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)).

2, 5, 6, 7, 8, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99

Offset: 1

Author

Antti Karttunen, Dec 17 2012

Keywords

• nonn

Comments

Used for computing A107346.

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

Examples

			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.)
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.
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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..5039

Crossrefs

Complement: A220695.

One less than A220656.

Cf. A000142, A003422, A007489, A007623, A084558, A212598, A153880, A231720.

Programs

Mathematica

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 *)

Scheme

;; Standalone iterative implementation (Nov 29 2013):
(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))))))
;; Alternative implementation:
(define (A220655 n) (+ n (A007489 (A084558 n))))

Formula

a(n) = A220656(n)-1 = A003422(A084558(n)+1) + A000142(A084558(n)) + A212598(n) - 1. [The original definition]

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

For n >= 1, A231720(n) = a(A153880(n)).

Extensions

Name changed by Antti Karttunen, Nov 29 2013