A139547 Triangle read by rows: T(n,k) = A003418(A010766).

1, 2, 1, 6, 1, 1, 12, 2, 1, 1, 60, 2, 1, 1, 1, 60, 6, 2, 1, 1, 1, 420, 6, 2, 1, 1, 1, 1, 840, 12, 2, 2, 1, 1, 1, 1, 2520, 12, 6, 2, 1, 1, 1, 1, 1, 2520, 60, 6, 2, 2, 1, 1, 1, 1, 1, 27720, 60, 6, 2, 2, 1, 1, 1, 1, 1, 1, 27720, 60, 12, 6, 2, 2, 1, 1, 1, 1, 1, 1, 360360, 60, 12, 6, 2, 2, 1, 1, 1, 1, 1

Offset: 0

Author

Mats Granvik, Apr 27 2008, May 07 2008

Keywords

• nonn
• tabl

Comments

This triangle fits the formula of I. Vardi in the Mathworld link about the von Mangoldt function. That formula is the basis for Chebyshev's estimate for the number of primes.

Examples

			Triangle begins:
1;
2,1;
6,1,1;
12,2,1,1;
60,2,1,1,1;
60,6,2,1,1,1;
420,6,2,1,1,1,1;
840,12,2,2,1,1,1,1;
2520,12,6,2,1,1,1,1,1;
2520,60,6,2,2,1,1,1,1,1;
27720,60,6,2,2,1,1,1,1,1,1;
27720,60,12,6,2,2,1,1,1,1,1,1;
360360,60,12,6,2,2,1,1,1,1,1,1,1;
...
		

References

• I. Vardi, Computational Recreations in Mathematica. Addison-Wesley, Redwood City, CA, 1991, p. 155.

Links

• Eric Weisstein's World of Mathematics, Mangoldt Function..

Crossrefs

Cf. A000142, A010766, A014963, A003418, A139550, A139552, A139554.

Programs

Mathematica

nn = 13; a = Exp[Accumulate[MangoldtLambda[Range[nn]]]]; Flatten[Table[Table[a[[Floor[n/k]]], {k, 1, n}], {n, 1, nn}]][[1 ;; 89]]
(*As a limit of a recurrence*)
Clear[t, s, n, k, z, nn, ss, a, aa];(*z=1 corresponds to Zeta[1],z=2 corresponds to Zeta[2],z=ZetaZero[1] corresponds to Zeta[ZetaZero[1]],etc.*) z = 1; a = Normal[Series[Zeta[s], {s, z, 0}]]; ss = 10^40; s = N[z + 1/ss, 10^2]; nn = 13; t[n_, k_] := t[n, k] = If[k == 1, n*Zeta[s] - Sum[t[n, i]/i^(s - 1), {i, 2, n}], If[n >= k, t[Floor[n/k], 1], 0], 0]; aa = Table[Table[If[n >= k, t[n, k] - a, 0], {k, 1, n}], {n, 1, nn}]; Flatten[Round[Exp[aa]]][[1 ;; 89]]
(* Mats Granvik, Jun 05 2016 *)

Formula

From Mats Granvik, Jun 05 2016: (Start)

T(n,k)=A003418(floor(n/k)).

Recurrence involving log(n!):

Let s=1.

T(n, k) = if k = 1 then log(n!) - Sum_{i=2..n} T(n, i)/i^(s - 1) else if n >= k then T(floor(n/k), 1) else 0 else 0.

Recurrence involving the Riemann zeta function:

Let z = 1.

Let a = the series expansion of zeta(s) at z.

Let ss -> Infinity.

Let s = z + 1/ss.

Then T(n,k) is generated by the recurrence:

a + Ts(n, k) = if k = 1 then n*zeta(s) - Sum_{i=2..n} Ts(n, i)/i^(s - 1) else if n >= k then Ts(floor(n/k), 1) else 0 else 0.

(End)

Extensions

Edited by Mats Granvik, Jun 28 2009

Further edits from N. J. A. Sloane, Jul 03 2009