A138239 Triangle read by rows: T(n,k) = A000040(k) if A002445(n) mod A000040(k) = 0, otherwise 1.

1, 2, 3, 2, 3, 5, 2, 3, 1, 7, 2, 3, 5, 1, 1, 2, 3, 1, 1, 11, 1, 2, 3, 5, 7, 1, 13, 1, 2, 3, 1, 1, 1, 1, 1, 1, 2, 3, 5, 1, 1, 1, 17, 1, 1, 2, 3, 1, 7, 1, 1, 1, 19, 1, 1, 2, 3, 5, 1, 11, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 23, 1, 1, 1, 2, 3, 5, 7, 1, 13, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset

0,2

Comments

Row products give A002445.

A prime number appears in a column at every A130290-th row from the (A130290+1)-th row onwards. The prime numbers are, so to speak, equidistantly distributed in the columns. A130290 is essentially A005097. Counting terms > 1 in the rows gives A046886.

Links

Michel Marcus, Rows n=0..100 of triangle, flattened

Example

First few rows of the triangle and row products are:
1 = 1
2*3 = 6
2*3*5 = 30
2*3*1*7 = 42
2*3*5*1*1 = 30
2*3*1*1*11*1 = 66
2*3*5*7*1*13*1 = 2730

Maple

T:= (n, k)-> (p-> `if`(irem(denom(bernoulli(2*n)), p)=0, p, 1))(ithprime(k)):
seq(seq(T(n, k), k=1..n+1), n=0..20);  # Alois P. Heinz, Aug 27 2017

Mathematica

t[n_, k_] := If[Mod[Denominator[BernoulliB[2n]], (p = Prime[k])] == 0, p, 1];
Flatten[Table[t[n, k], {n, 0, 13}, {k, 1, n+1}]][[1 ;; 102]] (* Jean-François Alcover, Jun 16 2011 *)

Prog

(PARI) tabl(nn) = {for (n=0, nn, dbn = denominator(bernfrac(2*n)); for (k=1, n+1, if (! (dbn % prime(k)), w = prime(k), w = 1); print1(w, ", "); ); print; ); } \\ Michel Marcus, Aug 27 2017

Crossrefs

Cf. A002445, A005097, A046886, A130290.

Sequence in context: A345873 A114868 A249049 * A361330 A112484 A174063

Adjacent sequences: A138236 A138237 A138238 * A138240 A138241 A138242

Keyword

nonn,tabl

Author

Mats Granvik, Mar 07 2008

Extensions

Definition edited by N. J. A. Sloane, Mar 18 2010

Offset corrected by Alois P. Heinz, Aug 27 2017

Status

approved