1988
DOI: 10.2307/2323062
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The Determination of Kaprekar Convergence and Loop Convergence of All Three-Digit Numbers

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Cited by 5 publications
(4 citation statements)
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“…Next, Klaus E. Eldridge and Seok Sagong in their paper [11] from 1988 described the convergence of {T n 3 (x)} ∞ n=1 for all three-digit numbers x for any base r ∈ N, r ≥ 2. They obtained, among others, the following result.…”
Section: Kaprekar's Transformationsmentioning
confidence: 99%
See 1 more Smart Citation
“…Next, Klaus E. Eldridge and Seok Sagong in their paper [11] from 1988 described the convergence of {T n 3 (x)} ∞ n=1 for all three-digit numbers x for any base r ∈ N, r ≥ 2. They obtained, among others, the following result.…”
Section: Kaprekar's Transformationsmentioning
confidence: 99%
“…Formulae (11) and (12) describe the following 45 numbers 111 × 9, 121 × 9, 222 × 9, 232 × 9, 242 × 9, 333 × 9, 343 × 9, 353 × 9, 363 × 9, 444 × 9, 454 × 9, . .…”
mentioning
confidence: 99%
“…In septenary, since 6 2 ×3=108=213 (7) , so calculating 48 positive integers within 108, in which a digit in high place is not greater than that in a low place, we know the iteration has five fixed points: 1, 10=13 (7) , 25=34 (7) ,32=44 (7) ,45=63 (7) , and two 4-circles (2,4, 16, 8)=(2, 4, 22, 11) (7) , (13, 37, 29, 17)=(16, 52, 41, 23) (7) . Proportions for converging to seven cases are about 1/48, 11/48, 3/16, 1/48, 1/8, 1/4, 1/6 respectively.…”
Section: Iteration Of Sum Of Squares Of Digitsmentioning
confidence: 99%
“…In septenary, since 6 3 ×4=864=2343 (7) , so computing 158 positive integers which their a digit in higher position is less or equal than a lower one within 864, we know the iteration has seven fixed points: 1, 9=12 (7) , 16=22 (7) , 133=250 (7) , 134=251 (7) , 152 = 305 (7) , 250 = 505 (7) , four 2-circles (2,8)=(2, 11) (7) , (91, 217)=(160, 430) (7) , (92, 218)=(2, 11) (7) , (244, 496)=(466, 1306) (7) , two 3-circles (10, 28, 64)=(13, 40, 121) (7) , (36, 126, 72)=(51, 240, 132) (7) , one 4-circles (258, 342, 648, 282)= (516, 666, 1614, 552) (7) , and one 9-circles ( 17, 35, 125, 251, 341, 557, 137, 197, 65)=(23, 50, 236, 506, 665, 1424, 254, 401, 122) (7) In octal, since 7 3 ×4=1372=2534 (8) , so computing 249 positive integers which their a digit in higher position is less or equal than a lower one within 1372, we know the iteration has six fixed points: 1, 92=134 (8) , 133=205 (8) , 307=463 (8) , 432=660 (8) , 433=661 (8) , and one 5-circle (434, 440, 559, 469, 476)=(662, 670, 1057, 725, 734) (8) . Proportions for converging to seven cases are about 59/249, 9/83, 4/249, 1/83, 5/83, 5/249, 136/249 respectively.…”
Section: Iteration Of Sum Of Cubes Of Digitsmentioning
confidence: 99%