%I
%S 0,1,1,2,3,2,3,7,7,3,5,15,24,15,5,7,31,62,62,31,7,13,63,161,212,161,
%T 63,13,19,127,381,635,635,381,127,19,35,255,900,1785,2244,1785,900,
%U 255,35,59,511,2044,4774,7154,7154,4774,2044,511,59,107,1023,4619
%N CIK transform of Pascal's triangle A007318.
%C From _Petros Hadjicostas_, Dec 06 2017: (Start)
%C Denote by b(n,k) the number of this sequence (double array) in row n and column k.
%C The CIK transform of double array (C(n,k): n>=1, k>=0) = (binomial(n,k): n>=1, k>=0), which has bivariate g.f. A(x,y) = Sum_{n>=1, k>=0} C(n,k)*x^n*y^k = x*(1+y)/(1x*(1+y)), is given by CIK(A(x,y)) = Sum_{s>=1} (phi(s)/s)*log(1A(x^s,y^s)). (Unlike in sequence A055891, here, 1 is not added to the formula of the CIK transform. The addition of 1 seems to be arbitrary.)
%C To find the auxiliary double array (e(n,k): n>=1, k>=0) used in the formula b(n,k) = (1/n)*Sum_{dgcd(n,k)} phi(d)*e(n/d, k/d), we use the formula E(x,y) = Sum_{n>=1, k>=0} e(n,k)*x^n*y^k = x*(dA(x,y)/dx)/(1A(x,y)). We may find E(x,y) = x*(1+y)/((12x*(1+y))*(1x*(1+y)), from which we can easily prove that e(n,k) = C(n,k)*(2^n1).
%C Letting y=1 in the bivariate g.f. for b(n,k), we get the univariate g.f. for the row sums in A055891. (For this, ignore row 0 here and the 0th element in sequence A055891.)
%C (End)
%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F From _Petros Hadjicostas_, Dec 06 2017: (Start)
%F Let b(n,k) be the number in row n and column k. We have b(0,0) = 0 and b(n,k) = (1/n)*Sum_{dgcd(n,k)} phi(d)*C(n/d, k/d)*(2^{n/d}1) for n>=1 and k>=0. Here, C(n,k) = binomial(n,k).
%F G.f. for b(n,k): Sum_{n>=1, k>=0} b(n,k)*x^n*y^k = Sum_{s>=1} (phi(s)/s)*log((12*x^s*(1+y^s))/(1x^s*(1+y^s)).
%F G.f. for row n>=1: Sum_{k>=0} b(n,k)*y^k = (1/n)*Sum_{dn} phi(d)*(2^{n/d}1)*(1+y^d)^{n/d}.
%F G.f. for column k = 0: Sum_{n>=1} b(n,k=0)*x^n = Sum_{s>=1} (phi(s)/s)*log((1x^s)/(12*x^s)) = x/(1x)  Sum_{s>=1} (phi(s)/s)*log(12*x^s).
%F G.f. for column k >= 1: Sum_{n>=1} b(n,k)*x^n = Sum_{dk} (phi(d)/d)*(g_{k/d}(2*x^d)  g_{k/d}(x^d)), where g_k(x) = Sum_{s=0..k1} C(k1, s)*(1)^s/((ks)*(1x)^{ks}).
%F (End)
%e 0;
%e 1, 1;
%e 2, 3, 2;
%e 3, 7, 7, 3;
%e 5, 15, 24, 15, 5;
%e 7, 31, 62, 62, 31, 7;
%e 13, 63, 161, 212, 161, 63, 13;
%e 19, 127, 381, 635, 635, 381, 127, 19;
%e ...
%Y Row sums give A055891.
%K nonn,tabl
%O 0,4
%A _Christian G. Bower_, May 16 2000
