cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A060130 Number of nonzero digits in factorial base representation (A007623) of n; minimum number of transpositions needed to compose each permutation in the lists A060117 & A060118.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 1, 2, 2, 3, 2, 3, 2, 3, 3
Offset: 0

Views

Author

Antti Karttunen, Mar 02 2001

Keywords

Examples

			19 = 3*(3!) + 0*(2!) + 1*(1!), thus it is written as "301" in factorial base (A007623). The count of nonzero digits in that representation is 2, so a(19) = 2.
		

Crossrefs

Cf. A227130 (positions of even terms), A227132 (of odd terms).
The topmost row and the leftmost column in array A230415, the left edge of triangle A230417.
Differs from similar A267263 for the first time at n=30.

Programs

  • Maple
    A060130(n) = count_nonfixed(convert(PermUnrank3R(n), 'disjcyc'))-nops(convert(PermUnrank3R(n), 'disjcyc')) or nops(fac_base(n))-nops(positions(0, fac_base(n)))
    fac_base := n -> fac_base_aux(n, 2); fac_base_aux := proc(n, i) if(0 = n) then RETURN([]); else RETURN([op(fac_base_aux(floor(n/i), i+1)), (n mod i)]); fi; end;
    count_nonfixed := l -> convert(map(nops, l), `+`);
    positions := proc(e, ll) local a, k, l, m; l := ll; m := 1; a := []; while(member(e, l[m..nops(l)], 'k')) do a := [op(a), (k+m-1)]; m := k+m; od; RETURN(a); end;
    # For procedure PermUnrank3R see A060117
  • Mathematica
    Block[{nn = 105, r}, r = MixedRadix[Reverse@ Range[2, -1 + SelectFirst[Range@ 12, #! > nn &]]]; Array[Count[IntegerDigits[#, r], k_ /; k > 0] &, nn, 0]] (* Michael De Vlieger, Dec 30 2017 *)
  • Scheme
    (define (A060130 n) (let loop ((n n) (i 2) (s 0)) (cond ((zero? n) s) (else (loop (quotient n i) (+ 1 i) (+ s (if (zero? (remainder n i)) 0 1)))))))
    ;; Two other implementations, that use memoization-macro definec:
    (definec (A060130 n) (if (zero? n) n (+ 1 (A060130 (A257687 n)))))
    (definec (A060130 n) (if (zero? n) n (+ (A257511 n) (A060130 (A257684 n)))))
    ;; Antti Karttunen, Dec 30 2017

Formula

a(0) = 0; for n > 0, a(n) = 1 + a(A257687(n)).
a(0) = 0; for n > 0, a(n) = A257511(n) + a(A257684(n)).
a(n) = A060129(n) - A060128(n).
a(n) = A084558(n) - A257510(n).
a(n) = A275946(n) + A275962(n).
a(n) = A275948(n) + A275964(n).
a(n) = A055091(A060119(n)).
a(n) = A069010(A277012(n)) = A000120(A275727(n)).
a(n) = A001221(A275733(n)) = A001222(A275733(n)).
a(n) = A001222(A275734(n)) = A001222(A275735(n)) = A001221(A276076(n)).
a(n) = A046660(A275725(n)).
a(A225901(n)) = a(n).
A257511(n) <= a(n) <= A034968(n).
A275806(n) <= a(n).
a(A275804(n)) = A060502(A275804(n)). [A275804 gives all the positions where this coincides with A060502.]
a(A276091(n)) = A260736(A276091(n)). [A276091 gives all the positions where this coincides with A260736.]

Extensions

Example-section added, name edited, the old Maple-code moved away from the formula-section, and replaced with all the new formulas by Antti Karttunen, Dec 30 2017

A230410 After a(0)=0, a(n) = A230415(A219666(n),A219666(n-1)).

Original entry on oeis.org

0, 1, 2, 2, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 1, 3, 1, 3, 1, 3, 3, 3, 1, 3, 4, 2, 2, 2, 4, 2, 2, 2, 2, 1, 3, 1, 3, 1, 3, 3, 3, 1, 3, 4, 2, 1, 3, 3, 3, 2, 4, 1, 3, 1, 3, 3, 3, 1, 3, 4, 2, 1, 2, 2, 2, 2, 3, 2, 2, 4, 3, 1, 3, 4, 2, 1, 2, 2, 2, 2, 3, 2, 1, 3, 2, 5, 2
Offset: 0

Views

Author

Antti Karttunen, Nov 10 2013

Keywords

Comments

After zero, a(n) = number of positions where digits in the factorial base representations of successive nodes A219666(n-1) and A219666(n) in the infinite trunk of the factorial beanstalk differ from each other.

Examples

			a(8) = 1, because A219666(8)=23, whose factorial base representation (A007623(23)) is '321', and A219666(7)=17, whose factorial base representation (A007623(17)) is '221', and they differ just in one digit position.
a(9) = 3, because A219666(9)=25, '...01001' in factorial base, which differs from '...0321' in three digit positions.
Note that A226061(4)=8 (A226061(n) tells the position of (n!)-1 in A219666), and 1+2+3 = 6 happens to be both a triangular number (A000217) and a factorial number (A000142).
The next time 1 occurs in this sequence because of this coincidence is at x=A226061(16) (whose value is currently not known), as at that point A219666(x) = 16!-1 = 20922789887999, whose factorial base representation is (15,14,13,12,11,10,9,8,7,6,5,4,3,2,1), and A000217(15) = 120 = A000142(5), which means that A219666(x-1) = A219651(20922789887999) = 20922789887879, whose factorial base representation is (15,14,13,12,11,10,9,8,7,6,4,4,3,2,1), which differs only in one position from the previous.
Of course 1's occur in this sequence for other reasons as well.
		

Crossrefs

Cf. A230415, A230406, A231717, A231719, A232094. A230422 gives the positions of ones.

Programs

  • Mathematica
    nn = 1200; m = 1; While[m! < nn, m++]; m; f[n_] := IntegerDigits[n, MixedRadix[Reverse@ Range[2, m]]]; Join[{0}, Function[w, Count[Subtract @@ Map[PadLeft[#, Max@ Map[Length, w]] &, w], k_ /; k != 0]]@ Map[f@ # &, {#1, #2}] & @@@ Partition[#, 2, 1] &@ TakeWhile[Reverse@ NestWhileList[# - Total@ f@ # &, nn, # > 0 &], # <= 500 &]] (* Michael De Vlieger, Jun 27 2016, Version 10 *)
  • Scheme
    (define (A230410 n) (if (zero? n) n (A230415bi (A219666 n) (A219666 (- n 1))))) ;; Where bi-variate function A230415bi has been given in A230415.

Formula

a(0)=0, and for n>=1, a(n) = A230415(A219666(n),A219666(n-1)).
For all n, a(A226061(n+1)) = A232094(n).

A232096 a(n) = largest m such that m! divides 1+2+...+n; a(n) = A055881(A000217(n)).

Original entry on oeis.org

1, 1, 3, 2, 1, 1, 2, 3, 1, 1, 3, 3, 1, 1, 5, 2, 1, 1, 2, 3, 1, 1, 3, 3, 1, 1, 3, 2, 1, 1, 2, 4, 1, 1, 3, 3, 1, 1, 3, 2, 1, 1, 2, 3, 1, 1, 4, 4, 1, 1, 3, 2, 1, 1, 2, 3, 1, 1, 3, 3, 1, 1, 4, 2, 1, 1, 2, 3, 1, 1, 3, 3, 1, 1, 3, 2, 1, 1, 2, 5, 1, 1, 3, 3, 1, 1, 3
Offset: 1

Views

Author

Antti Karttunen, Nov 18 2013

Keywords

Crossrefs

A042963 gives the positions of ones and A014601 the positions of larger terms.

Programs

Formula

a(n) = A055881(A000217(n)).
a(n) = A231719(A226061(n+1)). [Not a practical way to compute this sequence, but follows from the definitions]

A232095 Minimal number of factorials which add to 0+1+2+...+n; a(n) = A034968(A000217(n)).

Original entry on oeis.org

0, 1, 2, 1, 3, 4, 5, 3, 3, 6, 4, 5, 4, 7, 7, 1, 5, 5, 5, 8, 7, 9, 5, 5, 6, 8, 10, 6, 9, 8, 10, 8, 6, 10, 12, 7, 10, 11, 6, 5, 7, 7, 8, 9, 5, 8, 5, 6, 8, 7, 10, 7, 11, 14, 8, 8, 6, 11, 7, 10, 7, 12, 10, 10, 12, 14, 7, 12, 9, 9, 11, 9, 12, 12, 12, 14, 10, 7, 11, 11
Offset: 0

Views

Author

Antti Karttunen, Nov 18 2013

Keywords

Comments

1's occur at positions n=1, n=3 and n=15 as they are such natural numbers that A000217(n) is also one of the factorial numbers (A000142), as we have A000217(1) = 1 = 1!, A000217(3) = 1+2+3 = 6 = 3! and A000217(15) = 1 + 2 + ... + 15 = 120 = 5!
On the other hand, a(2)=2, as A000217(2) = 1+2 = 3 = 2! + 1!. Is this the only occurrence of 2?
Are some numbers guaranteed to occur an infinite number of times?

Crossrefs

Programs

Formula

a(n) = A034968(A000217(n)).
a(n) = A231717(A226061(n+1)). [Not a practical way to compute this sequence. Please see comments at A231717.]
For all n, a(n) >= A232094(n).
Showing 1-4 of 4 results.