cp's OEIS Frontend

A128546 Inrepfigit (INverse REPetitive FIbonacci-like diGIT) numbers (or Htiek numbers).

%I A128546 #12 Nov 09 2024 19:31:36
%S A128546 17,21,25,42,63,84,143,286,2355,5821,6618,11709,12482,33747,39571,
%T A128546 129109,466957,1162248,1565166,1968084,3636638,3853951,4898376,
%U A128546 6065280,13443745,13933175,17118698,22421197,24153462377
%N A128546 Inrepfigit (INverse REPetitive FIbonacci-like diGIT) numbers (or Htiek numbers).
%C A128546 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 A128546 42 is in the sequence because the terms of the sequence it creates are 2, 4, 6, 10, 16, 26, 42, ...
%t A128546 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 A128546 (C++)
%o A128546 #include <stdio.h>
%o A128546 // Here is a (messy) C++ code which finds the terms of the sequence below 100000000
%o A128546 int main() {
%o A128546   int k2;
%o A128546   for (int k = 10; k < 100000000; k++) {
%o A128546     k2 = k;
%o A128546     int array[9];
%o A128546     for (int i = 0; i <= 8; i++) {
%o A128546       array[i] = k2 % 10;
%o A128546       k2 /= 10;
%o A128546     }
%o A128546     bool c = true;
%o A128546     int check = 8;
%o A128546     for (int i = 0; i <= 8; i++) {
%o A128546       if ((array[8 - i] == 0) && c)
%o A128546         check--;
%o A128546       else
%o A128546         c = false;
%o A128546     }
%o A128546     bool b = false;
%o A128546     int n = 0;
%o A128546     while (n <= k && !b) {
%o A128546       n = 0;
%o A128546       for (int i = 0; i <= check; i++)
%o A128546         n += array[i];
%o A128546       if (n == k)
%o A128546         b = true;
%o A128546       for (int i = 0; i < check; i++)
%o A128546         array[i] = array[i + 1];
%o A128546       array[check] = n;
%o A128546     }
%o A128546     if (b)
%o A128546       printf("%d", k);
%o A128546   }
%o A128546   return 0;
%o A128546 }
%Y A128546 Cf. A007629.
%Y A128546 Cf. A097060 (reverse of these numbers).
%K A128546 base,nonn
%O A128546 1,1
%A A128546 Pierre Karpman (pierre.karpman(AT)laposte.net), Oct 23 2007