A065039 If n in base 10 is d_1 d_2 ... d_k then a(n) = d_1 + d_1d_2 + d_1d_2d_3 + ... + d_1...d_k.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70

Offset: 0

Author

Santi Spadaro, Nov 04 2001

Keywords

• nonn
• base
• easy
• nice

Comments

a(n) = (D(n) - sod(n))/9, for n >= 1, with sod(n) the sum of digits of n, and with D(n) any of the 10 numbers given in base 10 representation by d_(nod(n)-1) d_(nod(n)-2) ... d_0 b_0, where nod(n) is the number of digits of n = d_(nod(n)-1) d_(nod(n)-2) ... d_0 in base 10, and b_0 from {0, 1, ..., 9}. E.g., D(1234) stands for any number from {12340, 12341, ..., 12349}. This corresponds the well known (and easy to prove) rule that any number after subtraction of its sum of digits is divisible by 9. In this subtraction any of the last digit b_0 leads to the same result. Some mathematical tricks are based on this rule. See the Gardner reference. - Wolfdieter Lang, May 04 2010

Examples

			a(1234)=1370 because 1+12+123+1234=1370.
With repunits: a(1234) = 4*1 + 3*11 + 2*111 + 1*1111 = 1370. - _Wolfdieter Lang_, May 04 2010
		

References

• M. Gardner, Mathematische Zaubereien, Dumont, 2004, p. 39. German translation of: Mathematics, Magic and Mystery, Dover, 1956. [From Wolfdieter Lang, May 04 2010]

Links

• Harry J. Smith, Table of n, a(n) for n = 0..1000

Crossrefs

Complement of A065438.

Cf. A067079, A067080, A067082, A054899, A054861, A067080, A098844, A132027, A005187.

Programs

Haskell

import Data.List (inits)
a065039 n = sum $ map read $ tail $ inits $ show n
-- Reinhard Zumkeller, Mar 31 2011

Maple

A065039 := proc(n) local d,m: d:=convert(n,base,10): m:=nops(d): return add(op(convert(d[(m-k+1)..m], base, 10, 10^m)),k=1..m): end: seq(A065039(n),n=0..64); # Nathaniel Johnston, Jun 27 2011

Mathematica

a[n_] := Apply[Plus, Table[FromDigits[Take[IntegerDigits[n], k]], {k, 1, Length[IntegerDigits[n]]}]]
Table[d = IntegerDigits[n]; rd = 0; While[ Length[d] > 0, rd = rd + FromDigits[d]; d = Drop[d, -1]]; rd, {n, 0, 75} ]
f[n_] := Plus @@ NestList[ Quotient[ #, 10] &, n, Max[1, Floor@ Log[10, n]]]; Array[f, 70, 0] (* Robert G. Wilson v, Jun 29 2010 *)
Array[Total[Table[FromDigits[Take[IntegerDigits[#],x]],{x, IntegerLength[ #]}]]&,100,0](* Harvey P. Dale, Jan 02 2016 *)

PARI

{ for (n=0, 1000, a=0; k=n; until (k==0, a+=k; k\=10); write("b065039.txt", n, " ", a) ) } \\ Harry J. Smith, Oct 04 2009

Formula

a(n) = sum( k>=0, floor(n/10^ k)) = n+A054899(n). - Benoit Cloitre, Aug 03 2002

From Hieronymus Fischer, Aug 14 2007: (Start)

a(10*n)=10*n+a(n); a(n*10^m)=10*n*(10^m-1)/9+a(n).

a(k*10^m)=k*(10^(m+1)-1)/2, 0<=k<10, m>=0.

a(n)=10/9*n+O(log(n)), a(n+1)-a(n)=O(log(n)); this follows from the inequalities below.

a(n)<=(10*n-1)/9; equality holds for powers of 10.

a(n)>=(10*n-9)/9-floor(log_10(n)); equality holds for n=10^m-1, m>0.

lim inf (10*n/9-a(n))=1/9, for n-->oo.

lim sup (10*n/9-log_10(n)-a(n))=0, for n-->oo.

lim sup (a(n+1)-a(n)-log_10(n))=1, for n-->oo.

G.f.: sum{k>=0, x^(10^k)/(1-x^(10^k))}/(1-x).

(End)

a(n) = sum(d_(k)*RU(k+1),k=0..nod(n)-1), with the notation nod(n)and d_k given in a comment above, and RU(k)is the repunit (10^k-1)/9 (k times 1). - Wolfdieter Lang, May 04 2010