A068156 -id:A068156 - cp's OEIS frontend

A256549 Triangle read by rows, T(n,k) = {n,k}*h(k), where {n,k} are the Stirling set numbers and h(k) = hypergeom([-k+1,-k],[],1), for n>=0 and 0<=k<=n.

1, 0, 1, 0, 1, 3, 0, 1, 9, 13, 0, 1, 21, 78, 73, 0, 1, 45, 325, 730, 501, 0, 1, 93, 1170, 4745, 7515, 4051, 0, 1, 189, 3913, 25550, 70140, 85071, 37633, 0, 1, 381, 12558, 124173, 526050, 1077566, 1053724, 394353, 0, 1, 765, 39325, 567210, 3482451, 10718946, 17386446, 14196708, 4596553

Offset: 0

Peter Luschny, Apr 12 2015

			Triangle starts:
[1]
[0, 1]
[0, 1,  3]
[0, 1,  9,   13]
[0, 1, 21,   78,   73]
[0, 1, 45,  325,  730,  501]
[0, 1, 93, 1170, 4745, 7515, 4051]

Cf. A000110, A000262, A048993, A068156, A075729.

A000262 = lambda n: simplify(hypergeometric([-n+1, -n], [], 1))
A256549 = lambda n,k: A000262(k)*stirling_number2(n,k)
for n in range(7): [A256549(n,k) for k in (0..n)]

Row sums are A075729.
Alternating row sums are the signed Bell numbers (-1)^n*A000110(n).
T(n,k) = A048993(n,k)*A000262(k).
T(n,n) = A000262(n).
T(n+2,2) = A068156(n).

A266533 First differences of A266532.

Original entry on oeis.org

1, 3, 3, 9, 3, 9, 9, 21, 3, 9, 9, 21, 9, 21, 21, 45, 3, 9, 9, 21, 9, 21, 21, 45, 9, 21, 21, 45, 21, 45, 45, 93, 3, 9, 9, 21, 9, 21, 21, 45, 9, 21, 21, 45, 21, 45, 45, 93, 9, 21, 21, 45, 21, 45, 45, 93, 21, 45, 45, 93, 45, 93, 93, 189, 3, 9, 9, 21, 9, 21, 21, 45, 9, 21, 21, 45, 21, 45, 45, 93

Offset: 1

David Applegate and Omar E. Pol, Jan 18 2016

Number of Y-toothpicks added at n-th stage in the structure of A266532.

A simplified version of A160121.

			Written as an irregular triangle in which the row lengths are the terms of A011782 the sequence begins:
1;
3;
3, 9;
3, 9, 9, 21;
3, 9, 9, 21, 9, 21, 21, 45;
3, 9, 9, 21, 9, 21, 21, 45, 9, 21, 21, 45, 21, 45, 45, 93;
...
Observation: at least the first 11 terms of the right border coincide with A068156.

David Applegate, The movie version
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Index entries for sequences related to cellular automata
Index entries for sequences related to toothpick sequences

Cf. A000120, A011782, A038573, A068156. A139251, A147582, A160121, A151710, A160721, A266532, A267701.

a(1) = 1. It appears that a(n) = 3*A038573(n-1), n >= 2.

A275970 a(n) = 3*2^n + n - 1.

Original entry on oeis.org

2, 6, 13, 26, 51, 100, 197, 390, 775, 1544, 3081, 6154, 12299, 24588, 49165, 98318, 196623, 393232, 786449, 1572882, 3145747, 6291476, 12582933, 25165846, 50331671, 100663320, 201326617, 402653210, 805306395, 1610612764, 3221225501, 6442450974, 12884901919, 25769803808, 51539607585, 103079215138, 206158430243, 412316860452, 824633720869

Offset: 0

Miquel Cerda, Aug 15 2016

Carauleanu Marc, Table of n, a(n) for n = 0..400
S. W. Golomb, Properties of the sequence 3.2^n+1, Math. Comp., 30 (1976), 657-663.
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

LinearRecurrence[{4,-5,2},{2,6,13}, 25] (* or *) Table[3*2^n + n - 1, {n,0,25}] (* G. C. Greubel, Aug 18 2016 *)

a(n)=3*2^n+n-1 \\ Charles R Greathouse IV, Aug 27 2016

a(n) = 2*a(n-1) - n + 2.
a(n+1) - a(n) = A181565(n)
a(n) = A007283(n) + n - 1
a(n) = A083706(n) + A000079(n)
a(n) = A145071(n+1) - A000079(n)
a(n) = A079583(n) + A005408(n)
a(n) = A068156(n+1) - A079583(n)
a(n) = (A068156(n+1) + A005408(n)) / 2
a(n) = A000225(n) + A000325(n+1) + A005408(n)
a(n) = A068156(n+1) - A000225(n) - A000325(n+1)
a(n) = A068156(n+1) - A007283(n) + n + 2.
a(n) = A000079(n) + A000225(n) + A000295(n) + A005408(n)
From G. C. Greubel, Aug 18 2016: (Start)
O.g.f.: (2 - 2*x - x^2)/( (1-2*x)*(1-x)^2 ).
E.g.f.: 3*exp(2*x) + (x-1)*exp(x).
a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-2). (End)

A324548 Maximal value which A324543 attains among the divisors of n.

Original entry on oeis.org

0, 1, 3, 3, 7, 3, 15, 4, 9, 7, 31, 3, 63, 15, 8, 16, 127, 9, 255, 7, 21, 31, 511, 8, 21, 63, 12, 27, 1023, 8, 2047, 16, 31, 127, 20, 20, 4095, 255, 78, 32, 8191, 21, 16383, 31, 9, 511, 32767, 16, 45, 21, 127, 63, 65535, 12, 53, 27, 270, 1023, 131071, 8, 262143, 2047, 21, 72, 63, 31, 524287, 127, 511, 20, 1048575, 20, 2097151, 4095, 21

Offset: 1

Antti Karttunen, Mar 07 2019

nonn

			Divisors of 161051 are [1, 11, 121, 1331, 14641, 161051]. Applying A324543 to these gives the values [0, 31, 93, 124, 496, 248]. Of these 496 is the largest, thus a(161051) = 496.

Antti Karttunen, Table of n, a(n) for n = 1..4473

Cf. A324543.

A324548(n) = vecmax(apply(A324543,divisors(n))); \\ Needs also code from A324543.

a(n) = Max_{d|n} A324543(d).
a(A000040(n)) = A000225(n).
a(A001248(n)) = A068156(n) = 3*(2^n - 1).

A126269 Numbers n such that hcl(n,n) < hcl(n,n-1) where hcl(n,i) is the Huffman code length; see comments.

Original entry on oeis.org

3, 4, 9, 10, 21, 22, 45, 46, 93, 94, 189, 190, 381, 382, 765, 766, 1533, 1534, 3069, 3070, 6141, 6142, 12285, 12286, 24573, 24574, 49149, 49150

Offset: 3

Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Dec 22 2006

Consider a string which consists of n distinct symbols such that symbol(i) has frequency i (i=1,2,...,n). Then hcl(n,i) is the Huffman code length of symbol(i).

			Possible Huffman codes for n = 3,4,5 are:
1 : 00
2 : 01
3 : 1
--------
1 : 100
2 : 101
3 : 11
4 : 0
--------
1 : 000
2 : 001
3 : 01
4 : 10
5 : 11
hcl(3,3)=1 < 2=hcl(3,2) and hcl(4,4)=1 < 2=hcl(4,3); so 3,4 are in the sequence.
hcl(5,5)=2=hcl(5,4) so 5 is not in the sequence.

Sean A. Irvine, Java program (github)

Cf. A126268 A126014 A068156 A033484.

Conjecture: a(2k) = A033484(k-1) and a(2k-1) = A068156(k-1), k >= 2.
Conjectures from Colin Barker, Aug 06 2019: (Start)
G.f.: x^3*(3 + 4*x - 2*x^3) / ((1 - x)*(1 + x)*(1 - 2*x^2)).
a(n) = 3*a(n-2) - 2*a(n-4) for n>6.
a(n) = -5/2 + (-1)^n/2 + 3*2^((1/2)*(n-5))*(2-2*(-1)^n + sqrt(2) + (-1)^n*sqrt(2)) for n>2.
(End)

More terms from Sean A. Irvine, Aug 05 2019

A159290 A generalized Jacobsthal sequence.

Original entry on oeis.org

3, 5, 13, 25, 53, 105, 213, 425, 853, 1705, 3413, 6825, 13653, 27305, 54613, 109225, 218453, 436905, 873813, 1747625, 3495253, 6990505, 13981013, 27962025, 55924053, 111848105, 223696213, 447392425, 894784853, 1789569705, 3579139413

Offset: 0

Creighton Dement, Apr 08 2009

Sequence generated by the floretion: X*Y with X = 0.5('i + 'j + 'k + 'ee') and Y = 0.5(i' + j' + k' + 'ij' + 'ik' + 'ji' + 'jk' + 'ki' + 'kj' + 'ee')

G. C. Greubel, Table of n, a(n) for n = 0..1000
Creighton Dement, Online Floretion Multiplier [broken link]
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

A083943, A068156

[-1 + (2*(-1)^n + 5*2^(n+1))/3: n in [0..50]]; // G. C. Greubel, Jun 27 2018

LinearRecurrence[{2, 1, -2}, {3, 5, 13}, 50] (* or *) Table[-1 + (2*(-1)^n + 5*2^(n+1))/3, {n,0,30}] (* G. C. Greubel, Jun 27 2018 *)

x='x+O('x^50); Vec((3-x)/(-x^2+1-2*x+2*x^3)) \\ G. C. Greubel, Jun 27 2018

a(n) = -1 + (2*(-1)^n + 5*2^(n+1))/3.
G.f.: (3-x)/((1-x)*(1+x)*(1-2*x)).
a(n) = 3*A000975(n+1) - A000975(n). - R. J. Mathar, Sep 11 2019
a(n)+a(n+1) = A051633(n+1). - R. J. Mathar, Mar 23 2023

A178831 Rectangular array T(n,k) = binomial(n+1,2)*(n^k - (n-1)^k) read by antidiagonals.

Original entry on oeis.org

1, 1, 3, 1, 9, 6, 1, 21, 30, 10, 1, 45, 114, 70, 15, 1, 93, 390, 370, 135, 21, 1, 189, 1266, 1750, 915, 231, 28, 1, 381, 3990, 7810, 5535, 1911, 364, 36, 1, 765, 12354, 33670, 31515, 14091, 3556, 540, 45, 1, 1533, 37830, 141970, 172935, 97671, 30940, 6084, 765, 55

Offset: 1

Geoffrey Critzer, Dec 27 2010

T(n,k) is the sum of the elements in the image sets of all functions f:{1,2,...,k}->{1,2,...,n}.

Equivalently, the sum of the distinct entries in each length k sequence on {1,2,...,n}.

			Array begins
   1,   1,    1,     1,     1,      1, ...
   3,   9,   21,    45,    93,    189, ...
   6,  30,  114,   390,  1266,   3990, ...
  10,  70,  370,  1750,  7810,  33670, ...
  15, 135,  915,  5535, 31515, 172935, ...
  21, 231, 1911, 14091, 97671, 651651, ...

G. C. Greubel, Antidiagonals n = 1..100, flattened

Cf. A068156 the case for n=2.

T:=Flat(List([1..10], n->List([0..n-1], k-> Binomial(k+2, 2)*( (k+1)^(n-k) -k^(n-k)) ))); # G. C. Greubel, Jan 22 2019

[[Binomial(k+2,2)*((k+1)^(n-k) -k^(n-k)): k in [0..n-1]]: n in [1..10]]; // G. C. Greubel, Jan 22 2019

Table[Range[7]! Rest[CoefficientList[Series[Binomial[n+1,2] Exp[(n-1)x](Exp[x]-1),{x,0,7}],x]],{n,1,7}]//Grid
T[n_, k_]:= Binomial[n+2, 2]*((n+1)^k -n^k); Table[T[k, n-k], {n, 1, 10}, {k, 0, n-1}] (* G. C. Greubel, Jan 22 2019 *)

{T(n, k) = binomial(n+2, 2)*((n+1)^k -(n)^k)};
for(n=1,10, for(k=0,n-1, print1(T(k,n-k), ", "))) \\ G. C. Greubel, Jan 22 2019

[[binomial(k+2,2)*((k+1)^(n-k) -k^(n-k)) for k in (0..n-1)] for n in (1..10)] # G. C. Greubel, Jan 22 2019

E.g.f. for row n: binomial(n+1,2)*exp((n-1)*x)*(exp(x) - 1).

Terms a(29) onward added by G. C. Greubel, Jan 22 2019

A182659 A canonical permutation designed to thwart a certain naive attempt to guess whether sequences are permutations.

Original entry on oeis.org

0, 2, 3, 1, 5, 6, 7, 8, 9, 4, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 10, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 22, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68

Offset: 0

Sam Alexander, Nov 26 2010

A naive way to guess whether a function f:N->N is a permutation, based on just an initial subsequence (f(0),...,f(n)), is to guess "no" if (f(0),...,f(n)) contains a repeated entry or if there is some i in {0,...,n} such that i is not in {f(0),...,f(n)} and 2 i<=n; and guess "yes" otherwise. a(n) thwarts that method, causing it to change its mind infinitely often as n->infinity.
a(0)=0. Suppose a(0),...,a(n) have been defined.
1. If the above method guesses that (a(0),...,a(n)) is NOT an initial subsequence of a permutation, then unmark any "marked" numbers.
2. If the above method guesses that (a(0),...,a(n)) IS an initial subsequence of a permutation, then "mark" the smallest number not in {a(0),...,a(n)}.
3. Let a(n+1) be the least unmarked number not in {a(0),...,a(n)}.
A030301 can be derived by a similar method, where instead of trying to guess whether sequences are permutations, the naive victim is trying to guess whether sequences contain infinitely many 0s.

S. Alexander, On Guessing Whether A Sequence Has A Certain Property, arXiv:1011.6626 [math.LO], 2010-2012, J. Int. Seq. 14 (2011) # 11.4.4.

Cf. A030301, A082691, A182660, A068156, A033484.

a(0) = 0; if n = A068156(k+1) = 6*2^k - 3 for some k >= 0 then a(n) = A033484(k) = (n-1)/2; otherwise, a(n) = n+1. - Andrey Zabolotskiy, Feb 27 2025

a(22) corrected and further terms added by Andrey Zabolotskiy, Feb 27 2025

A279389 3 times Mersenne primes A000668.

Original entry on oeis.org

9, 21, 93, 381, 24573, 393213, 1572861, 6442450941, 6917529027641081853, 1856910058928070412348686333, 486777830487640090174734030864381, 510423550381407695195061911147652317181

Offset: 1

Omar E. Pol, Dec 20 2016

nonn

Also sum of n-th Mersenne prime and the radical of n-th even perfect number.

The binary representation of a(n) has only two zeros, starting with "10" and ending with "01". The sequence begins: 1001, 10101, 1011101, 101111101, 101111111111101,...

Cf. A000010, A000203, A000396, A000668, A051027, A147757.
Subsequence of A001748, and of A147758, and of A174055, and possibly of other sequences, see below:
Cf. A060482, A068156, A072087, A136252, A144482, A144856, A249780, A269633.

a(n) = 3*A000668(n) = A000668(n) + A139257(n).

a(n) = phi(M(n)) + sigma(sigma(M(n))) = A000010(A000668(n)) + A000203(A000203(A000668(n))) = A000010(A000668(n)) + A051027(A000668(n)).

A362809 Numbers k for which the area of the first part of the symmetric representation of sigma(k) equals sigma(k)/3 and its width is 1.

Original entry on oeis.org

15, 207, 1023, 2975, 5950, 19359, 147455, 294910, 1207359, 5017599, 2170814463

Offset: 1

Hartmut F. W. Hoft, May 04 2023

The symmetric representation of sigma(k), SRS(k), of every term k in this sequence consists of at least 3 parts, with a(1) = 15 and a(5) = 5950 being the only ones among the first 11 terms for which the SRS consists of exactly 3 parts. A251820 is a subsequence. a(12) > 5*10^9.
Suppose that a(n) = 2^i * q, i >= 0 and q odd. Because the first part of SRS(a(n)) has width 1, the smallest prime factor p of q satisfies p > 2^(i+1) -- see the locations of 1's in the triangle of A237048 and computation of widths in A249223. The area of the first part of SRS(a(n)) is (2^(i+1) - 1) * (q+1)/2 = (a(n) + 2^i) * (2^(i+1) - 1) / 2^(i+1) since the first part has 2^(i+1) - 1 legs (see the formula in A237591). Therefore, when 2^i * q is in the sequence then 2^j * q, for 0 <= j <= i is also. Sigma(a(n)) = A068156(i+1) * (a(n) + 2^i) / 2^(i+1).
Conjecture: The area of the first part of SRS(n) being equal to sigma(n)/3 implies that the first part has width 1.
This is true for all odd a(n) since their first part consists of a single leg of width 1. It also holds for even numbers through 10^7.
Observation: Consider the known 11 terms. Apart from 5950 and 294910 the rest are also in A063906. Question: Is A063906 a subsequence? - Omar E. Pol, Jul 09 2023
Answer: A063906 consists of the odd terms of this sequence since the first part of the symmetric representation of sigma for odd a(n) equals (a(n)+1)/2 which is equivalent to a(n) = 2*sigma(a(n))/3 - 1, i.e., a(n) is in A063906. - Hartmut F. W. Hoft, Jul 10 2023
A063906(10) = 58946212863 is a term here. - Omar E. Pol, Jul 12 2023

			15 belongs to the sequence since SRS(15) consists of the parts {8, 8, 8} of maximum widths {1, 2, 1} and sigma(15) = 24.
294910 belongs to the sequence since SRS(294910) consists of the 5 parts {221184, 109440, 2304, 109440, 221184} of maximum widths {1, 3, 2, 3, 1}, with 109440 + 2304 + 109440 = 211184 and sigma(294910) = 3 * 211184 = 663552.
From _Omar E. Pol_, Jul 07 2023: (Start)
Illustration of a(1) = 15.
The 14th row of triangle A237593 is [8, 3, 1, 2, 2, 1, 3, 8] and the 15th row of the same triangle is [8, 3, 2, 1, 1, 1, 1, 2, 3, 8] so the diagram of the symmetric representation of sigma(15) in the fourth quadrant is constructed as shown below:
.                                _
.                               | |
.                               | |
.                               | |
.                               | |
.                               | |
.                               | |
.                               | |
.                          _ _ _|_|
.                      _ _| |      8
.                     |    _|
.                    _|  _|
.                   |_ _|  8
.                   |
.    _ _ _ _ _ _ _ _|
.   |_ _ _ _ _ _ _ _|
.                    8
.
The area of the first part (or polygon) of the diagram equals sigma(15)/3 = 24/3 = 8 and its width is 1 so 15 is in the sequence. (End)

Cf. A068156, A235791, A237048, A237591, A237593, A249223, A251820.

Cf. A063906.

(* substitute code suggested by Andrey Zabolotskiy *)
cd[n_, k_] := Boole[Divisible[n, k]]
a237048[s_, j_] := If[OddQ[j], cd[s, j], cd[s-j/2, j]]
firstZeroQ[s_, a_] := Sum[(-1)^(j+1)a237048[s, j], {j, a}]==0
evenPart[n_] := 2^IntegerExponent[n, 2]
a362809[{m_, n_}] := Module[{a, b}, Select[Range[m, n], (a=evenPart[#]; b=(2a-1)/(2a); DivisorSigma[1, #]==3b(#+a)&&firstZeroQ[#, 2a])&]]
a362809[{1, 2170814463}] (* a(11) has a long computation time *)

cp's OEIS Frontend

A256549 Triangle read by rows, T(n,k) = {n,k}*h(k), where {n,k} are the Stirling set numbers and h(k) = hypergeom([-k+1,-k],[],1), for n>=0 and 0<=k<=n.

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A266533 First differences of A266532.

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A275970 a(n) = 3*2^n + n - 1.

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Mathematica

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A324548 Maximal value which A324543 attains among the divisors of n.

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A126269 Numbers n such that hcl(n,n) < hcl(n,n-1) where hcl(n,i) is the Huffman code length; see comments.

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A159290 A generalized Jacobsthal sequence.

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Magma

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PARI

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A178831 Rectangular array T(n,k) = binomial(n+1,2)*(n^k - (n-1)^k) read by antidiagonals.

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GAP

Magma

Mathematica

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Sage

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A182659 A canonical permutation designed to thwart a certain naive attempt to guess whether sequences are permutations.

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