A055891 -id:A055891 - cp's OEIS frontend

A055376 CIK transform of Pascal's triangle A007318.

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, 63, 13, 19, 127, 381, 635, 635, 381, 127, 19, 35, 255, 900, 1785, 2244, 1785, 900, 255, 35, 59, 511, 2044, 4774, 7154, 7154, 4774, 2044, 511, 59, 107, 1023, 4619

Offset: 0

Christian G. Bower, May 16 2000

From Petros Hadjicostas, Dec 06 2017: (Start)
Denote by b(n,k) the number of this sequence (double array) in row n and column k.
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)/(1-x*(1+y)), is given by CIK(A(x,y)) = -Sum_{s>=1} (phi(s)/s)*log(1-A(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.)
To find the auxiliary double array (e(n,k): n>=1, k>=0) used in the formula b(n,k) = (1/n)*Sum_{d|gcd(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)/(1-A(x,y)). We may find E(x,y) = x*(1+y)/((1-2*x*(1+y))*(1-x*(1+y))), from which we can easily prove that e(n,k) = C(n,k)*(2^n-1).
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.)
(End)

			Triangle begins:
   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,  63,  13;
  19, 127, 381, 635, 635, 381, 127,  19;
  ...

C. G. Bower, Transforms (2)
Index entries for triangles and arrays related to Pascal's triangle

Row sums give A055891.

From Petros Hadjicostas, Dec 06 2017: (Start)
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_{d|gcd(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).
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((1-2*x^s*(1+y^s))/(1-x^s*(1+y^s)).
G.f. for row n>=1: Sum_{k>=0} b(n,k)*y^k = (1/n)*Sum_{d|n} phi(d)*(2^{n/d}-1)*(1+y^d)^{n/d}.
G.f. for column k = 0: Sum_{n>=1} b(n,k=0)*x^n = Sum_{s>=1} (phi(s)/s)*log((1-x^s)/(1-2*x^s)) = -x/(1-x) - Sum_{s>=1} (phi(s)/s)*log(1-2*x^s).
G.f. for column k >= 1: Sum_{n>=1} b(n,k)*x^n = Sum_{d|k} (phi(d)/d)*(g_{k/d}(2*x^d) - g_{k/d}(x^d)), where g_k(x) = Sum_{s=0..k-1} C(k-1, s)*(-1)^s/((k-s)*(1-x)^{k-s}).
(End)

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A055376 CIK transform of Pascal's triangle A007318.

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