

A053392


a(n) is the concatenation of the sums of every pair of consecutive digits of n (with a(n) = 0 for 0 <= n <= 9).


21



0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 210
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OFFSET

0,12


COMMENTS

Let the decimal expansion of n be d_1 d_2 ... d_k; a(n) is formed by concatenating the decimal numbers d_1+d_2, d_2+d_3, ..., d_{k1}+d_k.  N. J. A. Sloane, Nov 01 2019
According to the Friedman link, 1496 is the smallest number whose trajectory increases without limit: see A328974
The Blomberg link asks whether a number n can have more than 10 predecessors, i.e. values of p such that a(p) = n. The answer is no, because there can be at most one predecessor ending with any given digit d. That can be proved by induction by observing that d and the last digit of n determine the last 1or2digit sum in the concatenation of sums forming n, and hence the penultimate digit of p. That is either incompatible with the known value of n, or it tells us what the last digit of p' is, where p' = p with its last digit removed. We also know that p' is the predecessor of n' = n with its known last digit sum removed, and so we know there is at most one solution for p' by inductive hypothesis, and hence at most one solution for p.  David J. Seal, Nov 06 2019


REFERENCES

Eric Angelini, Posting to Sequence Fans Mailing List, Oct 31 2019.


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Lars Blomberg, Notes (2011) on Problem of the Month (February 2000).
Erich Friedman, Problem of the Month (February 2000).
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625, Dec 08, 2020


EXAMPLE

For n=10 we have 10 > 1+0 = 1, hence a(10)=1;
987 > 9+8.8+7 > 17.15 > 1715, so a(987)=1715.


MAPLE

read("transforms") :
A053392 := proc(n)
if n < 10 then
0;
else
dgs := convert(n, base, 10) ;
dgsL := [op(1, dgs)+op(2, dgs)] ;
for i from 3 to nops(dgs) do
dgsL := [op(i, dgs)+op(i1, dgs), op(dgsL)] ;
end do:
digcatL(dgsL) ;
end if;
end proc: # R. J. Mathar, Nov 26 2019


MATHEMATICA

a[n_] := Total /@ Transpose[{Most[id = IntegerDigits[n]], Rest[id]}] // IntegerDigits // Flatten // FromDigits; Table[a[n], {n, 0, 119}] (* JeanFrançois Alcover, Apr 05 2013 *)


PROG

(Haskell)
a053392 :: Integer > Integer
a053392 n = if ys == "" then 0 else read ys where
ys = foldl (++) "" $ map show $ zipWith (+) (tail ds) ds
ds = (map (read . return) . show) n
 Reinhard Zumkeller, Nov 26 2013
(PARI) apply( {A053392(n)=if(n>9, n=digits(n); eval(concat(apply(i>Str(n[i1]+n[i]), [2..#n]))))}, [1..199]) \\ M. F. Hasler, Nov 01 2019


CROSSREFS

Cf. A053393 (periodic points), A060630, A103117, A194429, A328973 (a(n)>n), A328974 (trajectory of 1496), A328975 (numbers that blow up).
Sequence in context: A093882 A138953 A307629 * A114925 A043270 A089898
Adjacent sequences: A053389 A053390 A053391 * A053393 A053394 A053395


KEYWORD

nonn,easy,nice,base,look


AUTHOR

N. J. A. Sloane, Jan 07 2000


STATUS

approved



