[math-fun] penultimate divisor counts output
hihi, all - yes, well, there were some typos - now the program counts them, so i don't have to there will be a few more numbers tomorrow (hopefully, n=37 will appear), but for now, here is what i have 1 chains of length 1, 1 anchored, 1 cyclic, 1 both 1 chains of length 2, 1 anchored, 0 cyclic, 0 both 2 chains of length 3, 1 anchored, 2 cyclic, 1 both 2 chains of length 4, 1 anchored, 0 cyclic, 0 both 4 chains of length 5, 1 anchored, 2 cyclic, 1 both 5 chains of length 6, 1 anchored, 0 cyclic, 0 both 7 chains of length 7, 1 anchored, 3 cyclic, 1 both 7 chains of length 8, 5 anchored, 0 cyclic, 0 both 24 chains of length 9, 4 anchored, 5 cyclic, 4 both 22 chains of length 10, 3 anchored, 0 cyclic, 0 both 29 chains of length 11, 2 anchored, 6 cyclic, 2 both 39 chains of length 12, 8 anchored, 0 cyclic, 0 both 67 chains of length 13, 4 anchored, 6 cyclic, 4 both 55 chains of length 14, 6 anchored, 0 cyclic, 0 both 386 chains of length 15, 47 anchored, 147 cyclic, 47 both 235 chains of length 16, 44 anchored, 1 cyclic, 0 both 312 chains of length 17, 6 anchored, 22 cyclic, 6 both 347 chains of length 18, 37 anchored, 2 cyclic, 0 both 451 chains of length 19, 6 anchored, 27 cyclic, 6 both 1319 chains of length 20, 166 anchored, 165 cyclic, 0 both 5320 chains of length 21, 462 anchored, 519 cyclic, 462 both 3220 chains of length 22, 232 anchored, 0 cyclic, 0 both 4489 chains of length 23, 372 anchored, 516 cyclic, 372 both 20237 chains of length 24, 2130 anchored, 2021 cyclic, 0 both 36580 chains of length 25, 1589 anchored, 1912 cyclic, 1589 both 52875 chains of length 26, 9093 anchored, 506 cyclic, 0 both 197103 chains of length 27, 20896 anchored, 45658 cyclic, 20896 both 216562 chains of length 28, 20314 anchored, 514 cyclic, 0 both (for a sequence to be both cyclic and anchored, the first element must both divide n*(n+1)/2 and be equal to n, so n must be odd) one more tomorrow, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation cal@aero.org
participants (1)
-
Chris Landauer