A088493 a(n) = Sum_{k=1..8} floor(p(n, k)/p(n-1, k)), where p(n, k) = n!/( Product_{i=1..floor(n/2^k)} A004001(i) ).

16, 24, 32, 40, 45, 56, 60, 72, 73, 88, 81, 104, 101, 120, 108, 136, 129, 152, 129, 168, 157, 184, 141, 200, 185, 216, 178, 232, 213, 248, 188, 264, 241, 280, 226, 296, 269, 312, 222, 328, 297, 344, 273, 360, 325, 376, 237, 392, 353, 408, 321, 424, 381, 440

Offset

2,1

Links

G. C. Greubel, Table of n, a(n) for n = 2..5000

Formula

a(n) = Sum_{k=1..8} floor(p(n, k)/p(n-1, k)), where p(n, k) = n!/( Product_{i=1..floor(n/2^k)} A004001(i) ).

Mathematica

Conway[n_]:= Conway[n]= If[n<3, 1, Conway[Conway[n-1]] +Conway[n-Conway[n-1]]];
f[n_, k_]:= f[n, k]= Product[Conway[i], {i, Floor[n/2^k]}];
a[n_]:= a[n]= Sum[Floor[n*f[n-1, k]/f[n, k]], {k, 8}];
Table[a[n], {n, 2, 70}] (* modified by G. C. Greubel, Mar 27 2022 *)

Prog

(Sage)
@CachedFunction
def b(n): # A004001
    if (n<3): return 1
    else: return b(b(n-1)) + b(n-b(n-1))
def f(n, k): return product( b(j) for j in (1..(n//2^k)) )
def A088493(n): return sum( (n*f(n-1, k)//f(n, k)) for k in (1..8) )
[A088493(n) for n in (2..70)] # G. C. Greubel, Mar 27 2022

Crossrefs

Cf. A004001, A005185, A005229, A088491.

Sequence in context: A033987 A140349 A369353 * A074451 A253782 A247065

Adjacent sequences: A088490 A088491 A088492 * A088494 A088495 A088496

Keyword

nonn

Author

Roger L. Bagula, Nov 10 2003

Extensions

Edited by G. C. Greubel, Mar 27 2022

Status

approved