OFFSET
1,1
COMMENTS
Numbers such that the first half of digits have the same sum than the last half of digits are called balanced in the linked "Problem 217". (Note that here the meaning of "balanced" is neither that of A020492, nor that of A031443.)
Up to n=3 digits, the only balanced numbers are the palindromes, from n=4 on, there are non-palindromic balanced numbers, cf. A145808.
LINKS
Project Euler, Balanced numbers. Problem 217.
FORMULA
lim a(2n+1)/a(2n) = 100, lim a(2n)/a(2n-1) = 90 (as n -> oo).
EXAMPLE
a(1) = 1+2+...+9; a(2) = 11+22+...+99 = 11 a(1); a(3) = 101+111+121+....+191+202+...+989+999 = (101*10 + 10*9)*a(1); a(4) = 1001+1010+1102+1111+1120+1203+...+9889+9898+9999.
MATHEMATICA
balQ[n_]:=Module[{idn=IntegerDigits[n], len=Floor[IntegerLength[n]/2]}, Total[ Take[ idn, len]] == Total[Take[idn, -len]]]; Table[Total[ Select[ Range[ 10^n, 10^(n+1)-1], balQ]], {n, 0, 5}] (* This will generate the first six terms of the sequence. To generate more, (1) change the range of the Table from (0, 5) to (0, 6) or (0, 7), etc., but the program will take increasingly long to run. *) (* Harvey P. Dale, Apr 07 2013 *)
PROG
(PARI) A147808(n)={ local( t, c ); if( n==1, 45, /* global variable SC[sd] (used for n=2k and n=2k+1) stores [sum, count] of numbers with <= n\2 digits and digit sum = sd */ if( #SC != n\2*9, SC=vector( n\2*9, digsum, c=0; [sum( i=0, 10^(n\2)-1, if((i-digsum)%9==0 && digsum==sum(j=1, #t=Vecsmall(Str(i)), t[j])-48*#t, c++; i )), c] )); if( n%2==0, sum( i=10^((n\=2)-1), 10^n-1, SC[A007953(i)]*[1, i*10^n]~ ), t=10^(n\=2)*[100, 45]~; sum( i=10^(n-1), 10^n-1, SC[A007953(i)]*[10, [i, 1]*t]~ )))}
CROSSREFS
KEYWORD
base,nonn
AUTHOR
M. F. Hasler, Nov 23 2008
EXTENSIONS
a(13)-a(14) from Kevin P. Thompson, Dec 05 2021
STATUS
approved