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 1-9 of 9 results.

A014418 Representation of n in base of Catalan numbers (a classic greedy version).

Original entry on oeis.org

0, 1, 10, 11, 20, 100, 101, 110, 111, 120, 200, 201, 210, 211, 1000, 1001, 1010, 1011, 1020, 1100, 1101, 1110, 1111, 1120, 1200, 1201, 1210, 1211, 2000, 2001, 2010, 2011, 2020, 2100, 2101, 2110, 2111, 2120, 2200, 2201, 2210, 2211, 10000
Offset: 0

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Keywords

Comments

From Antti Karttunen, Jun 22 2014: (Start)
Also called "Greedy Catalan Base" for short.
Note: unlike A239903, this is a true base system, thus A244158(a(n)) = n holds for all n. See also A244159 for another, "less greedy" Catalan Base number system.
No digits larger than 3 will ever appear, because C(n+1)/C(n) approaches 4 from below, but never reaches it. [Where C(n) is the n-th Catalan number, A000108(n)].
3-digits cannot appear earlier than at the fifth digit-position from the right, the first example being a(126) = 30000.
The last digit is always either 0 or 1. (Cf. the sequences A244222 and A244223 which give the corresponding k for "even" and "odd" representations). No term ends as ...21.
No two "odd" terms (ending with 1) may occur consecutively.
A244217 gives the k for which a(k) starts with the digit 1, while A244216 gives the k for which a(k) starts with the digit 2 or 3.
A000108(n+1) gives the position of numeral where 1 is followed by n zeros.
A014138 gives the positions of repunits.
A197433 gives such k that a(k) = A239903(k). [Actually, such k, that the underlying strings of digits/numbers are same].
For the explanations, see the attached notes.
(End)

Examples

			A simple weighted sum of Sum_{k} digit(k)*C(k) [where C(k) = A000108(k), and digit(1) is the rightmost digit] recovers the natural number n (which the given numeral a(n) represents) as follows:
a(11) = 201, and indeed 2*C(3) + 0*C(2) + 1*C(1) = 2*5 + 0*2 + 1*1 = 11.
a(126) = 30000, and indeed, 3*C(5) = 3*42 = 126.
		

Crossrefs

Cf. A014420 (gives the sum of digits), A244221 (same sequence reduced modulo 2, or equally, the last digit of a(n)), A244216, A244217, A244222, A244223, A000108, A007623, A197433, A239903, A244155, A244158, A244320, A244318, A244159 (a variant), A244161 (in base-4), A014417 (analogous sequence for Fibonacci numbers).

Programs

  • Mathematica
    CatalanBaseIntDs[n_] := Module[{m, i, len, dList, currDigit}, i = 1; While[n > CatalanNumber[i], i++]; m = n; len = i; dList = Table[0, {len}]; Do[currDigit = 0; While[m >= CatalanNumber[j], m = m - CatalanNumber[j]; currDigit++]; dList[[len - j + 1]] = currDigit, {j, i, 1, -1}]; If[dList[[1]] == 0, dList = Drop[dList, 1]]; FromDigits@ dList]; Array [CatalanBaseIntDs, 50, 0] (* Robert G. Wilson v, Jul 02 2014 *)
  • Python
    from sympy import catalan
    def a244160(n):
        if n==0: return 0
        i=1
        while True:
            if catalan(i)>n: break
            else: i+=1
        return i - 1
    def a(n):
        if n==0: return 0
        x=a244160(n)
        return 10**(x - 1) + a(n - catalan(x))
    print([a(n) for n in range(51)]) # Indranil Ghosh, Jun 08 2017

Formula

From Antti Karttunen, Jun 23 2014: (Start)
a(0) = 0, a(n) = 10^(A244160(n)-1) + a(n-A000108(A244160(n))). [Here A244160 gives the index of the largest Catalan number that still fits into the sum].
a(n) = A007090(A244161(n)).
For all n, A000035(a(n)) = A000035(A244161(n)) = A244221(n).
(End)

Extensions

Description clarified by Antti Karttunen, Jun 22 2014

A028897 If n = Sum c_i 10^i then a(n) = Sum c_i 2^i.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 4
Offset: 0

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Author

Keywords

Comments

For n<100, this is the same result as "If n = Sum c_i 10^i then a(n) = Sum c_i (i+1)". - Henry Bottomley, Apr 20 2001
n_2 in the notation of A122618.
Left inverse of A007088 (binary numbers), cf. formula from Karttunen. - M. F. Hasler, Jun 13 2023

Crossrefs

Differs from A081594 and A244158 for the first time at n = 100, which here is a(100) = 4.
See A322000 for integers ordered according to the value of a(n).

Programs

  • Haskell
    a028897 0 = 0
    a028897 n = 2 * a028897 n' + d where (n', d) = divMod n 10
    -- Reinhard Zumkeller, Nov 06 2014
  • Mathematica
    a[n_ /; n < 10] := n; a[n_] := a[n] = If[Mod[n, 10] != 0, a[n-1] + 1, 2*a[n/10]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Apr 02 2016 *)
  • PARI
    a(n)=if(n<1,0,if(n%10,a(n-1)+1,2*a(n/10)))
    
  • PARI
    A028897(n)=fromdigits(digits(n),2) \\ M. F. Hasler, Feb 14 2019
    (MIT/GNU Scheme) (define (A028897 n) (let loop ((z 0) (i 0) (n n)) (if (zero? n) z (loop (+ z (* (modulo n 10) (expt 2 i))) (1+ i) (floor->exact (/ n 10)))))) ;; Antti Karttunen, Jun 22 2014
    

Formula

a(n) = 2*a(floor(n/10)) + (n mod 10). - Henry Bottomley, Apr 20 2001
a(0) = 0, a(n) = 2*a(n/10) if n == 0 (mod 10), a(n) = a(n-1)+1 otherwise. - Benoit Cloitre, Dec 21 2002
For all n, a(A007088(n)) = n. - Antti Karttunen, Jun 22 2014

Extensions

More terms from Erich Friedman.
Terms up to n = 100 added by Antti Karttunen, Jun 22 2014

A239903 List of Restricted-Growth Strings a_{k-1}a_{k-2}...a_{2}a_{1}, with k=2 and a_1 in {0,1} or k>2, a_{k-1}=1 and a_{j+1}>=1+a_j, for k-1>j>0.

Original entry on oeis.org

0, 1, 10, 11, 12, 100, 101, 110, 111, 112, 120, 121, 122, 123, 1000, 1001, 1010, 1011, 1012, 1100, 1101, 1110, 1111, 1112, 1120, 1121, 1122, 1123, 1200, 1201, 1210, 1211, 1212, 1220, 1221, 1222, 1223, 1230, 1231, 1232, 1233, 1234, 10000, 10001, 10010, 10011
Offset: 0

Views

Author

N. J. A. Sloane, Apr 06 2014

Keywords

Comments

We write the nonnegative integers as restricted growth strings (so called by J. Arndt in his book fxtbook.pdf, p. 325) in such a way that the Catalan numbers (cf. A000108) are expressed: 1=1, 10=2, 100=5, 1000=14, etc., 10...0 (with k zeros) = the k-th Catalan number. Once the entries of a restricted-growth string grow above 9, one would need commas or parentheses, say, to separate those entries. See Dejter (2017) for the precise definition.
In the paper "A system of numeration for middle-levels", restricted growth strings (RGSs) are defined as sequences that begin with either 0 or 1, with each successive number to the right being at least zero and at most one greater than its immediate left neighbor. Moreover, apart from case a(0), the RGSs are finite integer sequences of restricted growth which always start with 1 as their first element b_1 in position 1, and from then on, each successive element b_{i+1} in the sequence is restricted to be in range [0,(b_i)+1].
This sequence gives all such finite sequences in size-wise and lexicographic order, represented as decimal numbers by concatenating the integers of such finite sequences (e.g., from [1,2,0,1] we get 1201). The 58784th such sequence is [1, 2, 3, 4, 5, 6, 7, 8, 9, 9], thus a(58784) = 1234567899, after which comes the first RGS, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], where an element larger than 9 is present, which means that the decimal system employed here is unambiguous only up to n=58784. Note that 58785 = A000108(11)-1.
Also, if one considers Stanley's interpretation (u) of Catalan numbers, "sequences of a_1, a_2, ..., a_n of integers such that a_1 = 0 and 0 <= a_{i+1} <= a_{i} + 1" (e.g., 000, 001, 010, 011, 012 for C_3), and discards their initial zero, then one has a bijective correspondence with Dejter's RGSs of one element shorter length, which in turn are in bijective correspondence with the first C_n terms of this sequence (by discarding any leading zeros), from a(0) to a(C_n - 1). From this follows that the k-th Catalan number, A000108(k) (k>0), is represented in this system as 1 followed by k-1 zeros: a(1)=1, a(2)=10, a(5)=100, a(14)=1000, etc., and also that there exist exactly A000245(k) RGSs of length k.
Note how this differs from other number representations utilizing Catalan numbers, A014418 and A244159, in that while the latter are base-systems, where a simple weighted Sum_{k} digit(k)*C(k) recovers the natural number n (which the n-th numeral of such system represents), in contrast here it is the sum of appropriate terms in Catalan's Triangle (A009766, A030237), obtained by unranking a unique instance of a certain combinatorial structure (one of the Catalan interpretations), that gives a correspondence with a unique natural number. (Cf. also A014486.)
This sequence differs from "Semigreedy Catalan Representation", A244159, for the first time at n=10, where a(10) = 120, while A244159(10) = 121. That is also the first position where A244158(a(n)) <> n.
Please see Dejter's preprint for a more formal mathematical definition and how this number system is applied in relation to Havel's Conjecture on the existence of Hamiltonian cycles in the middle-levels graphs.
a(n) is given by the concatenation (with leading zeros removed) of the terms of row n + 23714 of A370222. - Paolo Xausa, Feb 17 2024

Examples

			Catalan's Triangle T(row,col) = A009766 begins with row n=0 and 0<=col<=n as:
  Row 0: 1
  Row 1: 1, 1
  Row 2: 1, 2,  2
  Row 3: 1, 3,  5,  5
  Row 4: 1, 4,  9, 14, 14
  Row 5: 1, 5, 14, 28, 42,  42
  Row 6: 1, 6, 20, 48, 90, 132, 132
  (the leftmost diagonal of 1s is "column 0").
  ...
For example, for n=38, we find that A081290(38)=14, which occurs on row A081288(n)-1 = 4, in columns A081288(n)-1 and A081288(n)-2, i.e., as T(4,4) and T(4,3). Thus we subtract 38-14 to get 24, and we see that the next term downward on the same diagonal, 28, is too large to accommodate into the same sum, so we go one diagonal up, starting now from T(3,2) = 5. This fits in, so we now have 24 - 5 = 19, and also the next term on the same diagonal, T(4,2) = 9, fits in, so we now have 19-9 = 10. The next term on the same diagonal, T(5,2) = 14, would not fit in anymore, so we rewind ourselves back to penultimate column, but one step up from where we started on this diagonal, so T(2,1) = 2, which fits in, 10 - 2 = 8, also the next one T(3,1) = 3, 8 - 3 = 5, and the next one T(4,1) = 4, 5 - 4 = 1, after which comes T(5,1) = 5 > 1, thus we jump to T(1,0) = 1, 1-1 = 0, and T(2,0)=1 would not fit anymore, thus next time the row would be zero, and the algorithm is ready with 1 (14), 2 (5+9), 3 (2+3+4) and 1 (1) terms collected, whose total sum 14+5+9+2+3+4+1 = 38, thus a(38) = 1231.
For n=20, the same algorithm results in 1 (14), 1 (5), 0 (not even the first tentative term T(2,1) = 2 from the column 1 would fit, so it is skipped), and from one row higher we get the needed 1 (1), so the total sum of these is 14+5+0+1 = 20, thus a(20) = 1101.
		

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 2: Seminumerical Algorithms, third edition, Addison-Wesley, 1977, p. 192.
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999, Exercise 19, interpretation (u).

Crossrefs

Cf. A000108 (Catalan numbers), A000245 (their first differences), A009766 (Catalan's triangle), A236855 (the sum of elements in k-th RGS), A236859 (for n>=1, gives the length of the initial ascent 123... in term a(n)), A244159 (different kinds of Catalan number systems).
Other Catalan combinatorial structures represented as integer sequences: A014486/A063171: Dyck words, parenthesizations, etc., A071156/A071158: Similar restricted words encoded with help of A007623 (Integers written in factorial base), A071153/A079436 (Łukasiewicz words).

Programs

  • Julia
    function CatalanNumerals(z)
        z == 0 && return 0
        f(n) = factorial(n)
        t(j, k) = div(f(k+j)*(k-j+1), f(j)*f(k+1))
        k, i = 2, 0
        while z >= t(i, i + 1) i += 1 end
        dig = fill(0, i); dig[1] = 1
        x = z - t(i - 1, i)
        m = i - 1
        while x > 0
            w, s, p = 0, 0, 0
            while w <= x
                p = w
                w += t(m - 1, m + s)
                s += 1
            end
            dig[k] = s - 1
            m -= 1; k += 1; x -= p
        end
        s = ""; for d in dig s *= string(d) end
        parse(Int, s)
    end
    [CatalanNumerals(n) for n in 0:42] |> println # Peter Luschny, Nov 10 2019
    
  • MATLAB
    function [ c ] = catrep(z)
    i=0; x=0; y=0; s=0;
    while z>=(factorial(2*i+1)*(2))/(factorial(i)*factorial(i+2))
    i=i+1;
    end
    y=(factorial(2*i-1)*(2))/(factorial(i-1)*factorial(i+1));
    a=zeros(1,i); a(1,1)=1; k=2; x=z-y; m=1;
    while x>0
    w=0; s=0; p=0;
    while w<=x
    p=w;
    w=w+(factorial(2*i-2*m+s-1)*(s+2))/(factorial(i-1-m)*factorial(i-m+s+1));
    s=s+1;
    end
    m=m+1; a(1,k)=s-1; k=k+1; x=x-p;
    end
    a
    end
    
  • Mathematica
    A239903full = With[{r = 2*Range[2, 11]-1}, Reverse[Map[FromDigits[r-#] &, Rest[Select[Subsets[Range[2, 21], {10}, 125477], Min[r-#] >= 0 &]]]]];
    A239903full[[;;100]] (* Paolo Xausa, Feb 17 2024 *)
  • Maxima
    define (t(j,k), (factorial(k+j)*(k-j+1))/(factorial(j)*factorial(k+1)));
    i:0;
    x:19;
    z:0;y:0;s:0;
    while x>=t(i,i+1) do (i:i+1);
    y:t(i-1,i);a:zeromatrix(1,i);a[1,1]:1;k:2;z:x-y;m:1;
    while (z>0) do (
    w:0,s:0,p=0,
    while (w<=z) do (
    p:w,
    w:w+t(i-1-m,i-m+s),
    s:s+1
    ),
    m:m+1,
    a[1,k]:s-1,k:k+1,
    z:z-p
    );
    print(a);
    
  • PARI
    \\ Valid for n<58786 (=A000108(11)).
    nxt(w)=if(w[1]==#w, vector(#w+1, i, i>#w), my(k=1); while(w[k]>w[k+1], w[k]=0; k++); w[k]++; w)
    seq(n)={my(a=vector(n), w=[1]); a[1]=0; for(i=2, #v, a[i]=fromdigits(Vecrev(w)); w=nxt(w)); a} \\ Andrew Howroyd, Jan 24 2023
  • Scheme
    (define (A239903_only_upto_16794 n) (if (zero? n) n (A235049 (A071159 (A081291 n))))) ;; Gives correct results only up to 16794.
    ;; The following gives correct results all the way up to n=58784.
    (define (A239903 n) (baselist-as-decimal (A239903raw n)))
    (definec (A239903raw n) (if (zero? n) (list) (let loop ((n n) (row (A244160 n)) (col (- (A244160 n) 1)) (srow (- (A244160 n) 1)) (catstring (list 0))) (cond ((or (zero? row) (negative? col)) (reverse! (cdr catstring))) ((> (A009766tr row col) n) (loop n srow (- col 1) (- srow 1) (cons 0 catstring))) (else (loop (- n (A009766tr row col)) (+ row 1) col srow (cons (+ 1 (car catstring)) (cdr catstring))))))))
    (define (baselist-as-decimal lista) (baselist->n 10 lista))
    (define (baselist->n base bex) (let loop ((bex bex) (n 0)) (cond ((null? bex) n) (else (loop (cdr bex) (+ (* n base) (car bex)))))))
    ;; From Antti Karttunen, Apr 14-19 2014
    

Formula

To find an RGS corresponding to natural number n, one first finds a maximum row index k such that T(k,k-1) <= n in the Catalan Triangle (A009766) illustrated in the Example section. Note that as the last two columns of this triangle consist of Catalan numbers (that is, T(k,k-1) = T(k,k) = A000108(k)), it means that the first number to be subtracted from n is A081290(n) which occurs as a penultimate element of the row A081288(n)-1, in the column A081288(n)-2. The unranking algorithm then proceeds diagonally downwards, keeping the column index the same, and incrementing the row index, as long as it will encounter terms such that their total sum stays less than or equal to n.
If the total sum of encountered terms on that diagonal would exceed n, the algorithm jumps back to the penultimate column of the triangle, but one row higher from where it started the last time, and again starts summing the terms as long as the total sum stays <= n.
When the algorithm eventually reaches either row zero or column less than zero, the result will be a list of numbers, each element being the number of terms summed from each diagonal, so that the diagonal first traversed appears as the first 1 (as that first diagonal will never allow more than one term), and the number of terms summed from the last traversed diagonal appears the last number in the list. These lists of numbers are then concatenated together as decimal numbers.
These steps can also be played backwards in order to recover the corresponding decimal integer n from such a list of numbers, giving a "ranking function" which will be the inverse to this "unranking function".
For n=1..16794 (where 16794 = A000108(10)-2), a(n) = A235049(A071159(A081291(n))). - Antti Karttunen, Apr 14 2014
Alternative, simpler description of the algorithm from Antti Karttunen, Apr 21 2014: (Start)
Consider the following square array, which is Catalan triangle A009766 without its rightmost, "duplicate" column, appropriately transposed (cf. also tables A030237, A033184 and A054445):
Row| Terms on that row
---+--------------------------
1 | 1 1 1 1 1 ...
2 | 2 3 4 5 6 ...
3 | 5 9 14 20 27 ...
4 | 14 28 48 75 110 ...
5 | 42 90 165 275 429 ...
6 | 132 297 572 1001 1638 ...
To compute the n-th RGS, search first for the greatest Catalan number C_k which is <= n (this is A081290(n), found as the first term of row A081288(n)-1). Then, by a greedy algorithm, select from each successive row (moving towards the top of table) as many terms from the beginning of that row as will still fit into n, subtracting them from n as you go. The number of terms selected from the beginning of each row gives each element of the n-th RGS, so that the number of terms selected from the topmost row (all 1's) appears as its last element.
(End)

Extensions

Description, formula and examples edited/rewritten by Italo J Dejter, Apr 13 2014 and Antti Karttunen, Apr 18 2014

A244159 Semigreedy Catalan Representation of nonnegative integers.

Original entry on oeis.org

0, 1, 10, 11, 12, 100, 101, 110, 111, 112, 121, 122, 123, 211, 1000, 1001, 1010, 1011, 1012, 1100, 1101, 1110, 1111, 1112, 1121, 1122, 1123, 1211, 1212, 1221, 1222, 1223, 1232, 1233, 1234, 1322, 2111, 2112, 2121, 2122, 2123, 2211, 10000
Offset: 0

Views

Author

Antti Karttunen, Jun 23 2014

Keywords

Comments

Algorithm for constructing the sequence: Define a(0) as 0, and for larger values of n, find first the largest Catalan number which is less than or equal to n [which is A081290(n)], and the index k = A244160(n), of that Catalan number. Initialize a vector of k zeros, [0, 0, ..., 0]. Set n_remaining = n - A000108(k) and add 1 to the leftmost element of vector, so that it will become [1, 0, ..., 0]. Then check whether the previous Catalan number, C(m) = A000108(m), where m = k-1, exceeds the n_remaining, and provided that C(m) <= n_remaining, then set n_remaining = n_remaining - C(m) and increment by one the m-th element of the vector (where the 1st element is the rightmost), otherwise just decrement m by one and keep on doing the same with lesser and lesser Catalan numbers, and whenever it is possible to subtract them from n_remaining (without going less than zero), do so and increment the corresponding m-th element of the vector, as long as either n_remaining becomes zero, or after subtracting C(1) = 1 from n_remaining, it still has not reached zero. In the latter case, find again the largest Catalan number which is less than or equal to n_remaining, and start the process again. However, after a finite number of such iterations, n_remaining will finally reach zero, and the result of a(n) is then the vector of numbers constructed, concatenated together and represented as a decimal number.
This shares with "Greedy Catalan Base" (A014418) the property that a simple weighted sum of Sum_{k=1..} digit(k)*C(k) recovers the natural number n, which the given numeral string like A014418(n) or here, a(n), represents. (Here C(k) = the k-th Catalan number, A000108(k), and digit(1) = the digit in the rightmost, least significant digit position.)
In this case, A244158(a(n)) = n holds for only up to 33603, after which comes the first representation containing a "digit" larger than nine, at a(33604), where the underlying string of numbers is [1,2,3,4,5,6,7,8,9,10] but the decimal system used here can no more unambiguously represent them.
On the other hand, with the given Scheme-functions, we always get n back with: (CatBaseSumVec (A244159raw n)).
For n >= 1, A014138(n) gives the positions of repunits: 1, 11, 111, 1111, ...
The "rep-2's": 22222, 222222, 2222222, 22222222, 222222222, ..., etc., occur in positions 128, 392, 1250, 4110, 13834, ... i.e. 2*A014138(n) for n >= 5.

Examples

			For n = 18, the largest Catalan number <= 18 is C(4) = 14.
Thus we initialize a vector of four zeros [0, 0, 0, 0] and increment the first element to 1: [1, 0, 0, 0] and subtract 14 from 18 to get the remainder 4.
We see that the next smaller Catalan number, C(3) = 5 is greater than 4, so we cannot subtract it without going negative, so the second leftmost element of the vector stays as zero.
We next check C(2) = 2, which is less than 4, thus we increment the zero at that point to 1, and subtract 4 - 2 to get 2.
We compare 2 to C(1) = 1, and as 1 <= 2, it is subtracted 2-1 = 1, and the corresponding element in the vector incremented, thus after the first round, the vector is now [1, 0, 1, 1], and n remaining is 1.
So we start the second round because n has not yet reached the zero, and look for the largest Catalan number <= 1, which in this case is C(1) = 1, so we subtract it from remaining n, and increment the element in the position 1, after which n has reached zero, and the vector is now [1, 0, 1, 2], whose concatenation as decimal numbers thus yields a(18) = 1012.
		

Crossrefs

Cf. A014418 (a classical greedy variant), A244231 (maximum "digit value"), A244232 (sum of digits), A244233 (product of digits), A244314 (positive terms which have at least one zero digit), A244316 (the one-based position of digit incremented last in the described process).
Differs from A239903 for the first time at n=10, where a(10) = 121, while A239903(10) = 120.

Formula

If A176137(n) = 1, a(n) = A007088(A244230(n)), otherwise a(n) = A007088(A244230(n)-1) + a(n-A197433(A244230(n)-1)).
For all n, a(A197433(n)) = A007088(n).
For all n >= 1, a(A000108(n)) = 10^(n-1).
Each a(A014143(n)) has a "triangular" representation [1, 2, 3, ..., n, n+1].

A197433 Sum of distinct Catalan numbers: a(n) = Sum_{k>=0} A030308(n,k)*C(k+1) where C(n) is the n-th Catalan number (A000108). (C(0) and C(1) not treated as distinct.)

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 8, 14, 15, 16, 17, 19, 20, 21, 22, 42, 43, 44, 45, 47, 48, 49, 50, 56, 57, 58, 59, 61, 62, 63, 64, 132, 133, 134, 135, 137, 138, 139, 140, 146, 147, 148, 149, 151, 152, 153, 154, 174, 175, 176, 177, 179, 180, 181, 182, 188, 189, 190, 191, 193, 194, 195, 196
Offset: 0

Views

Author

Philippe Deléham, Oct 15 2011

Keywords

Comments

Replace 2^k with A000108(k+1) in binary expansion of n.
From Antti Karttunen, Jun 22 2014: (Start)
On the other hand, A244158 is similar, but replaces 10^k with A000108(k+1) in decimal expansion of n.
This sequence gives all k such that A014418(k) = A239903(k), which are precisely all nonnegative integers k whose representations in those two number systems contain no digits larger than 1. From this also follows that this is a subsequence of A244155.
(End)

Crossrefs

Characteristic function: A176137.
Subsequence of A244155.
Cf. also A060112.
Other sequences that are built by replacing 2^k in binary representation with other numbers: A022290 (Fibonacci), A029931 (natural numbers), A059590 (factorials), A089625 (primes), A197354 (odd numbers).

Programs

  • Mathematica
    nmax = 63;
    a[n_] := If[n == 0, 0, SeriesCoefficient[(1/(1-x))*Sum[CatalanNumber[k+1]* x^(2^k)/(1 + x^(2^k)), {k, 0, Log[2, n] // Ceiling}], {x, 0, n}]];
    Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Nov 18 2021, after Ilya Gutkovskiy *)

Formula

For all n, A244230(a(n)) = n. - Antti Karttunen, Jul 18 2014
G.f.: (1/(1 - x))*Sum_{k>=0} Catalan number(k+1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017

Extensions

Name clarified by Antti Karttunen, Jul 18 2014

A081594 Let n = 10x + y where 0 <= y <= 9, x >= 0. Then a(n) = 2x+y.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 20
Offset: 0

Views

Author

N. J. A. Sloane, Apr 22 2003

Keywords

Crossrefs

Cf. A081502. Differs from A028897, A156230 and A244158 for the first time at n=100, which here is a(100) = 20.

Programs

  • Magma
    [(n+4*y)/5 where y is n mod 10: n in [0..100]]; // Bruno Berselli, Jun 24 2014
    
  • Maple
    A081594:=n->n-8*floor(n/10); seq(A081594(n), n=0..100); # Wesley Ivan Hurt, Jun 25 2014
  • Mathematica
    CoefficientList[Series[-x (7 x^9 - x^8 - x^7 - x^6 - x^5 - x^4 - x^3 - x^2 - x - 1)/((x - 1)^2 (x + 1) (x^4 - x^3 + x^2 - x+1) (x^4 + x^3 + x^2 + x + 1)), {x, 0, 150}], x] (* Vincenzo Librandi, Jun 25 2014 *)
    LinearRecurrence[{1,0,0,0,0,0,0,0,0,1,-1},{0,1,2,3,4,5,6,7,8,9,2},110] (* or *) Table[Range[n,n+9],{n,0,26,2}]//Flatten (* Harvey P. Dale, Jul 22 2021 *)
  • PARI
    my(n, x, y); vector(200, n, y=(n-1)%10; x=(n-1-y)\10; 2*x+y) \\ Colin Barker, Jun 24 2014
    
  • Sage
    [n-8*floor(n/10) for n in (0..100)] # Bruno Berselli, Jun 24 2014

Formula

a(n) = (2 * floor(n/10)) + (n modulo 10). - Antti Karttunen, Jun 22 2014
G.f.: -x*(7*x^9 -x^8 -x^7 -x^6 -x^5 -x^4 -x^3 -x^2 -x -1) / ((x -1)^2*(x +1)*(x^4 -x^3 +x^2 -x +1)*(x^4 +x^3 +x^2 +x +1)). - Colin Barker, Jun 23 2014
a(n) = n - 8*floor(n/10). [Bruno Berselli, Jun 24 2014]

Extensions

Terms up to n=100 added by Antti Karttunen, Jun 22 2014
G.f. revised by Vincenzo Librandi, Jun 25 2014

A244155 Numbers n such that when the n-th Catalan restricted growth string [b_k, b_{k-1}, ..., b_2, b_1] (see A239903) is viewed as a simple numeral in Catalan Base: b_k*C(k) + b_{k-1}*C(k-1) + ... + b_2*C(2) + b_1*C(1) it is equal to n. Here C(m) = A000108(m).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 174, 175
Offset: 0

Views

Author

Antti Karttunen, Jun 22 2014

Keywords

Comments

In range 0 .. 58784, these are numbers k such that A244158(A239903(n)) = k. (see comments at A244157).

Crossrefs

Complement of A244156. Positions of zeros in A244157.
A197433 is a subsequence.

A244157 a(n) = difference between n and the n-th Catalan restricted growth string [b_k, b_{k-1}, ..., b_2, b_1] (see A239903) when it is viewed as a simple numeral in Catalan Base: b_k*C(k) + b_{k-1}*C(k-1) + ... + b_2*C(2) +b_1*C(1). Here C(m) = A000108(m).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 4, 4, 4, 4, 4, 5, 5, 5, 5, 7, 7, 7, 7, 7, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 4, 4, 4, 4, 4, 5, 5, 5, 5, 7, 7, 7, 7, 7, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 18
Offset: 0

Views

Author

Antti Karttunen, Jun 22 2014

Keywords

Crossrefs

A244155 gives the positions of zeros, A244156 the positions of nonzeros.

Programs

  • Scheme
    (define (A244157 n) (- n (CatBaseSum (A239903raw n)))) ;; A239903raw given in A239903.
    (define (CatBaseSum lista) (let loop ((digits (reverse lista)) (i 1) (s 0)) (if (null? digits) s (loop (cdr digits) (+ i 1) (+ s (* (car digits) (A000108 i)))))))

Formula

a(n) = n - A244158(A239903(n)) up to 58784, after which the "digits" in Catalan restricted growth strings grow larger than 9 and their decimal representation used in A239903 starts corrupting the results.
At n=58785 (= C(11)-1, where C(k) = the k-th Catalan number, A000108(k)), the correct value for this sequence is a(58785) = 58785 - ((1*C(10)) + (2*C(9)) + (3*C(8)) + (4*C(7)) + (5*C(6)) + (6*C(5)) + (7*C(4)) + (8*C(3)) + (9*C(2)) + (10*C(1))) = 25181.
Use the Scheme-program given in the Program sections of this entry and A239903 (the function A239903raw) to get correct results for all n.

A322001 Digits of n interpreted in factorial base: a(Sum d_k*10^k) = Sum d_k*k!

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 6
Offset: 0

Views

Author

M. F. Hasler, Nov 27 2018

Keywords

Comments

More terms than usual are given to distinguish the sequence from A081594, A028897 and A244158, which agree up to a(99). The last two correspond to k! replaced by 2^k resp. Catalan(k).
This is a left inverse to A007623 (factorial base representation of n): A322001(A007623(n)) = n for all n >= 0. One could imagine variants which have a(n) = 0 or a(n) = -1 if n is not a term of A007623. Restricted to the range of A007623, it is also a right inverse to A007623, at least up to the 10 digit terms, beyond which A007623 becomes non-injective.

Crossrefs

Cf. A007623 (right inverse), A081594, A028897, A244158.

Programs

  • Mathematica
    a[n_] := Module[{d=Reverse@IntegerDigits[n]}, Sum[d[[i]]*i!, {i,1,Length[d]}]]; Array[a, 100, 0] (* Amiram Eldar, Nov 28 2018 *)
  • PARI
    A322001(n)=sum(i=1,#n=Vecrev(digits(n)),n[i]*i!) \\ M. F. Hasler, Nov 27 2018
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