%I #71 Sep 05 2021 02:31:06
%S 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,
%T 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,
%U 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
%N a(n) is the concatenation of the sums of every pair of consecutive digits of n (with a(n) = 0 for 0 <= n <= 9).
%C 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_{k-1}+d_k. - _N. J. A. Sloane_, Nov 01 2019
%C According to the Friedman link, 1496 is the smallest number whose trajectory increases without limit: see A328974
%C 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 1-or-2-digit 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
%D Eric Angelini, Posting to Sequence Fans Mailing List, Oct 31 2019.
%H Reinhard Zumkeller, <a href="/A053392/b053392.txt">Table of n, a(n) for n = 0..10000</a>
%H Lars Blomberg, <a href="https://erich-friedman.github.io/mathmagic/0200/PDSC.pdf">Notes (2011) on Problem of the Month (February 2000).</a>
%H Erich Friedman, <a href="https://erich-friedman.github.io/mathmagic/0200.html">Problem of the Month (February 2000).</a>
%H Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, <a href="https://arxiv.org/abs/2012.04625">Finding structure in sequences of real numbers via graph theory: a problem list</a>, arXiv:2012.04625 [math.CO], 2020.
%e For n=10 we have 10 -> 1+0 = 1, hence a(10)=1;
%e 987 -> 9+8.8+7 -> 17.15 -> 1715, so a(987)=1715.
%p read("transforms") :
%p A053392 := proc(n)
%p if n < 10 then
%p 0;
%p else
%p dgs := convert(n,base,10) ;
%p dgsL := [op(1,dgs)+op(2,dgs)] ;
%p for i from 3 to nops(dgs) do
%p dgsL := [op(i,dgs)+op(i-1,dgs),op(dgsL)] ;
%p end do:
%p digcatL(dgsL) ;
%p end if;
%p end proc: # _R. J. Mathar_, Nov 26 2019
%t a[n_] := Total /@ Transpose[{Most[id = IntegerDigits[n]], Rest[id]}] // IntegerDigits // Flatten // FromDigits; Table[a[n], {n, 0, 119}] (* _Jean-François Alcover_, Apr 05 2013 *)
%o (Haskell)
%o a053392 :: Integer -> Integer
%o a053392 n = if ys == "" then 0 else read ys where
%o ys = foldl (++) "" $ map show $ zipWith (+) (tail ds) ds
%o ds = (map (read . return) . show) n
%o -- _Reinhard Zumkeller_, Nov 26 2013
%o (PARI) apply( {A053392(n)=if(n>9,n=digits(n);eval(concat(apply(i->Str(n[i-1]+n[i]),[2..#n]))))}, [1..199]) \\ _M. F. Hasler_, Nov 01 2019
%o (Python)
%o def A053392(n):
%o if n < 10: return 0
%o d = list(map(int, str(n)))
%o return int("".join(str(d[i] + d[i+1]) for i in range(len(d)-1)))
%o print([A053392(n) for n in range(120)]) # _Michael S. Branicky_, Sep 04 2021
%Y Cf. A053393 (periodic points), A060630, A103117, A194429, A328973 (a(n)>n), A328974 (trajectory of 1496), A328975 (numbers that blow up).
%K nonn,easy,nice,base,look
%O 0,12
%A _N. J. A. Sloane_, Jan 07 2000