%I #4 Mar 30 2012 18:52:05

%S 1,2,1,3,1,1,5,1,1,1,1,7,1,1,1,1,1,1,11,1,1,1,1,1,1,1,1,1,1,13,1,1,1,

%T 1,1,1,1,1,1,1,1,1,17,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,19,1,1,1,1,1,1,

%U 1,1,1,1,1,1,1,1,1,1,1,1,23,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,29

%N Triangle T(n,m) read by rows: T(n,0)= prime(n); T(n,m)=1 if m>=1.

%C The sequence reflects a conjecture on the denominator of inverse Bernoulli polynomials in A178340: if the row index is one less than one of the primes in A008578, the row of denominators starts with that prime and contains 1's in the remaining entries.

%C [Row sums in A178252 are A159069(n+1), unless there is a common factor in numerator and denominator. The row sum over columns with index of the same parity as the row index in the table of fractions of the [x^m] B^{-1}(n,x) in A178252 are: 1, 1, 1/3+1=4/3, 1+1=2, 1/5+2+1=16/5, 1+10/3+1=16/3, 1/7+3+5+1=64/7, 16, 256/9, 256/5, 1024/11, 512/3, 496/13, ... =A084623(n+1)/A000265(n+1).]

%F T(n,0) = A008578(n+1). T(n,m) =1, 1<=m<=A008578(n+1)-1.

%e 1;

%e 2,1;

%e 3,1,1;

%e 5,1,1,1,1;

%e 7,1,1,1,1,1,1;

%e 11,1,1,1,1,1,1,1,1,1,1;

%e 13,1,1,1,1,1,1,1,1,1,1,1,1;

%e 17,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;

%e 19,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;

%e 23,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;

%e 29,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;

%Y Cf. A076274 (row sums).

%K nonn,tabf,easy,less

%O 0,2

%A _Paul Curtz_, May 31 2010