Cumulative concatenation of A000032 Lucas numbers (beginning at 2).

[Jonathan Post]

Cumulative concatenation of A000032 Lucas numbers (beginning at 2).

a(n+1) = Concatenate(a(n),A000032(n+1)).

This is to A000032 as A130774 is to A000204.

Like these Lucas numbers, a(n) cycles even, odd, odd, even, odd, odd, ...

n  a(n)  factors
1  2      prime
2  21     3 * 7 semiprime
3  213   3 * 71 semiprime
4  2134  2 * 11 * 97
5  21347 is prime
6  2134711  719 * 2969 semiprime
7  213471118 = 2 * 7 * 19 * 802523
8  21347111829 = 3 * 12743 * 558401
9  2134711182947 = 7 * 491 * 621097231
10 213471118 294776 = 2^3 * 3^2 * 41 * 7349 * 9839987
11 213471118294 776123 = 3 * 41 * 785903 * 2208335567
12 213471118294776123199 = 11 * 23 * 349 * 2417648598420967
13 213471118294776123199322 = 2 * 23 * 37 * 3929 * 6991 * 4566234789049
14 213471118294776123199322521 = 61950375139 * 3445840607353939 semiprime
...
no more prime nor semiprime values through n = 60, which has a 381
digit composite factor.



Jonathan Post:

> So OEIS does not yet have the sequnce where a(n) is the smallest
> integer whose n-th power begins with n identical digits, base 10?

I think not. If I did this correctly, it begins:

10 683405939
11 7074698775
12 26331754107
13 844494314469
...

The next term, 1247955519394, is the first whose nth power does not  
begin with *precisely* n identical digits. *That* number would be  
11303028458639.

Hans


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