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%I #20 Feb 24 2019 17:45:30
%S 1,0,2,-3,2,0,2,-14,15,0,2,-22,2,0,84,-93,2,0,2,-38,172,0,2,-508,323,
%T 0,292,-54,2,0,2,-1212,444,0,2580,-1753,2,0,628,-2396,2,0,2,-86,8142,
%U 0,2,-18554,8991,0,1092,-102,2,0,16724,-6716,1372,0,2,-44844,2,0,81846,-58747,33284,0,2,-134,2028,0,2,-127882,2,0,62326,-150,268492,0,2,-468282,268541,0,2,-249196,100100,0,3252,-26748,2,0,738612,-182,3724,0,157780,-2117880,2,0,2517158,-1462761,2,0,2,-44508,2003576,0,2,-897068,2,0,5332,-4736762,2,0,344084,-230,9352622,0,3769924,-15721720,8097455
%N G.f.: Sum_{n>=0} x^n * (1 - x^(n+1))^n / (1 + x^(n+1))^(n+1).
%C Odd terms occur only at n = k*(k+2) for k >= 0 (conjecture);
%C a(n*(n+2)) = A323696(n).
%H Paul D. Hanna, <a href="/A323695/b323695.txt">Table of n, a(n) for n = 0..10201</a>
%F G.f.: Sum_{n>=0} x^n * (1 - x^(n+1))^n / (1 + x^(n+1))^(n+1).
%F G.f.: Sum_{n>=0} (-x)^n * (1 + x^(n+1))^n / (1 - x^(n+1))^(n+1).
%F G.f.: Sum_{n>=0} x^n * (1 - x^(n+1))^(2*n+1) / (1 - x^(2*n+2))^(n+1).
%F G.f.: Sum_{n>=0} (-x)^n * (1 + x^(n+1))^(2*n+1) / (1 - x^(2*n+2))^(n+1).
%F FORMULAS INVOLVING TERMS.
%F a(n*(n+2)) = 1 (mod 2) for n >= 0.
%F a(4*n+1) = 0 for n >= 0.
%F a(p-1) = 2 for odd primes p.
%e G.f.: A(x) = 1 + 2*x^2 - 3*x^3 + 2*x^4 + 2*x^6 - 14*x^7 + 15*x^8 + 2*x^10 - 22*x^11 + 2*x^12 + 84*x^14 - 93*x^15 + 2*x^16 + 2*x^18 - 38*x^19 + 172*x^20 + 2*x^22 - 508*x^23 + 323*x^24 + 292*x^26 - 54*x^27 + 2*x^28 + 2*x^30 - 1212*x^31 + 444*x^32 + 2580*x^34 - 1753*x^35 + ...
%e such that
%e A(x) = 1/(1+x) + x*(1-x^2)/(1+x^2)^2 + x^2*(1-x^3)^2/(1+x^3)^3 + x^3*(1-x^4)^3/(1+x^4)^4 + x^4*(1-x^5)^4/(1+x^5)^5 + x^5*(1-x^6)^5/(1+x^6)^6 + x^6*(1-x^7)^6/(1+x^7)^7 + x^7*(1-x^8)^7/(1+x^8)^8 + x^8*(1-x^9)^8/(1+x^9)^9 + ...
%e Also,
%e A(x) = 1/(1-x) - x*(1+x^2)/(1-x^2)^2 + x^2*(1+x^3)^2/(1-x^3)^3 - x^3*(1+x^4)^3/(1-x^4)^4 + x^4*(1+x^5)^4/(1-x^5)^5 - x^5*(1+x^6)^5/(1-x^6)^6 + x^6*(1+x^7)^6/(1-x^7)^7 - x^7*(1+x^8)^7/(1-x^8)^8 + x^8*(1+x^9)^8/(1-x^9)^9 + ...
%e TRIANGLE FORM.
%e This sequence may be written as a triangle like so
%e 1;
%e 0, 2, -3;
%e 2, 0, 2, -14, 15;
%e 0, 2, -22, 2, 0, 84, -93;
%e 2, 0, 2, -38, 172, 0, 2, -508, 323;
%e 0, 292, -54, 2, 0, 2, -1212, 444, 0, 2580, -1753;
%e 2, 0, 628, -2396, 2, 0, 2, -86, 8142, 0, 2, -18554, 8991;
%e 0, 1092, -102, 2, 0, 16724, -6716, 1372, 0, 2, -44844, 2, 0, 81846, -58747;
%e 33284, 0, 2, -134, 2028, 0, 2, -127882, 2, 0, 62326, -150, 268492, 0, 2, -468282, 268541;
%e 0, 2, -249196, 100100, 0, 3252, -26748, 2, 0, 738612, -182, 3724, 0, 157780, -2117880, 2, 0, 2517158, -1462761;
%e 2, 0, 2, -44508, 2003576, 0, 2, -897068, 2, 0, 5332, -4736762, 2, 0, 344084, -230, 9352622, 0, 3769924, -15721720, 8097455; ...
%e in which the right border consists of all the odd numbers in this sequence:
%e [1, -3, 15, -93, 323, -1753, 8991, -58747, 268541, -1462761, ..., A323696(n), ...].
%e RELATED INFINITE SERIES.
%e At x = 1/2, g.f. A(x) yields the following convergent infinite series:
%e A(1/2) = Sum_{n>=0} 1/2^n * (1 - 1/2^(n+1))^n / (1 + 1/2^(n+1))^(n+1), and
%e A(1/2) = Sum_{n>=0} (-1/2)^n * (1 + 1/2^(n+1))^n / (1 - 1/2^(n+1))^(n+1).
%e Equivalently,
%e A(1/2) = Sum_{n>=0} 2*(2^(n+1) - 1)^n / (2^(n+1) + 1)^(n+1), and
%e A(1/2) = Sum_{n>=0} (-1)^n * 2*(2^(n+1) + 1)^n / (2^(n+1) - 1)^(n+1) ;
%e written explicitly,
%e A(1/2) = 2/(2+1) + 2*(2^2-1)/(2^2+1)^2 + 2*(2^3-1)^2/(2^3+1)^3 + 2*(2^4-1)^3/(2^4+1)^4 + 2*(2^5-1)^4/(2^5+1)^5 + 2*(2^6-1)^5/(2^6+1)^6 + 2*(2^7-1)^6/(2^7+1)^7 + ...
%e also,
%e A(1/2) = 2/(2-1) - 2*(2^2+1)/(2^2-1)^2 + 2*(2^3+1)^2/(2^3-1)^3 - 2*(2^4+1)^3/(2^4-1)^4 + 2*(2^5+1)^4/(2^5-1)^5 - 2*(2^6+1)^5/(2^6-1)^6 + 2*(2^7+1)^6/(2^7-1)^7 + ...
%e where
%e A(1/2) = 1.22454925222028301885538820406968501668119591425752205551983366491...
%o (PARI) {a(n) = my(SUM = sum(m=0, n, x^m*(1 - x^(m+1) +x*O(x^n))^m / (1 + x^(m+1) +x*O(x^n))^(m+1) ) ); polcoeff(SUM, n)}
%o for(n=0, 120, print1(a(n), ", "))
%o (PARI) {a(n) = my(SUM = sum(m=0, n, (-x)^m*(1 + x^(m+1) +x*O(x^n))^m / (1 - x^(m+1) +x*O(x^n))^(m+1) ) ); polcoeff(SUM, n)}
%o for(n=0, 120, print1(a(n), ", "))
%Y Cf. A323696 (a(n*(n+2)).
%Y Cf. A323675, A324300, A323680.
%K sign
%O 0,3
%A _Paul D. Hanna_, Feb 23 2019
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