A250202 The "sum of squares of digits" problem in base 12, start with 6 (written in base 10).

6, 36, 9, 81, 117, 162, 38, 13, 2, 4, 16, 17, 26, 8, 64, 41, 34, 104, 128, 164, 66, 61, 26, 8, 64, 41, 34, 104, 128, 164, 66, 61, 26, 8, 64, 41, 34, 104, 128, 164, 66, 61, 26, 8, 64, 41, 34, 104, 128, 164, 66, 61, 26, 8, 64, 41, 34, 104, 128, 164, 66, 61, 26, 8, 64, 41

Offset: 1

Author

Eric Chen, Mar 13 2015

Keywords

• nonn
• easy
• base

Comments

Periodic with period 10.

In base 12, there are 3 fixed points and 4 cycles (only 1 fixed point and 1 cycle in base 10, see A161772):

1 -> 1 (length 1);

5 -> 21 -> 5 (length 2);

8 -> 54 -> 35 -> 2a -> 88 -> a8 -> 118 -> 56 -> 51 -> 22 -> 8 (length 10);

18 -> 55 -> 42 -> 18 (length 3);

25 -> 25 (length 1);

68 -> 84 -> 68 (length 2);

a5 -> a5 (length 1);

Notice 25 (decimal 29) and a5 (decimal 125) are Armstrong numbers in base 12 (A161949), there are no 2-digit Armstrong numbers in base 10.

In base 12, there are only few happy numbers (no such between 10 (decimal 12) and 100 (decimal 144)), but in base 10, there are 20 happy numbers less than or equal to 100 (see A007770).

Links

• Eric Chen, Table of n, a(n) for n = 1..1000
• Eric Chen, List of A250202 starts with n for n up to 40 (decimal 48)
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Crossrefs

Cf. A000216, A000218, A080709, A000221, A008460, A008461, A008462, A008463, A161772.

Programs

Mathematica

NestList[Total[IntegerDigits[#, 12]^2]&, 6, 144]
Join[{6, 36, 9, 81, 117, 162, 38, 13, 2, 4, 16, 17},LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 1},{26, 8, 64, 41, 34, 104, 128, 164, 66, 61},54]] (* Ray Chandler, Aug 26 2015 *)
PadRight[{6,36,9,81,117,162,38,13,2,4,16,17},80,{66,61,26,8,64,41,34,104,128,164}] (* Harvey P. Dale, Aug 06 2017 *)

PARI

a(n) = [6, 36, 9, 81, 117, 162, 38, 13, 2, 4, 16, 17, 26, 8, 64, 41, 34, 104, 128, 164, 66, 61][n%10+10*(n>=10)+10*(n%10<3 & n>=20)]