cp's OEIS Frontend

A220655 For n with a unique factorial base representation n = du*u! + ... + d2*2! + d1*1! (each di in range 0..i, cf. A007623), a(n) = (du+1)*u! + ... + (d2+1)*2! + (d1+1)*1!; a(n) = n + A007489(A084558(n)).

%I A220655 #29 Jan 21 2020 00:28:16
%S A220655 2,5,6,7,8,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,57,
%T A220655 58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,
%U A220655 81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99
%N A220655 For n with a unique factorial base representation n = du*u! + ... + d2*2! + d1*1! (each di in range 0..i, cf. A007623), a(n) = (du+1)*u! + ... + (d2+1)*2! + (d1+1)*1!; a(n) = n + A007489(A084558(n)).
%C A220655 Used for computing A107346.
%C A220655 Term a(n) can be obtained by adding one to each digit of factorial base representation of n (A007623(n)) and then reinterpreting it as a kind of pseudo-factorial base representation, ignoring the fact that now some of the digits might be over the maximum allowed in that position. Please see the example section. - _Antti Karttunen_, Nov 29 2013
%H A220655 Antti Karttunen, <a href="/A220655/b220655.txt">Table of n, a(n) for n = 1..5039</a>
%F A220655 a(n) = A220656(n)-1 = A003422(A084558(n)+1) + A000142(A084558(n)) + A212598(n) - 1. [The original definition]
%F A220655 a(n) = n + A007489(A084558(n)). [The above formula reduces to this, which proves that the original Dec 17 2012 description and the new main description produce the same sequence. Essentially, we are adding to n a factorial base repunit '1...111' with as many fact.base digits as n has.] - _Antti Karttunen_, Nov 29 2013
%F A220655 For n >= 1, A231720(n) = a(A153880(n)).
%e A220655 1 has a factorial base representation A007623(1) = '1', as 1 = 1*1!. Incrementing the digit 1 with 1, we get 2*1! = 2, thus a(1) = 2. (Note that although '2' is not a valid factorial base representation, it doesn't matter here.)
%e A220655 2 has a factorial base representation '10', as 2 = 1*2! + 0*1!. Incrementing the digits by one, we get 2*2! + 1*1! = 5, thus a(2) = 5.
%e A220655 3 has a factorial base representation '11', as 3 = 1*2! + 1*1!. Incrementing the digits by one, we get 2*2! + 2*1! = 6, thus a(3) = 6.
%t A220655 Block[{nn = 66, m = 1}, While[Factorial@ m < nn, m++]; m = MixedRadix[Reverse@ Range[2, m]]; Array[FromDigits[1 + IntegerDigits[#, m], m] &, nn]] (* _Michael De Vlieger_, Jan 20 2020 *)
%o A220655 (Scheme)
%o A220655 ;; Standalone iterative implementation (Nov 29 2013):
%o A220655 (define (A220655 n) (let loop ((n n) (z 0) (i 2) (f 1)) (cond ((zero? n) z) (else (loop (quotient n i) (+ (* f (+ 1 (remainder n i))) z) (+ 1 i) (* f i))))))
%o A220655 ;; Alternative implementation:
%o A220655 (define (A220655 n) (+ n (A007489 (A084558 n))))
%Y A220655 Complement: A220695.
%Y A220655 One less than A220656.
%Y A220655 Cf. A000142, A003422, A007489, A007623, A084558, A212598, A153880, A231720.
%K A220655 nonn
%O A220655 1,1
%A A220655 _Antti Karttunen_, Dec 17 2012
%E A220655 Name changed by _Antti Karttunen_, Nov 29 2013