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).

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

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_{k-1}+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 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

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 [math.CO], 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(i-1, 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}] (* Jean-François 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[i-1]+n[i]), [2..#n]))))}, [1..199]) \\ M. F. Hasler, Nov 01 2019
(Python)
def A053392(n):
    if n < 10: return 0
    d = list(map(int, str(n)))
    return int("".join(str(d[i] + d[i+1]) for i in range(len(d)-1)))
print([A053392(n) for n in range(120)]) # Michael S. Branicky, Sep 04 2021

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