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

%I A139547 #13 Feb 16 2025 08:33:08
%S A139547 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 A139547 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 A139547 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 A139547 Triangle read by rows: T(n,k) = A003418(A010766).
%C A139547 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 A139547 I. Vardi, Computational Recreations in Mathematica. Addison-Wesley, Redwood City, CA, 1991, p. 155.
%H A139547 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/MangoldtFunction.html">Mangoldt Function.</a>.
%F A139547 From _Mats Granvik_, Jun 05 2016: (Start)
%F A139547 T(n,k)=A003418(floor(n/k)).
%F A139547 Recurrence involving log(n!):
%F A139547 Let s=1.
%F A139547 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 A139547 Recurrence involving the Riemann zeta function:
%F A139547 Let z = 1.
%F A139547 Let a = the series expansion of zeta(s) at z.
%F A139547 Let ss -> Infinity.
%F A139547 Let s = z + 1/ss.
%F A139547 Then T(n,k) is generated by the recurrence:
%F A139547 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 A139547 (End)
%e A139547 Triangle begins:
%e A139547 1;
%e A139547 2,1;
%e A139547 6,1,1;
%e A139547 12,2,1,1;
%e A139547 60,2,1,1,1;
%e A139547 60,6,2,1,1,1;
%e A139547 420,6,2,1,1,1,1;
%e A139547 840,12,2,2,1,1,1,1;
%e A139547 2520,12,6,2,1,1,1,1,1;
%e A139547 2520,60,6,2,2,1,1,1,1,1;
%e A139547 27720,60,6,2,2,1,1,1,1,1,1;
%e A139547 27720,60,12,6,2,2,1,1,1,1,1,1;
%e A139547 360360,60,12,6,2,2,1,1,1,1,1,1,1;
%e A139547 ...
%t A139547 nn = 13; a = Exp[Accumulate[MangoldtLambda[Range[nn]]]]; Flatten[Table[Table[a[[Floor[n/k]]], {k, 1, n}], {n, 1, nn}]][[1 ;; 89]]
%t A139547 (*As a limit of a recurrence*)
%t A139547 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 A139547 (* _Mats Granvik_, Jun 05 2016 *)
%Y A139547 Cf. A000142, A010766, A014963, A003418, A139550, A139552, A139554.
%K A139547 nonn,tabl
%O A139547 0,2
%A A139547 _Mats Granvik_, Apr 27 2008, May 07 2008
%E A139547 Edited by _Mats Granvik_, Jun 28 2009
%E A139547 Further edits from _N. J. A. Sloane_, Jul 03 2009