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O N A P R O B L E M O F F . S M A R A N D A C H E *

Z H N G v V E N P E N G N D Y r Y U N

R e s e a r c h C e n te r fo r B a s ic S c ie n c e , X i ' a n J i a o t o n g U n iv e r s i ty X i 'a n , S haan ...a , P .R . C h i n a

A B ST R A C T. L e t d s ( n ) d e n ot es t h e s u m o f t h e b a s e 10 d ig it s o f n E N . F o r n a tu ra lx 2 a n d a rb i t ra r y f ix e d e x p o n e n t m E N , le t Ar n ( x ) = L d ; C n ) . T h e m a i n

n zp u r p o s e o f t h i s p a p e r i s to g iv e tw o ex a ct c a :lc u la t in g fo r m u la s fo r A l x ) a n d Az (x ) .

1. IN T R O D U C T IO N

F o r a n y p o s i t iv e i n t e g e r n , l e t d s (n ) d e n o te s t h e s u m o f t h e b a s e 10 d ig i ts o f n .

F o r e x a m p l e , ds(O ) = 0 , d s ( l ) = 1, d s(2 ) = 2 , d s( l l) = 2 , d A1 2 ) = 3 , . In pr o b le m 21 o f b o o k [1], P r o f e s s o r F .S m a r a da c h e ask us to st u d y t h e pr o p e rt ie s o f s e q u e n c e {d s ( n ) }. F o r n a tu r a l n u m b e r x 2 a n d a r b i t r a r y fi x e d e x p o n e n t m. E N , l e t

Am ( x) = L d : (n ) . 1 )n < x

T h e m a in p u r p o s e o f t h is p a p e r is t o s tu d y th e c a l c u l a t i n g p r o b le m o f Am x ) , a n d u s e e l em en t a r y m e t h o d s to d e d u c e tw o e x a c t c a lc u la t i n g f o r m u la s fo r A l (x ) a n d A2 x ) . T h a t is , w e s h a l l p ro v e t h e fo ll ow in g :

T h e o r e m . F or a n y p o s i t ive in t e g e r x , l e t x = a 1 10 k + a 2 1O k2 + .. . + as 10 k , with k1 > k2 > .. , > ks 0 a n d 1 ; a i ; 9 , i = 2 ,3 , , s . T h e n we h a ve th e c a l w l a ti n g fo r m u l a s

A x ) = a (~k + a _ a i + 1 . O ki.. . J z . z L ) , i = l j=

A L

[ k i (8 1 k i + 3 3 ) 9 k i (. 1 ) i 2 (4a i - l ) a i + l J ] k ( x = a ; a ; + a - 1 0 . 4 2 ) 6 i=1 j l

+ t ai [C 9ki - a i -1 ) lO k i + 2 t ajlokij ~ai ) . = 2 ; = z ; =

F o r g e n e r a l i n te g e r m 3, u s in g o u r m e th o d s w e c a n a lso g iv e a n e x a c t c a lc ul a ti ng f o r m u la fo r A m ;r. ). B u t i n t h e s e c a se s , t h e c o m p u ta ti o n s a r e m o r e c o m p le x .

K e y w o r d s a n d ph ra tj es . F . S m a r a n d a c h e p r o b l e m : S u m o f b a s e 10 d i g i t s : C a l cu la ti n g fo rm u la .* T h i s w o r k i s su p p or te d b y th e N .S .F . a n d t.h e P .S .F . o f P . R . C h i l l a .

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2. PROOF OF TH TH OR M

In this section, we complete the proof of the Theorem. First we need followingtwo simple Lemmas.

Lemma 1. For any integer k ~ 0, we have the identities

a) A1 10k) = ~ k .10k;

k k a-I kb) A d a 10 = 2 : + . a l O , 1 ~ a ~ 9.Proof We first prove a) of Lemma 1 by induction. For k = 0 and 1, we haveA1 100) = .11 1) = 0, .11 10 1) = .11 10) = 45. So that the identity

A_1 10k) = L ds n) = ~ k 10 k 2)n < 1 0 ~

holds for k = 0 and 1. Assume 2) is true for k = m - 1. Then by the inductiveassumption we have

= A1 9 . 10 m - I ) + 9. 10 m - 1 + A1 lO m - 1 )= A 1 8 . 10

m- I ) + 8 + 9) . lO m - 1 + 2A I lOm-l)

=

= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) . l O m - l + 10A 1 10 m - I = ~ . 10 m + 10 . ~ . m - 1) . lOm-l

2 2

= ~ mlOm.2

That is, 2) is true for k = m. This proves the first part of Lemma 1.The second part b) follows from a) of Lemma 1 and the recurrence formula

k1 a lO = L dAn) + L ds n)n a - l ) 1 0 a - 1 )10 n < a l 0

- L ds{n) + L ds n + a - 1 - 10 k )n a - 1 - 1 0 k 0 ~ n < 1 0

L a n) + a - 1 - 10 k + L dsCn)n a - 1 ) -10 n

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This completes the proof of Lemma 1.

Lemma 2 For any integer k ~ 0 and 1 S a S 9. we have the identities. 81k 33 .

c) A2 10k = . k 1 0 k ;

4

d) 4 Ok) [k 81k 33) 9k 1) a - 1) 2a - 1 ]k

2 a 1 = 4 2 6 a .

Proof. These results can be deduced by Lemma 1, induction and the recurrence

formula

d ~ n )

= L d; n)+ L d ; n+9 10 k )n k2 > > ks ~ 0 under the base 10. Then applying Lemma 1 repeatedly wehave

= A1 a1 . lOki ) a1 x - a1 . lOki ) A1 x - a1 . lOki )

= Adal lOkl) Ada2 lOk2) al ;1; - a1 . lOki )

+ CL2 ;r - a l . lOki - 0 2 . 10k2) A.I x - al . 10 k, - a2 . lOk2)

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s s i

= L A 1 a i . lO ki) + L a i X - L a j .1 0 k j ) i = l i = l j l

= t ~ . ; + a i ; 1 . j . lOki + t a j . lOki ~ l . j) 1 =1 1 = 2 J l

~ 9k

~ a

i

+1) k L : i + L a j - - . - . a j . 1 I . i = l - j l -T h i s p ro v e s th e fi rs t p a r t o f t h e T he o r e m .

A p p l y in g L e m m a 2 a n d t h e fi rs t p a r t o f t h e T h e o r e m r e p e a t e d ly w e h a v e A 2 (x ) = L d ; n ) + L d ; n )

= A 2 a 1 1 0kI

+ L (d s (n ) + a d 2 O ~n < x a I 10 l : I

= A 2 (a 1 . 1 0k

+ L d ; n ) + 2a1 . d s (n ) + a i = A 2 (a 1 . 10 k + i . x - a . lO k I

+ 2 a 1 A1 x - a . lO k I + A 2 x - a . lO k I =

s s s

i = L A 2 a j . lO k i) + L aUx - L j .1 O k j + L 2 a i A 1 X - L j . 1 0 k j)

i= 1 i = l j= 1 i = l

T h i s c o m p le te s th e p r o o f o f t h e s e c o n d p a r t o f t h e T h e o r e m .

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j=

Ok j . 1 I

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R E F E R E N C E S

1. F . S m a r a n d a c h e , O n l y p ro b le m s n o t S o l u ti o n s X i q u a n P u b . H o u s e , C h i c g o , 1 9 9 3 , p p . 22 . 2. T o m M . A po s to l, I n t ro d u c t i o n t o n a l y t i c N u .m b e r T h e o r y S p r in g e r -V e r l a g , N e w Y o rk , 1 9 7 6 .3 . " S m a r a n d a c h e S e q u e n c e s a t h tt p: // w w w .g a l lu p .u n m .e d u/ "s m a ra n da c he / s n aq i nt .t xt .4 . " S m a r a n d a c h e S e q u e n c e s at h tt p : // w w w .g a ll u p .u n m .e d u / " s m a ra n da c h e / s n a q i nt 2 . tx t . 5 . " S m a r an d a c h e S e q u e n c e s a t h tt p : // w w w . g a l l u p . u n m . e d u / " s m a r a n d a c h e / s n a . q i n t3 . tx t .

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