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 1-10 of 12 results. Next

A244320 a(n) = n - A014420(n).

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 4, 5, 5, 6, 8, 8, 9, 9, 13, 13, 14, 14, 15, 17, 17, 18, 18, 19, 21, 21, 22, 22, 26, 26, 27, 27, 28, 30, 30, 31, 31, 32, 34, 34, 35, 35, 41, 41, 42, 42, 43, 45, 45, 46, 46, 47, 49, 49, 50, 50, 54, 54, 55, 55, 56, 58, 58, 59, 59, 60, 62, 62, 63, 63, 67, 67, 68, 68, 69
Offset: 0

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Author

Antti Karttunen, Jul 02 2014

Keywords

Crossrefs

Programs

Formula

a(n) = n - A014420(n).

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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Author

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

A265745 a(n) is the number of Jacobsthal numbers (A001045) needed to sum to n using the greedy algorithm.

Original entry on oeis.org

0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 4, 3, 4, 5, 4, 5, 4, 5, 6, 5, 6, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2
Offset: 0

Views

Author

Antti Karttunen, Dec 17 2015

Keywords

Comments

Sum of digits in "Jacobsthal greedy base", A265747.
It would be nice to know for sure whether this sequence gives also the least number of Jacobsthal numbers that add to n, i.e., that there cannot be even better nongreedy solutions.
The integer 63=21+21+21 has 3 for its 'non-greedy' solution, and a(63) = 5 for its greedy solution 63=43+11+5+3+1. - Yuriko Suwa, Jul 11 2021
Positions where a(n) is different from A372555(n) are n=63, 84, 148, 169, 191, 212, 234, 255, etc. See A372557. - Antti Karttunen, May 07 2024

Examples

			a(0) = 0, because no numbers are needed to form an empty sum, which is zero.
For n=1 we need just A001045(2) = 1, thus a(1) = 1.
For n=2 we need A001045(2) + A001045(2) = 1 + 1, thus a(2) = 2.
For n=4 we need A001045(3) + A001045(2) = 3 + 1, thus a(4) = 2.
For n=6 we form the greedy sum as A001045(4) + A001045(2) = 5 + 1, thus a(6) = 2. Alternatively, we could form the sum as A001045(3) + A001045(3) = 3 + 3, but the number of summands in that case is no less.
For n=7 we need A001045(4) + A001045(2) + A001045(2) = 5 + 1 + 1, thus a(7) = 3.
For n=8 we need A001045(4) + A001045(3) = 5 + 3, thus a(8) = 2.
For n=10 we need A001045(4) + A001045(4) = 5 + 5, thus a(10) = 2.
		

Crossrefs

Cf. A054111 (apparently the positions of the first occurrence of each n > 0).

Programs

  • Mathematica
    jacob[n_] := (2^n - (-1)^n)/3; maxInd[n_] := Floor[Log2[3*n + 1]]; A265745[n_] := A265745[n] = 1 + A265745[n - jacob[maxInd[n]]]; A265745[0] = 0; Array[A265745, 100, 0] (* Amiram Eldar, Jul 21 2023 *)
  • PARI
    A130249(n) = floor(log(3*n + 1)/log(2));
    A001045(n) = (2^n - (-1)^n) / 3;
    A265745(n) = {if(n == 0, 0, my(d = n - A001045(A130249(n))); if(d == 0, 1, 1 + A265745(d)));} \\ Amiram Eldar, Jul 21 2023
  • Python
    def greedyJ(n): n1 = (3*n+1).bit_length() - 1; return (2**n1 - (-1)**n1)//3
    def a(n): return 0 if n == 0 else 1 + a(n - greedyJ(n))
    print([a(n) for n in range(107)]) # Michael S. Branicky, Jul 11 2021
    

Formula

a(0) = 0; for n >= 1, a(n) = 1 + a(n - A001045(A130249(n))). [This formula uses a simple greedy algorithm.]

A265744 a(n) is the number of Pell numbers (A000129) needed to sum to n using the greedy algorithm (A317204).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Dec 17 2015

Keywords

Comments

a(0) = 0, because no numbers are needed to form an empty sum, which is zero.
It would be nice to know for sure whether this sequence also gives the least number of Pell numbers that add to n, i.e., that there cannot be even better nongreedy solutions.

References

  • A. F. Horadam, Zeckendorf representations of positive and negative integers by Pell numbers, Applications of Fibonacci Numbers, Springer, Dordrecht, 1993, pp. 305-316.

Crossrefs

Similar sequences: A007895, A116543, A278043.

Programs

  • Mathematica
    pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; a[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[pell[k] <= m, k++]; k--; AppendTo[s, k]; m -= pell[k]; k = 1]; Plus @@ IntegerDigits[Total[3^(s - 1)], 3]]; Array[a, 100, 0] (* Amiram Eldar, Mar 12 2022 *)

Formula

a(n) = A007953(A317204(n)). - Amiram Eldar, Mar 12 2022

A352508 Catalan-Niven numbers: numbers that are divisible by the sum of the digits in their representation in terms of the Catalan numbers (A014418).

Original entry on oeis.org

1, 2, 4, 5, 6, 9, 10, 12, 14, 16, 18, 21, 24, 28, 30, 32, 33, 40, 42, 44, 45, 48, 55, 56, 57, 60, 65, 72, 78, 80, 84, 88, 95, 100, 105, 112, 126, 128, 130, 132, 134, 135, 138, 140, 144, 145, 146, 147, 152, 155, 156, 168, 170, 174, 180, 184, 185, 195, 210, 216
Offset: 1

Views

Author

Amiram Eldar, Mar 19 2022

Keywords

Comments

Numbers k such that A014420(k) | k.
All the Catalan numbers (A000108) are terms.
If k is an odd Catalan number (A038003), then k+1 is a term.

Examples

			4 is a term since its Catalan representation, A014418(4) = 20, has the sum of digits A014420(4) = 2 + 0 = 2 and 4 is divisible by 2.
9 is a term since its Catalan representation, A014418(9) = 120, has the sum of digits A014420(9) = 1 + 2 + 0 = 3 and 9 is divisible by 3.
		

Crossrefs

Programs

  • Mathematica
    c[n_] := c[n] = CatalanNumber[n]; q[n_] := Module[{s = {}, m = n, i}, While[m > 0, i = 1; While[c[i] <= m, i++]; i--; m -= c[i]; AppendTo[s, i]]; Divisible[n, Plus @@ IntegerDigits[Total[4^(s - 1)], 4]]]; Select[Range[216], q]

A352509 Numbers k such that k and k+1 are both Catalan-Niven numbers (A352508).

Original entry on oeis.org

1, 4, 5, 9, 32, 44, 55, 56, 134, 144, 145, 146, 155, 184, 234, 324, 329, 414, 426, 429, 434, 455, 511, 512, 603, 636, 930, 1004, 1014, 1160, 1183, 1215, 1287, 1308, 1448, 1472, 1505, 1562, 1595, 1808, 1854, 1967, 1985, 1995, 2051, 2075, 2096, 2135, 2165, 2255
Offset: 1

Views

Author

Amiram Eldar, Mar 19 2022

Keywords

Examples

			4 is a term since 4 and 5 are both Catalan-Niven numbers: the Catalan representation of 4, A014418(20) = 20, has the sum of digits 2+0 = 2 and 4 is divisible by 2, and the Catalan representation of 5, A014418(5) = 100, has the sum of digits 1+0+0 = 1 and 5 is divisible by 1.
		

Crossrefs

Programs

  • Mathematica
    c[n_] := c[n] = CatalanNumber[n]; q[n_] := Module[{s = {}, m = n, i}, While[m > 0, i = 1; While[c[i] <= m, i++]; i--; m -= c[i]; AppendTo[s, i]]; Divisible[n, Plus @@ IntegerDigits[Total[4^(s - 1)], 4]]]; Select[Range[2300], q[#] && q[#+1] &]

A352510 Starts of runs of 3 consecutive Catalan-Niven numbers (A352508).

Original entry on oeis.org

4, 55, 144, 145, 511, 2943, 6950, 7734, 9470, 9750, 15630, 15631, 35034, 35464, 41590, 41986, 64735, 68523, 68870, 77510, 81150, 90958, 106063, 118264, 119043, 135970, 139403, 163188, 164862, 164863, 171346, 181510, 200759, 202761, 202762, 208024, 209230, 209586
Offset: 1

Views

Author

Amiram Eldar, Mar 19 2022

Keywords

Examples

			4 is a term since 4, 5 and 6 are all Catalan-Niven numbers: the Catalan representation of 4, A014418(20) = 20, has the sum of digits 2+0 = 2 and 4 is divisible by 2, the Catalan representation of 5, A014418(5) = 100, has the sum of digits 1+0+0 = 1 and 5 is divisible by 1, and the Catalan representation of 6, A014418(6) = 101, has the sum of digits 1+0+1 = 2 and 6 is divisible by 2.
		

Crossrefs

Programs

  • Mathematica
    c[n_] := c[n] = CatalanNumber[n]; catNivQ[n_] := Module[{s = {}, m = n, i}, While[m > 0, i = 1; While[c[i] <= m, i++]; i--; m -= c[i]; AppendTo[s, i]]; Divisible[n, Plus @@ IntegerDigits[Total[4^(s - 1)], 4]]]; seq[count_, nConsec_] := Module[{cn = catNivQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {catNivQ[k]}]; k++]; s]; seq[30, 3]

A352511 Starts of runs of 4 consecutive Catalan-Niven numbers (A352508).

Original entry on oeis.org

144, 15630, 164862, 202761, 373788, 450189, 753183, 1403961, 1779105, 2588415, 2673774, 2814229, 2850880, 3009174, 3013722, 3045870, 3091023, 3702390, 3942519, 4042950, 4432128, 4725432, 4938348, 5718942, 5907312, 6268248, 6519615, 6592752, 6791379, 7095492, 8567802
Offset: 1

Views

Author

Amiram Eldar, Mar 19 2022

Keywords

Comments

Conjecture: There are no runs of 5 consecutive Catalan-Niven numbers (checked up to 10^9).

Examples

			144 is a term since 144, 145, 146 and 147 are all divisible by the sum of the digits in their Catalan representation:
    k  A014418(k)  A014420(k)  k/A014420(k)
  ---  ----------  ----------  ------------
  144      100210           4            36
  145      100211           5            29
  146      101000           2            73
  147      101001           3            49
		

Crossrefs

Programs

  • Mathematica
    c[n_] := c[n] = CatalanNumber[n]; catNivQ[n_] := Module[{s = {}, m = n, i}, While[m > 0, i = 1; While[c[i] <= m, i++]; i--; m -= c[i]; AppendTo[s, i]]; Divisible[n, Plus @@ IntegerDigits[Total[4^(s - 1)], 4]]]; seq[count_, nConsec_] := Module[{cn = catNivQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {catNivQ[k]}]; k++]; s]; seq[5, 4]

A244232 Sum of "digit values" in Semigreedy Catalan Representation of n, A244159.

Original entry on oeis.org

0, 1, 1, 2, 3, 1, 2, 2, 3, 4, 4, 5, 6, 4, 1, 2, 2, 3, 4, 2, 3, 3, 4, 5, 5, 6, 7, 5, 6, 6, 7, 8, 8, 9, 10, 8, 5, 6, 6, 7, 8, 6, 1, 2, 2, 3, 4, 2, 3, 3, 4, 5, 5, 6, 7, 5, 2, 3, 3, 4, 5, 3, 4, 4, 5, 6, 6, 7, 8, 6, 7, 7, 8, 9, 9, 10, 11, 9, 6, 7, 7, 8, 9, 7, 8, 8, 9, 10, 10, 11, 12, 10, 11, 11, 12, 13, 13, 14, 15, 13, 10, 11, 11, 12, 13, 11, 6, 7, 7, 8, 9, 7, 8, 8, 9, 10, 10, 11, 12, 10, 7, 8, 8, 9, 10, 8, 9, 9, 10, 11, 11, 12, 1
Offset: 0

Views

Author

Antti Karttunen, Jun 25 2014

Keywords

Comments

Note that a(33604) = A000217(10) = 55 because the sum is computed from the underlying list (vector) of numbers, and thus is not subject to any corruption by decimal representation as A244159 itself is.
Equivalent description: partition n "greedily" as terms of A197433, i.e. n = A197433(i) + A197433(j) + ... + A197433(k), always using the largest term of A197433 that still "fits in" (i.e. is <= n remaining). Then a(n) = A000120(i) + A000120(j) + ... + A000120(k).

Examples

			For n=18, using the alternative description, we see that it is partitioned  into the terms of A197433 as a greedy sum A197433(11) + A197433(1) = 17 + 1. Thus a(18) = A000120(11) + A000120(1) = 3+1 = 4.
For n=128, we see that is likewise represented as A197433(31) + A197433(31) = 64 + 64. Thus a(128) = 2*A000120(31) = 10.
		

Crossrefs

Formula

If A176137(n) = 1, a(n) = A000120(A244230(n)), otherwise a(n) = A000120(A244230(n)-1) + a(n-A197433(A244230(n)-1)).
For all n, a(A000108(n)) = 1. [And moreover, Catalan numbers, A000108, give all such k that a(k) = 1].
For all n, a(A014138(n)) = n and a(A014143(n)) = A000217(n+1).

A236855 a(n) is the sum of digits in A239903(n).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Apr 18 2014

Keywords

Examples

			As the 0th Catalan String is empty, indicated by A239903(0)=0, a(0)=0.
As the 18th Catalan String is [1,0,1,2] (A239903(18)=1012), a(18) = 1+0+1+2 = 4.
Note that although the range of validity of A239903 is inherently limited by the decimal representation employed, it doesn't matter here: We have a(58785) = 55, as the corresponding 58785th Catalan String is [1,2,3,4,5,6,7,8,9,10], even though A239903 cannot represent that unambiguously.
		

Crossrefs

Programs

  • Mathematica
    A236855list[m_] := With[{r = 2*Range[2, m]-1}, Reverse[Map[Total[r-#] &, Select[Subsets[Range[2, 2*m-1], {m-1}], Min[r-#] >= 0 &]]]];
    A236855list[6] (* Generates C(6) terms *) (* Paolo Xausa, Feb 19 2024 *)
  • Scheme
    (define (A236855 n) (apply + (A239903raw n)))
    (define (A239903raw n) (if (zero? n) (list) (let loop ((n n) (row (- (A081288 n) 1)) (col (- (A081288 n) 2)) (srow (- (A081288 n) 2)) (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))))))))
    ;; Alternative definition:
    (define (A236855 n) (let ((x (A071155 (A081291 n)))) (- (A034968 x) (A060130 x))))

Formula

a(n) = A034968(x) - A060130(x), where x = A071155(A081291(n)).
For up to n = A000108(11)-2 = 58784, a(n) = A007953(A239903(n)).
Catalan numbers, A000108, give the positions of ones, and the n-th triangular number occurs for the first time at the position immediately before that, i.e., a(A001453(n)) = A000217(n-1).
For each n, a(n) >= A000217(A236859(n)).
Showing 1-10 of 12 results. Next