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%I #10 Jul 28 2015 12:41:28
%S 0,1,2,3,10,11,12,13,20,21,22,23,30,31,32,33,40,41,42,43,50,51,52,53,
%T 60,61,62,63,70,71,72,73,80,81,82,83,163,170,171,172,173,180,181,182,
%U 183,263,270,271,272,273,280,281,282,283,363,370,371,372,373,380,381
%N Efficient representation of n in "square base" where xyz means 9x+4y+z and z<4, y<9 and x<16 etc.
%C Efficient means the smallest possible a(n), cf. example. From n = 9*9+8*4+3 = 116 on, the terms (coded in base 10) become ambiguous because digits may be larger than 9, e.g., 1000 could mean 1*16 or 10*9. One possible convention to avoid ambiguity would be to reserve as many digits as might be required for the largest possible coefficient: 2 digits for the coefficients of 9 (which may reach 16-1 = 15) through 81; 3 digits for the coefficients of 100 through 30^2, 4 digits for the coefficients of 31^2 (which may reach 32^2-1 = 1023) etc. - _M. F. Hasler_, Jul 25 2015
%D F. Smarandache, Definitions solved and unsolved problems, conjectures and theorems in number theory and geometry, edited by M. Perez, Xiquan Publishing House, 2000.
%H F. Smarandache, <a href="http://www.gallup.unm.edu/~smarandache/Definitions-book.pdf">Definitions, Solved and Unsolved Problems, Conjectures, ... </a>
%e a(50)=280 since 2*9+8*4+0*1=50; writing 20000 for 2*25 or 3xyz (for 3*16+x*9+y*4+z) or 5yz or 4yz or 3yz would be less efficient (larger "result" when read in base 10), and it is not possible to write 50 as 1*9+y*4+z*1 with y<9 and z<4.
%o (PARI) a(n,s=0)={v=[3];until(v[#v]>=n,v=concat(v,v[#v]+((2+#v)^2-1)*(1+#v)^2)); for(i=1,#v-1,s=s*10+t=max(ceil((n-v[#v-i])/(#v-i+1)^2),0);n-=t*(#v-i+1)^2);s*10+n} \\ _M. F. Hasler_, Jul 25 2015
%Y Cf. A007961 for greedy representation of n in "square base".
%K base,nonn
%O 0,3
%A _Henry Bottomley_, Jun 07 2000
%E Corrected and edited by _M. F. Hasler_, Jul 25 2015
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