OFFSET
1,2
COMMENTS
Write the natural numbers in groups, with group sizes incremented by one, each time: 1; 2,3; 4,5,6; 7,8,9,10; ... and add the numbers in each group, then take the i-th power for the i-th group to get the i-th entry in the sequence.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..158
FORMULA
a(n) = A006003(n)^n.
a(n) = ((n + n^3)/2)^n. - Harvey P. Dale, Apr 24 2022
MATHEMATICA
Module[{nn=15}, #[[1]]^#[[2]]&/@Thread[{Total/@TakeList[Range[(nn(nn+1))/2], Range[ nn]], Range[nn]}]] (* or *) Table[((n+n^3)/2)^n, {n, 20}] (* Harvey P. Dale, Apr 24 2022 *)
PROG
(Python)
num = 100
n = 0
a = range(num+1)
for i in range(1, num):
....sum = 0
....for j in range(1, i+1):
........sum = sum + (n+j)
....n = n + i
....a[i] = sum**i
CROSSREFS
KEYWORD
nonn
AUTHOR
Amir H. Farrahi, Feb 05 2011
EXTENSIONS
Edited by N. J. A. Sloane, Feb 05 2011
STATUS
approved