A253902 Write numbers 1, then 2^2 down to 1, then 3^2 down to 1, then 4^2 down to 1 and so on.
1, 4, 3, 2, 1, 9, 8, 7, 6, 5, 4, 3, 2, 1, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
Offset: 1
Examples
From _Omar E. Pol_, Jan 20 2015: (Start) Written as an irregular triangle in which row lengths are successive squares, the sequence begins: 1; 4, 3, 2, 1; 9, 8, 7, 6, 5, 4, 3, 2, 1; 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1; ... (End)
Programs
-
Maple
T:= n-> (t-> seq(t-i, i=0..t-1))(n^2): seq(T(n), n=1..6); # Alois P. Heinz, Nov 05 2024
-
Mathematica
a253902[n_] := Flatten@ Table[Reverse[Range[i^2]], {i, n}]; a253902[6] (* Michael De Vlieger, Jan 19 2015 *)
-
PARI
lista(nn=10) = {for (n=1, nn, forstep(k=n^2, 1, -1, print1(k, ", ");););} \\ Michel Marcus, Jan 20 2015
-
Python
from sympy import integer_nthroot def A253902(n): return (k:=(m:=integer_nthroot(3*n,3)[0])+(6*n>m*(m+1)*((m<<1)+1))+2)*(k*((k<<1)-9)+13)//6-n # Chai Wah Wu, Nov 05 2024
Formula
For 1 <= n <= 650, a(n) = -n + (t + 2)*(2*t^2 - t + 3)/6, where t = floor((3*n)^(1/3)+1/2).
a(n) = k*(2*k^2-9*k+13)/6-n where k = floor((3*n)^(1/3))+3 if n>A000330(floor((3*n)^(1/3))) and k = floor((3*n)^(1/3))+2 otherwise. - Chai Wah Wu, Nov 05 2024
Comments