A010552 Multiply successively by 1 (once), 2 (twice), 3 (thrice), etc.
1, 1, 2, 4, 12, 36, 108, 432, 1728, 6912, 27648, 138240, 691200, 3456000, 17280000, 86400000, 518400000, 3110400000, 18662400000, 111974400000, 671846400000, 4031078400000, 28217548800000, 197522841600000, 1382659891200000, 9678619238400000, 67750334668800000
Offset: 0
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..300
Programs
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Maple
a:= proc(n) option remember; `if`(n=0, 1, a(n-1) * floor(sqrt(2*n)+1/2)) end: seq(a(n), n=0..30); # Alois P. Heinz, May 20 2013
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Mathematica
t = Table[n, {n, 9}, {n}] // Flatten; b[n_] := t[[n]]; a[0] = 1; a[n_] := a[n] = b[n] a[n-1]; Table[a[n], {n, 0, t // Length}] (* Jean-François Alcover, Feb 21 2016 *)
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PARI
lista(nn) = my(na = 1, i = 0, im = 1, list=List()); listput(list, na); while (i < nn, for (j=1, im, na *= im; listput(list, na); i++; ); im++; ); Vec(list); \\ Michel Marcus, May 20 2013; Jun 19 2025
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Python
from math import isqrt, comb, prod def A010552(n): return (a:=(m:=isqrt(k:=n<<1))+(k>m*(m+1)))**(n-comb(a,2))*prod(i**i for i in range(2,a)) # Chai Wah Wu, Jun 19 2025
Formula
a(0)=1, a(n) = a(n-1)*floor(sqrt(2*n)+1/2) for n>0. - Alois P. Heinz, May 20 2013