A131836 Multiplicative persistence of the Sierpinski numbers of the first kind (n^n + 1).
0, 0, 2, 2, 3, 2, 2, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1
Examples
For n=4 we have A014566(4) = Sierpinski number 257 --> 2*5*7 = 70 --> 7*0 = 0 thus persistence = 2, and a(4) = 2. - Edited by _Antti Karttunen_, Oct 08 2017
Links
- Antti Karttunen, Table of n, a(n) for n = 1..2048
Programs
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Maple
P:=proc(n) local i,k,w,ok,cont; for i from 1 by 1 to n do w:=1; k:=i^i+1; ok:=1; if k<10 then print(0); else cont:=1; while ok=1 do while k>0 do w:=w*(k-(trunc(k/10)*10)); k:=trunc(k/10); od; if w<10 then ok:=0; print(cont); else cont:=cont+1; k:=w; w:=1; fi; od; fi; od; end: P(100);
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Mathematica
Table[-1 + Length@ NestWhileList[Times @@ IntegerDigits@ # &, If[n == 0, 2, n^n + 1], # > 9 &], {n, 105}] (* Michael De Vlieger, Oct 08 2017 *)
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Scheme
;; The whole program follows: (define (A131836 n) (A031346 (A014566 n))) (define (A014566 n) (+ 1 (expt n n))) (define (A031346 n) (let loop ((n n) (k 0)) (if (< n 10) k (loop (A007954 n) (+ 1 k))))) (define (A007954 n) (if (zero? n) n (let loop ((n n) (m 1)) (if (zero? n) m (let ((d (modulo n 10))) (loop (/ (- n d) 10) (* d m))))))) ;; Antti Karttunen, Oct 08 2017
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