A007618 a(n) = a(n-1) + sum of digits of a(n-1), a(1) = 5.
5, 10, 11, 13, 17, 25, 32, 37, 47, 58, 71, 79, 95, 109, 119, 130, 134, 142, 149, 163, 173, 184, 197, 214, 221, 226, 236, 247, 260, 268, 284, 298, 317, 328, 341, 349, 365, 379, 398, 418, 431, 439, 455, 469, 488, 508, 521, 529, 545, 559, 578, 598, 620, 628, 644
Offset: 1
References
- N. Agronomof, Problem 4421, L'Intermédiaire des mathématiciens, v. 21 (1914), p. 147. (Mentions sequence starting at 11.) - N. J. A. Sloane, Nov 22 2013.
- D. R. Kaprekar, Puzzles of the Self-Numbers. 311 Devlali Camp, Devlali, India, 1959.
- D. R. Kaprekar, The Mathematics of the New Self Numbers, Privately Printed, 311 Devlali Camp, Devlali, India, 1963.
- J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 65.
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- D. R. Kaprekar, The Mathematics of the New Self Numbers [annotated and scanned]
- Index entries for Colombian or self numbers and related sequences
Programs
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Haskell
a007618 n = a007618_list !! (n-1) a007618_list = iterate a062028 5 -- Reinhard Zumkeller, Oct 14 2013
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Python
from itertools import accumulate def f(an, _): return an + sum(int(d) for d in str(an)) print(list(accumulate([5]*55, f))) # Michael S. Branicky, May 10 2021
Formula
a(n) = A062028(a(n-1)) for n > 1. - Reinhard Zumkeller, Oct 14 2013
Comments