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a(0)=1, after which, for any n uniquely written as du*u! + ... + d2*2! + d1*1! (each di in range 0..i), a(n) = (du+1)*(u+1)! + ... + (d2+1)*3! + (d1+1)*2! + 1; the natural numbers with their factorial base representation (A007623) shifted left one step and each digit incremented by one, converted back to decimal.
4

%I #8 Nov 29 2013 22:20:24

%S 1,5,15,17,21,23,57,59,63,65,69,71,81,83,87,89,93,95,105,107,111,113,

%T 117,119,273,275,279,281,285,287,297,299,303,305,309,311,321,323,327,

%U 329,333,335,345,347,351,353,357,359,393,395,399,401,405,407,417,419

%N a(0)=1, after which, for any n uniquely written as du*u! + ... + d2*2! + d1*1! (each di in range 0..i), a(n) = (du+1)*(u+1)! + ... + (d2+1)*3! + (d1+1)*2! + 1; the natural numbers with their factorial base representation (A007623) shifted left one step and each digit incremented by one, converted back to decimal.

%H Antti Karttunen, <a href="/A231720/b231720.txt">Table of n, a(n) for n = 0..10080</a>

%F a(0)=1, and for n>=1, a(n) = A220655(A153880(n)).

%e 1 has a factorial base representation A007623(1) = '1'. This shifted once left is '10', and when each digit is incremented by one, this will be '21', and 2*2! + 1*1! = 5 (also A007623(5) = '21'), thus a(1)=5.

%e 2 has a factorial base representation '10'. This shifted once left is '100', and with each digit incremented, makes '211'. 2*3! + 1*2! + 1*1! = 15, thus a(2)=15.

%o (Scheme)

%o ;; Standalone iterative implementation:

%o (define (A231720 n) (let loop ((n n) (z 1) (i 2) (f 2)) (cond ((zero? n) z) (else (loop (quotient n i) (+ (* f (+ 1 (remainder n i))) z) (+ 1 i) (* f (+ i 1)))))))

%Y Subset of A227157. Cf. A007623, A153880, A220655.

%K nonn,base

%O 0,2

%A _Antti Karttunen_, Nov 29 2013