A286146 Lower triangular region of square array A286101.
1, 2, 5, 4, 16, 13, 7, 12, 67, 25, 11, 46, 106, 191, 41, 16, 23, 31, 80, 436, 61, 22, 92, 211, 379, 596, 862, 85, 29, 38, 277, 59, 781, 302, 1541, 113, 37, 154, 58, 631, 991, 193, 1954, 2557, 145, 46, 57, 436, 212, 96, 467, 2416, 822, 4006, 181, 56, 232, 529, 947, 1486, 2146, 2927, 3829, 4852, 5996, 221, 67, 80, 94, 109, 1771, 142, 3487, 355, 706, 1832, 8647
Offset: 1
Examples
The first twelve rows of the triangle: 1, 2, 5, 4, 16, 13, 7, 12, 67, 25, 11, 46, 106, 191, 41, 16, 23, 31, 80, 436, 61, 22, 92, 211, 379, 596, 862, 85, 29, 38, 277, 59, 781, 302, 1541, 113, 37, 154, 58, 631, 991, 193, 1954, 2557, 145, 46, 57, 436, 212, 96, 467, 2416, 822, 4006, 181, 56, 232, 529, 947, 1486, 2146, 2927, 3829, 4852, 5996, 221, 67, 80, 94, 109, 1771, 142, 3487, 355, 706, 1832, 8647, 265 ---------------------------------------------------------------- For T(4,3) we have gcd(4,3) = 1 and lcm(4,3) = 12, thus T(4,3) = (1/2)*(2 + (12+1)^2 - 1 - 3*12) = 67. For T(6,4) we have gcd(6,4) = 2 and lcm(6,4) = 12, thus T(6,4) = (1/2)*(2 + (12+2)^2 - 2 - 3*12) = 80. For T(12,1) we have gcd(12,1) = 1 and lcm(12,1) = 12, thus T(12,1) = T(4,3) = 67. For T(12,2) we have gcd(12,2) = 2 and lcm(12,1) = 12, thus T(12,1) = T(6,4) = 80. For T(12,8) we have gcd(12,8) = 4 and lcm(12,8) = 24, thus T(12,8) = (1/2)*(2 + (24+4)^2 - 4 - 3*24) = 355.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10585; the first 145 rows of triangle
Crossrefs
Programs
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Python
from sympy import lcm, gcd def t(n, k): return (2 + ((gcd(n, k) + lcm(n, k))**2) - gcd(n, k) - 3*lcm(n, k))/2 for n in range(1, 21): print([t(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, May 11 2017
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Scheme
(define (A286146 n) (A286101bi (A002024 n) (A002260 n))) ;; For A286101bi see A286101.
Formula
As a triangle (with n >= 1, 1 <= k <= n):
T(n,k) = (1/2)*(2 + ((gcd(n,k)+lcm(n,k))^2) - gcd(n,k) - 3*lcm(n,k)).