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A178623
Triangle T(n,m) read by rows: T(n,0)= prime(n); T(n,m)=1 if m>=1.
0
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, 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, 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
OFFSET
0,2
COMMENTS
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.
[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).]
FORMULA
T(n,0) = A008578(n+1). T(n,m) =1, 1<=m<=A008578(n+1)-1.
EXAMPLE
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;
13,1,1,1,1,1,1,1,1,1,1,1,1;
17,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
19,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
23,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
29,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
CROSSREFS
Cf. A076274 (row sums).
Sequence in context: A353430 A353391 A141412 * A210765 A362372 A160183
KEYWORD
nonn,tabf,easy,less
AUTHOR
Paul Curtz, May 31 2010
STATUS
approved