A147808 Sum of n-digit numbers which are balanced: the first [n/2] digits have the same sum as the last [n/2] digits.
45, 495, 49500, 3314850, 331431000, 27336542310, 2733612983100, 238305122029260, 23830484311542600, 2140037814262627400, 214003761418373774000, 19587943639318412097360, 1958794348735327250973600, 181693537570273520779480800
Offset: 1
Examples
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.
Links
- Project Euler, Balanced numbers. Problem 217.
Programs
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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 *) -
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]~ )))}
Formula
lim a(2n+1)/a(2n) = 100, lim a(2n)/a(2n-1) = 90 (as n -> oo).
Extensions
a(13)-a(14) from Kevin P. Thompson, Dec 05 2021
Comments