Smarandache - Fibonacci Triplets H.lbstcdl
We recall the definition of the Smarandache Function S(n): S(n)
= the smallest positive integer such that S(n)! is divisible by n.
and the Fibonacci recurrence formula:
which for F 0 = F 1 = 1 defines the Fibonacci series. This article is concerned with isolated occurrencies of triplets n, n-l, n-2 for which S(n)=S(n-l)+S(n-2). Are there infinitly many such triplets? Is there a method of finding such triplets that would indicate that there are in fact infinitely many of them. A straight forward search by applying the definition of the Smarandache Function to consecutive integers was used to identify the first eleven triplets which are listed in table 1. As often in empirical number theory this merely scratched the surface of the ocean of integers. As can be seen from diagram 1 the next triplet may occur for a value of n so large that a sequential search may be impractical and will not make us much wiser. Table 1. The first 11 Smarandache-Fibonacci Triplets #
n
1 2 3 4 5 6 7 8 9 10 11
11 121 4902 26245 32112 64010 368140 415664 2091206 2519648 4573053
Stn)
S(n-1)
11 2*11
5 5
43
29
181
18 197 2*23 2*41 2*73 2*101 2*101 2*53
223 173
233 313 269 1109 569
S(n-2) 2*3 17 2*7 163 2*13 127 151 167 67 907
463
However, an interesting observation can be made from the triplets already found. Apart from n=26245 the Smarandache-Fibonacci triplets have in common that one member is two times a prime number while the other two members are prime numbers. This observation 130
Smarandache-Fibonacci Triplets. H. Ibstedt
Diagram1. The values of n for which the first 11 Smarandache-Fibonacci triplets occur
5 liE
4.5
/ /
4
3.5
_
3
j
1/1
C
C
~ 2.5 ~
-
/
L
2
1.5
I
0.5
I
/
~
o
2
3
4
5
6
7
8
9
10
11
Solution number
leads to a method to search for Smarandache-Fibonacci triplets in which the following two theorems play a role:
I.
IT n=ab with (a,b) = 1 and S(a) <S(b) then S(n)=S(b).
II.
IT n = pa where p is a prime number and
~p
then S(pa) = ap.
The search for Smarandache-Fibonacci triplets will be restricted to integers which meet the following requirements: n = xpa with
~p
(1)
and S(x)<ap
n-1 = yqb with b5q and S(y)<bq
(2)
n-2 = zrC with csr and S(z)<cr
(3)
p,q and r are primes. We then have S(n)=ap, S(n-1)=bq and S(n-2) = cr. From this and by subtracting (2) from (1) and (3) from (2) we get
= bq +
cr
(4)
xpa _ yqb = 1
(5)
ap
(6) 131
TABLE 2. Smarandache - Fibonacci Triplets_ #
N
SeN)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59
4 11 121 4902 32112 64010 368140 415664 2091206 2519648 4573053
4 11 22 43 223 173 233 313 269 1109 569 2591 2861 3433 7589 1714 1129 6491 9859 2213 10501 2647 2467 5653 3671 6661 2861 5801 2066 3506 3394 3049 2161 3023 12821 2174 4778
60
61 62 63 64 65 66 67 68 69 70 71 72
7783364
79269727 136193976 321022289 445810543 559199345 670994143 836250239 893950202 937203749 1041478032 1148788154 1305978672 1834527185 2390706171 2502250627 3969415464 3970638169 4652535626 6079276799 6493607750 6964546435 11329931930 11695098243 11777879792 13429326313 13849559620 14298230970 14988125477 17560225226 18704681856 23283250475 25184038673 29795026m 69481145903 107456166733 10m2646054 122311664350 126460024832 155205225351 196209376292 210621762776 211939749997 344645609138 484400122414 533671822944 620317662021 703403257356 859525157632 898606860813 972733721905 1185892343342 1225392079121 1294530625810 1517767218627 1905302845042 2679220490034 3043063820555 6098616817142 6505091986039 13666465868293
SeN-1) * *
*
* *
6883
2038 3209 4241 3046 4562 5582 11278 6301 10562 8222 20626 6917 8317 7246 6914 16774 7226 16811 21089 21929 13147 14158 19973 10267 18251 12202 17614 11617 22079 11402 14951 24767 31729 28099
3 5 5 29 197 46* 82* 167 202 * 202 * 106 * 202 * 2719 554 * 178 * 761 662
* * *
*
838 482 2062 10223 1286 746 1514 634 2642 2578 1198 643 3307 2837 1262 1814 2026 1294 1597 1597 2474 1847 2986
* * * * * * * * * * *
*
* * *
132
SeN-2)
*
* * * *
* * * * * *
* * * * *
*
3118 * 1823 463
1951 8819 3722 6043 6673 10463 2578 4034 3257 1567 11273 2803 12658 18118 20302 10874 3557 13402 10214 12022 9293 5807 8318 21478 7459 12202 20206 19862 16442
*
* *
* * * * * * * * * * * * *
2 6 17 14 26 127 151 146 67 907
t * * * * *
463
2389 142 * 2879 7411 953 467 5653 9377 151 278 * 1361 1721 4139 3037 4019 283 4603 1423 199 557 1787 347 997 11527 577 3181 4409 191 223 1123 1223 4099 3631 2459 2579 4519 1549 10163 4339 4283 3989 5347 5501 4423 4153 2971 1627 2273 10601 6571 53 6229 2909
11807 3299 601 3943 2749 4561 11867 11657
0 0 0 -4 -1 -1 -1 -8 -1 0 -3 0 10 -1 5 -1 -5 -1 1 0 -9 -1 3 0 -5 0 -1 -2 -6 0 -1 5 -4 -4 2 6 1 1 7 2 -2 4 -10 -2 0 3 -1 -1 0 11 -5 -5 11 0 9 -1 0 0 -2 -5 1 -4 -2 -4 -3 -8 -1 11 5 2 2 7
Smarandache-Fibonacci Triplets. H. Ibstedt
The greatest common divisor (pa, qb)= 1 obviously divides the right hand side of (5). This is the condition for (5) to have infinitely many solutions for each solution (p,q) to (4)_ These are found using Euclid's algorithm and can be written in the form: (5')
where t is an integer and (Xo.Yo) is the principal solution. Solutions to (5') are substituted in (6') in order to obtain integer solutions for z. (6')
Solutions were generated for (a,b,c)=(2,I,I), (a,b,c)=(1,2,1) and (a,b,c)=(I,I,2) with the parameter t restricted to the interval -11 =:; t =:; 11. The result is shown in table 2. Since the correctness of these calculations are easily verfied from factorisations of S(n), S(n-l), and S(n-2) these are given in table 3 for two large solutions taken from an extension of table 2. Table 3. Factorisation of two Smarandache-Fibonacci Triplets. 16,738,688,950,356 = 22.3-3H93路15.26f n= n-1= 16,738,688,950,355 =5-197-1,399-1,741-6.977 n-2= 16,738,688,950,354=2-72-19-23-53-313-23.561
S(n) = S(n-1)= S(n-2)=
2-15.2f22 f2.977 23_561
19,448,047,080,036=22.3~32.17.0932 n= n-1= 19,448,047,080,035 =5-7-19-37-61-761-17.027 n-2= 19,448,047,080,034 = 2-97-1,609-3,63H7. 159
S(n) = S(n-1)= S(n-1)=
2-17.093 17.Q27 1Z.152
Conjecture. There are infinitely many triplets n, n-1, n-2 such that S(n)=S(n-1)+S(n-2). Questions: 1. It is interesting to note that there are only 7 cases in table 2 where S(n-2) is two times a prime number and that they all occur for relatively small values of n. Which is the next one? 2. The solution for n = 26245 stands out as a very interesting one. Is it a unique case or is it a member of family of Smarandache-Fibonacci triplets different from those studied in this article? References: C. Ashbacher and M_ Mudge, Personal Computer World, October 1995, page 302_
M_ Mudge, in a Letter to R_ Muller (05/14/96), states that: "John Humphries of Hulse Ground Farm, Little Faringdo, Lechlade, Glovcester, GL7 3QR, U.K., has found a set of three numbers, greater than 415662, whose Smarandache Function satisfies the Fibonacci Recurrence, i.e. S(2091204) = 67, S(2091205) = 202, S(2091206) = 269, and 67 + 202 = 269. 133