A328974 -id:A328974 - cp's OEIS frontend

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

N. J. A. Sloane, Jan 07 2000

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

			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.

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

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.

Cf. A053393 (periodic points), A060630, A103117, A194429, A328973 (a(n)>n), A328974 (trajectory of 1496), A328975 (numbers that blow up).

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

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

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 *)

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

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

A328975 Numbers whose trajectory under repeated application of the map in A053392 increases without limit.

Original entry on oeis.org

1496, 1497, 1498, 1499, 1587, 1588, 1589, 1691, 1692, 1693, 1694, 1695, 1696, 1697, 1698, 1699, 1719, 1728, 1729, 1783, 1784, 1785, 1786, 1787, 1788, 1789, 1791, 1792, 1793, 1794, 1795, 1796, 1797, 1798, 1799, 1819, 1867, 1868, 1869, 1874, 1875, 1876, 1877, 1878, 1879, 1880, 1881, 1882, 1883, 1884, 1885

Offset: 1

N. J. A. Sloane, Nov 02 2019

Computed by Hans Havermann, Nov 01 2019.
There is a simple technique, due to Hans Havermann, that proves that many terms that in this sequence blow up: find a term in the trajectory that has an internal substring of three 9's. See A328974 for the case 1496.
The same reasoning shows that almost all numbers belong to the sequence.
From Scott R. Shannon, Nov 23 2019: (Start)
Although many of the numbers in this sequence eventually reach a term that contains three 9's, some do not. The first such example is 4949 which leads to 44444 after two steps, and starts a sequence of values that leads back to a much larger all-4's number every nine steps. No 9's appear in any of the intermediate values. Similar series found which lead to limitless loops are values with all 4's with a final 5, or all 3's with a final digit of 1 to 6. Of the first 33909 starting values that increase without limit 209 lead into one of these non-9 loops. (End)
From Scott R. Shannon, Nov 24 2019: (Start)
One can show that strings of repeated digits longer than a certain length will increase without limit by calculating the number of digits when a new value containing only the same digit appears again in the iterative sequence. Below shows the details of this for digits 1 to 9. The number of cycles is the number of iterations of A053392 required before a new value containing only the starting digit is seen. The minimum starting value is the smallest same-digit number such that subsequent iterations will produce another value with only the same digit of equal or longer length. The number of digits in the subsequence reoccurrence shows the number of digits in this value given the starting value has n digits. All sufficiently long single-digit numbers, with digits 1 to 8, increase 8-fold in length, minus a constant, after 9 iterations.
.
digit | # of cycles | min start value | # digits in reoccurrence
  1   |      9      |     1111111     |      8*n - 45
  2   |      9      |      222222     |      8*n - 38
  3   |      3      |       33333     |      2*n -  5
  4   |      9      |       44444     |      8*n - 31
  5   |      9      |       55555     |      8*n - 33
  6   |      3      |        6666     |      2*n -  4
  7   |      9      |       77777     |      8*n - 27
  8   |      9      |        8888     |      8*n - 28
  9   |      2      |         999     |      2*n -  3
(End)

Scott R. Shannon, Table of n, a(n) for n = 1..33909 (terms 1..109 from Hans Havermann).
Hans Havermann, Proof of correctness of first 109 terms

Cf. A053392, A328974.

cp's OEIS Frontend

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).

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