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A128546 Inrepfigit (INverse REPetitive FIbonacci-like diGIT) numbers (or Htiek numbers). 3

%I #9 Jul 19 2021 01:20:08

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

%o #include <stdio.h>

%o int main()

%o {

%o int k2;

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

%o {

%o k2 = k;

%o int array [9];

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

%o {

%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 {

%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 {

%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

%o ", 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

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Last modified March 29 04:23 EDT 2024. Contains 371264 sequences. (Running on oeis4.)