

A290223


Algorithm: s(k) = n. s(k+1) = s(k)  digitsum(s(k))^2 if s(k) >= 0 and s(k+1) = s(k)+digitsum(abs(s(k)))^2 if s(k) < 0. Below gives the end behavior for each number n.


3



0, 2, 3, 6, 6, 6, 3, 11, 9, 9, 3, 3, 6, 6, 6, 3, 11, 9, 0, 3, 3, 6, 2, 6, 3, 11, 9, 9, 11, 3, 6, 3, 6, 3, 11, 9, 9, 11, 3, 6, 3, 6, 3, 6, 9, 9, 11, 3, 6, 3, 6, 3, 6, 9, 9, 11, 3, 6, 6, 6, 3, 2, 9, 9, 11, 3, 6, 6, 6, 3, 3, 9, 9, 11, 3, 6, 2, 6, 3, 3, 0, 9, 11, 3, 6, 6, 6, 3, 6, 9, 9, 11, 3, 6, 6, 6, 3, 6, 9, 9, 3, 3, 6, 3, 6, 3, 3, 9, 9
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OFFSET

1,2


COMMENTS

0 means the sequence s(k) becomes the 0 sequence.
2 means the sequence s(k) becomes 2, 2, 2, 2, ...
3 means the sequence s(k) becomes 3, 6, 30, 21, 12, 3, ...
6 means the sequence s(k) becomes 6, 30, 21, 12, 3, 6, ...
9 means the sequence s(k) oscillates between two numbers, each of which have a digit sum of 9. For example, 18 > 63 > 18 > 63 > ... so a(18) = 9.
11 means the sequence s(k) oscillates between two numbers, each of which have a digit sum of 11. For example, 65 > 56 > 65 > ... so a(65) = 11.
a(n) = 2 for n = 2, 23, 62, 77, 119, 194, 287, 398. The next number n such that a(n) = 2 is over 10^5. This is believed to be finite.
a(n) = 11 for n = 8, 17, 26, 29, 35, 38, 47, 56, 65, 74, 83, 92, 149, 158, 167, 197. The next number n is over 10^5. This is believed to be finite.
The subsequences when a(n)=0, 3, 6, and 9 are believed to be infinite.


LINKS



EXAMPLE

a(19) = 0 because 19  (1+9)^2 = 81. Then 81 + (8+1)^2 = 0.
a(13) = 6 because 13  (1+3)^2 = 3. Then 3 + (3)^2 = 6.
a(17) = 11 because 17  (1+7)^2 = 47. Then 47 + (4+7)^2 = 74. Then 74  (7+4)^2 = 47, and so on.
a(23) = 2 because 23  (2+3)^2 = 2. Then 2 + (2)^2 = 2.
a(25) = 3 because 25  (2+5)^2 = 24. Then 24 + (2+4)^2 = 12. Then 12  (1+2)^2 = 3.
a(28) = 9 because 28  (2+8)^2 = 72. Then 72 + (7+2)^2 = 9. Then 9(9)^2 = 72, and so on.


PROG

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


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



