(x) = sqrt( 1 + 2*x + Integral Integral 8*A(x)^4 dx dx ).
%F A381361 (4) A(x) = exp( x + Integral Integral (4*A(x)^6 + 2)/(3*A(x)^4) dx dx ).
%F A381361 (5) A(x)^4 = (A(x)^2)''/8.
%F A381361 (6.a) 0 = A'(x)^2 + A(x)*A''(x) - 4*A(x)^4.
%F A381361 (6.b) 0 = 1 + 3*A(x)^2*A'(x)^2 - 4*A(x)^6.
%F A381361 (6.c) 0 = 1 - 3*A(x)^3*A''(x) + 8*A(x)^6.
%F A381361 (6.d) 0 = 1 + 2*A(x)^2*A'(x)^2 - A(x)^3*A''(x).
%F A381361 Define the trisections of A(x) = T0 + T1 + T2 by
%F A381361 _ T0 = Sum_{n>=0} a(3*n)*x^(3*n)/(3*n)!,
%F A381361 _ T1 = Sum_{n>=0} a(3*n+1)*x^(3*n+1)/(3*n+1)!,
%F A381361 _ T2 = Sum_{n>=0} a(3*n+2)*x^(3*n+2)/(3*n+2)!,
%F A381361 then these series obey the following formulas.
%F A381361 (7.a) (T2^2 + 2*T0*T1)^2 = (T0^2 + 2*T1*T2) * (T1^2 + 2*T0*T2).
%F A381361 (7.b) (T2^2 - T0*T1)^2  = -2 * (T0^2 - T1*T2) * (T1^2 - T0*T2).
%F A381361 (8) A(x) = 1/sqrt(1 - 2*Integral (T0 - T1) * (A(x) - 3*T2) dx).
%F A381361 (9) A(x) = exp( x + 2*Integral Integral A(x)^2 - (T2^2 + 2*T0*T1) dx dx ).
%F A381361 (10.a) T0^2 + 2*T1*T2 = ((A(x)*A'(x) - 1)/A(x)^2)'/4.
%F A381361 (10.b) T2^2 + 2*T0*T1 = A(x)^2 - (A'(x)/A(x))'/2.
%F A381361 (10.c) T1^2 + 2*T0*T2 = ((A(x)*A'(x) + 1)/A(x)^2)'/4.
%F A381361 (11.a) T0^2 + 2*T1*T2 = A(x)^2/3 + (1 + 3*A(x)*A'(x))/(6*A(x)^4).
%F A381361 (11.b) T2^2 + 2*T0*T1 = A(x)^2/3 - 1/(3*A(x)^4).
%F A381361 (11.c) T1^2 + 2*T0*T2 = A(x)^2/3 + (1 - 3*A(x)*A'(x))/(6*A(x)^4).
%F A381361 (12) A(x) = B(2*x)^(1/2) where B(x) is the e.g.f. of A381360.
%e A381361 E.g.f: A(x) = 1 + x/1! + 3*x^2/2! + 7*x^3/3! + 41*x^4/4! + 225*x^5/5! + 1435*x^6/6! + 11815*x^7/7! + 108945*x^8/8! + 1062145*x^9/9! + 12367475*x^10/10! + ...
%e A381361 where A(x) = exp( Integral abs(1/A(x)^2) dx ).
%e A381361 RELATED SERIES.
%e A381361 Compare the signed coefficients of 1/A(x)^2 given by
%e A381361 1/A(x)^2 = 1 - 2*x + 16*x^3/3! - 64*x^4/4! + 2560*x^6/6! - 20480*x^7/7! + 1884160*x^9/9! - 21299200*x^10/10! + 3604480000*x^12/12! + ...
%e A381361 to the unsigned coefficients in log(A(x)) given by
%e A381361 log(A(x)) = x + 2*x^2/2! + 16*x^4/4! + 64*x^5/5! + 2560*x^7/7! + 20480*x^8/8! + 1884160*x^10/10! + 21299200*x^11/11! + 3604480000*x^13/13! + ...
%e A381361 to see that log(A(x)) = Integral abs(1/A(x)^2) dx.
%e A381361 Let B(x) be the e.g.f. of A381360, which starts as
%e A381361 B(x) = A(x) = 1 + x + 2*x^2/2! + 4*x^3/3! + 12*x^4/4! + 40*x^5/5! + 160*x^6/6! + 720*x^7/7! + 3680*x^8/8! + ... + A381360(n)*x^n/n! + ...
%e A381361 then A(x) = B(2*x)^(1/2).
%e A381361 SERIES TRISECTIONS.
%e A381361 The series trisections of A(x) = T0 + T1 + T2 begin
%e A381361 T0 = 1 + 7*x^3/3! + 1435*x^6/6! + 1062145*x^9/9! + 2052583225*x^12/12! + 8162349262375*x^15/15! + 58124967786683875*x^18/18! + ...
%e A381361 T1 = x + 41*x^4/4! + 11815*x^7/7! + 12367475*x^10/10! + 30729193825*x^13/13! + 149689173742625*x^16/16! + 1262035025301370375*x^19/19! + ...
%e A381361 T2 = 3*x^2/2! + 225*x^5/5! + 108945*x^8/8! + 154967175*x^11/11! + 489419016075*x^14/14! + 2891193923066625*x^17/17! + 28655826402568055625*x^20/20! + ...
%e A381361 where (T2^2 + 2*T0*T1)^2 = (T0^2 + 2*T1*T2) * (T1^2 + 2*T0*T2),
%e A381361 also, (T2^2 - T0*T1)^2  = -2 * (T0^2 - T1*T2) * (T1^2 - T0*T2).
%o A381361 (PARI) {a(n) = my(A = sqrt(1 + 2*serreverse( intformal( 1/sqrt(1 + 8*x + 16*x^2 + 32*x^3/3 + x*O(x^n))) ) )); n!*polcoef(A,n)}
%o A381361 for(n=0,25, print1(a(n),", "))
%Y A381361 Cf. A381360, A104133, A104134.
%K A381361 nonn
%O A381361 0,3
%A A381361 _Paul D. Hanna_, Feb 26 2025