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%I #12 Jun 27 2016 12:10:39
%S 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,
%T 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,
%U 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
%N Triangle read by rows: T(n,k) = A003418(A010766).
%C 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.
%D I. Vardi, Computational Recreations in Mathematica. Addison-Wesley, Redwood City, CA, 1991, p. 155.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MangoldtFunction.html">Mangoldt Function.</a>.
%F From _Mats Granvik_, Jun 05 2016: (Start)
%F T(n,k)=A003418(floor(n/k)).
%F Recurrence involving log(n!):
%F Let s=1.
%F 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.
%F Recurrence involving the Riemann zeta function:
%F Let z = 1.
%F Let a = the series expansion of zeta(s) at z.
%F Let ss -> Infinity.
%F Let s = z + 1/ss.
%F Then T(n,k) is generated by the recurrence:
%F 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.
%F (End)
%e Triangle begins:
%e 1;
%e 2,1;
%e 6,1,1;
%e 12,2,1,1;
%e 60,2,1,1,1;
%e 60,6,2,1,1,1;
%e 420,6,2,1,1,1,1;
%e 840,12,2,2,1,1,1,1;
%e 2520,12,6,2,1,1,1,1,1;
%e 2520,60,6,2,2,1,1,1,1,1;
%e 27720,60,6,2,2,1,1,1,1,1,1;
%e 27720,60,12,6,2,2,1,1,1,1,1,1;
%e 360360,60,12,6,2,2,1,1,1,1,1,1,1;
%e ...
%t nn = 13; a = Exp[Accumulate[MangoldtLambda[Range[nn]]]]; Flatten[Table[Table[a[[Floor[n/k]]], {k, 1, n}], {n, 1, nn}]][[1 ;; 89]]
%t (*As a limit of a recurrence*)
%t 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]]
%t (* _Mats Granvik_, Jun 05 2016 *)
%Y Cf. A000142, A010766, A014963, A003418, A139550, A139552, A139554.
%K nonn,tabl
%O 0,2
%A _Mats Granvik_, Apr 27 2008, May 07 2008
%E Edited by _Mats Granvik_, Jun 28 2009
%E Further edits from _N. J. A. Sloane_, Jul 03 2009
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