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

17, 21, 25, 42, 63, 84, 143, 286, 2355, 5821, 6618, 11709, 12482, 33747, 39571, 129109, 466957, 1162248, 1565166, 1968084, 3636638, 3853951, 4898376, 6065280, 13443745, 13933175, 17118698, 22421197, 24153462377

Offset

1,1

Comments

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.

Links

Table of n, a(n) for n=1..29.

Example

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

Mathematica

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 *)

Prog

Here is a (messy) C++ code which finds the terms of the sequence below 100000000
#include <stdio.h>
int main()
{
int k2;
for ( int k = 10 ; k < 100000000 ; k++ )
{
k2 = k;
int array [9];
for ( int i = 0 ; i <= 8; i++ )
{
array[i] = k2 % 10;
k2 /= 10;
}
bool c = true;
int check=8;
for ( int i = 0; i <=8; i++ )
{
if ((array[8-i]==0)&&c)
check--;
else
c=false;
}
bool b = false;
int n = 0;
while ( n <= k && !b )
{
n = 0;
for ( int i = 0; i <= check; i++ )
n += array[i];
if ( n == k )
b = true;
for ( int i = 0 ; i < check ; i++ )
array[i] = array[i+1];
array[check] = n;
}
if ( b )
printf("%d
", k);
}
return 0;
}

Crossrefs

Cf. A007629.

Cf. A097060 (reverse of these numbers).

Sequence in context: A282109 A373583 A307863 * A188200 A060875 A259555

Adjacent sequences: A128543 A128544 A128545 * A128547 A128548 A128549

Keyword

base,nonn

Author

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

Status

approved