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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A055376 CIK transform of Pascal's triangle A007318.

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%I A055376 #40 Apr 19 2025 19:37:02
%S A055376 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 A055376 63,13,19,127,381,635,635,381,127,19,35,255,900,1785,2244,1785,900,
%U A055376 255,35,59,511,2044,4774,7154,7154,4774,2044,511,59,107,1023,4619
%N A055376 CIK transform of Pascal's triangle A007318.
%C A055376 From _Petros Hadjicostas_, Dec 06 2017: (Start)
%C A055376 Denote by b(n,k) the number of this sequence (double array) in row n and column k.
%C A055376 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.)
%C A055376 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).
%C A055376 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 A055376 (End)
%H A055376 C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%H A055376 <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F A055376 From _Petros Hadjicostas_, Dec 06 2017: (Start)
%F A055376 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).
%F A055376 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)).
%F A055376 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}.
%F A055376 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).
%F A055376 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}).
%F A055376 (End)
%e A055376 Triangle begins:
%e A055376    0;
%e A055376    1,   1;
%e A055376    2,   3,   2;
%e A055376    3,   7,   7,   3;
%e A055376    5,  15,  24,  15,   5;
%e A055376    7,  31,  62,  62,  31,   7;
%e A055376   13,  63, 161, 212, 161,  63,  13;
%e A055376   19, 127, 381, 635, 635, 381, 127,  19;
%e A055376   ...
%Y A055376 Row sums give A055891.
%K A055376 nonn,tabl
%O A055376 0,4
%A A055376 _Christian G. Bower_, May 16 2000