A193583 Number of fixed points under iteration of sum of squares of digits in base b.

1, 3, 1, 3, 1, 5, 3, 3, 1, 3, 3, 7, 1, 3, 1, 7, 5, 3, 1, 7, 3, 7, 1, 3, 1, 7, 3, 3, 3, 7, 5, 7, 3, 3, 1, 7, 5, 3, 1, 5, 3, 11, 3, 3, 3, 15, 3, 3, 3, 3, 3, 7, 1, 7, 1, 15, 3, 3, 3, 3, 3, 7, 3, 3, 1, 7, 7, 3, 5, 3, 7, 15, 1, 7, 3, 7, 3, 3, 3, 7, 5, 15, 1, 3, 3

Offset: 2

Author

Martin Renner, Jul 31 2011

Keywords

• nonn
• base

Comments

If b>=2 and a>=b^2 then S(a,2,b)

From Christian N. K. Anderson, Apr 22 2013: (Start)

It can be shown that no fixed point has more than 2 digits in base b, and that the two-digit number A+Bb must satisfy the condition that (2A-1)^2+(2B-b)^2=1+b^2. The number of ways of writing (1+b^2) as the sum of two squares is d(1+b^2)-1, where d(n) is the number of divisors of n. (Beardon, 1998, Theorem 3.1)

From the above chain of logic follows:

- The value of the fixed points can be determined by investigating only 8*A002654(n^2+1) pairs of possibilities.

- a(n) = A000005(n^2+1)-1

- a(n) = A193432(n)-1

a(n)=1 iff n^2+1 is prime, and the value of that single fixed point is 1.

The only odd value of n for which a(n)=9 is n=239.

Several values of a(n) occur very infrequently. For example, a(1068)=13 is the only occurrence of 13 for n < 10000. (End)

Examples

			In the decimal system all integers go to (1) or (4, 16, 37, 58, 89, 145, 42, 20) under the iteration of sum of squares of digits, hence there is one fixed point and one cycle. Therefore a(10) = 1.
a(5)=3 because 1 is always a fixed point; also in base 5, decimal 13 -> 23 and 2^2+3^2 = 13; decimal 18 -> 33 and 3^2+3^2 = 18. - _Christian N. K. Anderson_, Apr 22 2013
		

Links

• Martin Renner and Christian N. K. Anderson, Table of n, a(n) for n = 2..10000 (first 299 values from Martin Renner)
• Alan F. Beardon, Sums of Squares of Digits, The Mathematical Gazette, 82(1998), 379-388.
• H. G. Grundman, E. A. Teeple, Generalized Happy Numbers, Fibonacci Quarterly 39 (2001), nr. 5, p. 462-466.

Crossrefs

Equals A193432-1.

Cf. A007770.

Cf. A002654, A000005.

Programs

R

library(gmp); y=rep(0,10000)
for(B in 1:10000) y[B]=prod(table(as.numeric(factorize(1+as.bigz(B)^2)))+1)-1; y # Christian N. K. Anderson, Apr 22 2013

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