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
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..
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
Comments