A128546 - OEIS

%I #12 Nov 09 2024 19:31:36

%S 17,21,25,42,63,84,143,286,2355,5821,6618,11709,12482,33747,39571,

%T 129109,466957,1162248,1565166,1968084,3636638,3853951,4898376,

%U 6065280,13443745,13933175,17118698,22421197,24153462377

%N Inrepfigit (INverse REPetitive FIbonacci-like diGIT) numbers (or Htiek numbers).

%C This sequence is similar to A007629 (Keith numbers). It consists of the numbers n>9 with the following property: n is a term of the sequence S whose first k terms are the k digits of n (with the first term equal to the units digit) and with S(n+1)=sum of the k previous terms.

%e 42 is in the sequence because the terms of the sequence it creates are 2, 4, 6, 10, 16, 26, 42, ...

%t iKeithQ[n_Integer] := Module[{b = Reverse[IntegerDigits[n]], s, k = 0}, s = Total[b]; While[s < n, AppendTo[b, s]; k++; s = 2*s - b[[k]]]; s == n]; Select[Range[10, 100000], iKeithQ] (* _T. D. Noe_, Mar 15 2011 *)

%o (C++)

%o #include <stdio.h>

%o // Here is a (messy) C++ code which finds the terms of the sequence below 100000000

%o int main() {

%o   int k2;

%o   for (int k = 10; k < 100000000; k++) {

%o     k2 = k;

%o     int array[9];

%o     for (int i = 0; i <= 8; i++) {

%o       array[i] = k2 % 10;

%o       k2 /= 10;

%o     }

%o     bool c = true;

%o     int check = 8;

%o     for (int i = 0; i <= 8; i++) {

%o       if ((array[8 - i] == 0) && c)

%o         check--;

%o       else

%o         c = false;

%o     }

%o     bool b = false;

%o     int n = 0;

%o     while (n <= k && !b) {

%o       n = 0;

%o       for (int i = 0; i <= check; i++)

%o         n += array[i];

%o       if (n == k)

%o         b = true;

%o       for (int i = 0; i < check; i++)

%o         array[i] = array[i + 1];

%o       array[check] = n;

%o     }

%o     if (b)

%o       printf("%d", k);

%o   }

%o   return 0;

%o }

%Y Cf. A007629.

%Y Cf. A097060 (reverse of these numbers).

%K base,nonn

%O 1,1

%A Pierre Karpman (pierre.karpman(AT)laposte.net), Oct 23 2007