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Numbers n such that A290223(n) = 11.
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%I #4 Jul 24 2017 12:31:50

%S 8,17,26,29,35,38,47,56,65,74,83,92,149,158,167,197

%N Numbers n such that A290223(n) = 11.

%C This sequence is believed to be finite. If it exists, a(17) > 10^5.

%e 26 is in this sequence because 26 - (2+6)^2 = -38. Then -38 + (3+8)^2 = 83. Then 83 - (8+3)^2 = -38, and so on.

%o (PARI)

%o a(n)=k=n; c=1; v=List(); listput(v, k); while(c, if(k>=0, k-=sumdigits(k)^2; c+=1; if(k==2||k==3||k==0||k==6||k==9, return(k)); if(vecsearch(Vec(v), k), return(sumdigits(abs(k)))); listput(v, k)); if(k<0, k+=sumdigits(-k)^2; c+=1; if(k==2||k==3||k==0||k==6||k==9, return(k)); if(vecsearch(Vec(v), k), return(sumdigits(abs(k)))); listput(v, k)); c+=1)

%o for(n=1,10^5,if(a(n)==11,print1(n,", ")))

%Y Cf. A007953, A003132, A290223.

%K nonn,base,fini,full

%O 1,1

%A _Derek Orr_, Jul 24 2017