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A007618
a(n) = a(n-1) + sum of digits of a(n-1), a(1) = 5.
(Formerly M3792)
23
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
1,1
COMMENTS
a(n) = A004207(n+5) for n > 52. - Reinhard Zumkeller, Oct 14 2013
a(2) = 10 and a(590) = 10000 are the first two powers of 10 in this sequence; there are no others below a(19017393928) = 1000000000093. Conjecture: the sequence contains infinitely many powers of 10. - Charles R Greathouse IV, Mar 29 2022
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).
FORMULA
a(n) = A062028(a(n-1)) for n > 1. - Reinhard Zumkeller, Oct 14 2013
PROG
(Haskell)
a007618 n = a007618_list !! (n-1)
a007618_list = iterate a062028 5 -- Reinhard Zumkeller, Oct 14 2013
(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
CROSSREFS
KEYWORD
nonn,base,easy
STATUS
approved