A122793 Connell sum sequence (partial sums of the Connell sequence).
1, 3, 7, 12, 19, 28, 38, 50, 64, 80, 97, 116, 137, 160, 185, 211, 239, 269, 301, 335, 371, 408, 447, 488, 531, 576, 623, 672, 722, 774, 828, 884, 942, 1002, 1064, 1128, 1193, 1260, 1329, 1400, 1473, 1548, 1625, 1704, 1785, 1867, 1951, 2037, 2125, 2215, 2307, 2401, 2497, 2595, 2695, 2796, 2899, 3004, 3111, 3220
Offset: 1
Links
- Grady D. Bullington, The Connell Sum Sequence, J. Integer Seq. 10 (2007), Article 07.2.6. (includes direct formula for a(n))
- Ian Connell, Elementary Problem E1382, American Mathematical Monthly, v. 66, no. 8 (October, 1959), p. 724.
- Douglas E. Iannucci and Donna Mills-Taylor, On Generalizing the Connell Sequence, J. Integer Sequences, Vol. 2, 1999, #99.1.7.
Crossrefs
Programs
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Python
from math import isqrt def A122793(n): return n*(n+1)-(r:=(k:=isqrt(m:=n<<1))+int((m<<2)>(k<<2)*(k+1)+1))*((6*n+1)-r**2)//6 # Chai Wah Wu, Jul 26 2022
Formula
a(n) = (n-th triangular number) - n + (n-th partial sum of A122797).
a(n) = n^2 + n - R*((6*n+1)-R^2)/6, where R = round(sqrt(2*n)). - Gerald Hillier, Nov 29 2009
Comments