A000346 -id:A000346 - cp's OEIS frontend

A345916 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum <= 0.

0, 3, 5, 9, 10, 13, 15, 17, 18, 23, 25, 29, 33, 34, 36, 39, 41, 43, 45, 46, 49, 50, 53, 55, 57, 58, 61, 63, 65, 66, 68, 71, 75, 77, 78, 81, 85, 89, 90, 95, 97, 98, 103, 105, 109, 113, 114, 119, 121, 125, 129, 130, 132, 135, 136, 139, 141, 142, 145, 147, 149

Offset: 1

Gus Wiseman, Jul 08 2021

nonn

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The sequence of terms together with the corresponding compositions begins:
     0: ()
     3: (1,1)
     5: (2,1)
     9: (3,1)
    10: (2,2)
    13: (1,2,1)
    15: (1,1,1,1)
    17: (4,1)
    18: (3,2)
    23: (2,1,1,1)
    25: (1,3,1)
    29: (1,1,2,1)
    33: (5,1)
    34: (4,2)
    36: (3,3)

The version for Heinz numbers of partitions is A000290.
These compositions are counted by A058622.
These are the positions of terms <= 0 in A344618.
The opposite (k >= 0) version is A345914.
The version for unreversed alternating sum is A345915.
The strictly negative (k < 0) version is A345920.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A008549, A025047, A027187, A028260, A032443, A114121, A163493, A344607, A344610, A345908.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Select[Range[0,100],sats[stc[#]]<=0&]

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A002664 a(n) = 2^n - C(n,0)- ... - C(n,4).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 7, 29, 93, 256, 638, 1486, 3302, 7099, 14913, 30827, 63019, 127858, 258096, 519252, 1042380, 2089605, 4185195, 8377705, 16764265, 33539156, 67090962, 134196874, 268411298, 536843071, 1073709893

Offset: 0

N. J. A. Sloane

From Gary W. Adamson, Jul 24 2010: (Start)
Starting with "1" = eigensequence of a triangle with binomial C(n,5):
(1, 6, 21, 56, ...) as the left border and the rest 1's. (End)
The Kn26 sums, see A180662, of triangle A065941 equal the terms (doubled) of this sequence minus the five leading zeros. - Johannes W. Meijer, Aug 15 2011
Starting (0, 0, 0, 0, 1, 7, 29, ...), this is the binomial transform of (0, 0, 0, 0, 1, 2, 2, 2, ...). Starting (1, 7, 29, ...), this is the binomial transform of (1, 6, 16, 26, 31, 32, 32, 32, ...). - Gary W. Adamson, Jul 28 2015

J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, 1995, Chapter 3, pp. 76-79.
J. Eckhoff, Der Satz von Radon in konvexen Productstrukturen II, Monat. f. Math., 73 (1969), 7-30.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. K. Guy, Letter to N. J. A. Sloane
Ângela Mestre, José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
H. P. Robinson, Letter to N. J. A. Sloane, Mar 21 1985
Index entries for linear recurrences with constant coefficients, signature (7,-20,30,-25,11,-2).

a(n) = A055248(n, 5). Partial sums of A002663.
Cf. A000079, A000225, A000295, A002662, A002663, A035038-A035042.
Cf. A007318.

a002664 n = a002664_list !! n
a002664_list = map (sum . drop 5) a007318_tabl
-- Reinhard Zumkeller, Jun 20 2015

[2^n-n^4/24+n^3/12-11*n^2/24-7*n/12-1: n in [0..35]]; // Vincenzo Librandi, May 20 2011

a:=n->sum(binomial(n+1,2*j),j=3..n+1): seq(a(n), n=0..30); # Zerinvary Lajos, May 12 2007
A002664:=1/(2*z-1)/(z-1)**5; # conjectured by Simon Plouffe in his 1992 dissertation

a=1;lst={};s1=s2=s3=s4=s5=0;Do[s1+=a;s2+=s1;s3+=s2;s4+=s3;s5+=s4;AppendTo[lst,s5];a=a*2,{n,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Jan 10 2009 *)
Table[Sum[ Binomial[n, k + 5], {k, 0, n}], {n, 0, 30}] (* Zerinvary Lajos, Jul 08 2009 *)
Table[2^n-Total[Binomial[n,Range[0,4]]],{n,0,30}] (* or *) LinearRecurrence[ {7,-20,30,-25,11,-2},{0,0,0,0,0,1},40] (* Harvey P. Dale, Sep 03 2016 *)

G.f.: x^5/((1-2*x)*(1-x)^5).
a(n) = Sum_{k=0..n} C(n, k+5) = Sum_{k=5..n} C(n, k); a(n) = 2a(n-1) + C(n-1, 4). - Paul Barry, Aug 23 2004
a(n) = 2^n - n^4/24 + n^3/12 - 11*n^2/24 - 7*n/12 - 1.  - Bruno Berselli, May 19 2011 [Robinson (1985) gives an alternative version of this formula, for a different offset. - N. J. A. Sloane, Oct 20 2015]
E.g.f.: exp(x)*(24*(exp(x) - 1) - 24*x - 12*x^2 - 4*x^3 - x^4)/24. - Stefano Spezia, Mar 09 2025

A269920 Triangle read by rows: T(n,f) is the number of rooted maps with n edges and f faces on an orientable surface of genus 0.

Original entry on oeis.org

1, 1, 1, 2, 5, 2, 5, 22, 22, 5, 14, 93, 164, 93, 14, 42, 386, 1030, 1030, 386, 42, 132, 1586, 5868, 8885, 5868, 1586, 132, 429, 6476, 31388, 65954, 65954, 31388, 6476, 429, 1430, 26333, 160648, 442610, 614404, 442610, 160648, 26333, 1430

Offset: 0

Gheorghe Coserea, Mar 14 2016

Row n contains n+1 terms.

			Triangle starts:
n\f    [1]     [2]     [3]     [4]     [5]     [6]     [7]     [8]
[0]    1;
[1]    1,      1;
[2]    2,      5,      2;
[3]    5,      22,     22,     5;
[4]    14,     93,     164,    93,     14;
[5]    42,     386,    1030,   1030,   386,    42;
[6]    132,    1586,   5868,   8885,   5868,   1586,   132;
[7]    429,    6476,   31388,  65954,  65954,  31388,  6476,   429;
[8]    ...

Gheorghe Coserea, Rows n = 0..200, flattened
Sean R. Carrell, Guillaume Chapuy, Simple recurrence formulas to count maps on orientable surfaces, arXiv:1402.6300 [math.CO], 2014.

Columns k=1-6 give: A000108, A000346, A000184, A000365, A000473, A000502.
Row sums give A000168 (column 0 of A269919).
Cf. A006294 (row maxima).

Q[0, 1, 0] = 1; Q[n_, f_, g_] /; n<0 || f<0 || g<0 = 0;
Q[n_, f_, g_] := Q[n, f, g] = 6/(n+1) ((2n-1)/3 Q[n-1, f, g] + (2n-1)/3 Q[n - 1, f-1, g] + (2n-3) (2n-2) (2n-1)/12 Q[n-2, f, g-1] + 1/2 Sum[l = n-k; Sum[v = f-u; Sum[j = g-i; Boole[l >= 1 && v >= 1 && j >= 0] (2k-1) (2l-1) Q[k-1, u, i] Q[l-1, v, j], {i, 0, g}], {u, 1, f}], {k, 1, n}]);
Table[Q[n, f, 0], {n, 0, 8}, {f, 1, n+1}] // Flatten (* Jean-François Alcover, Aug 10 2018 *)

N = 8; G = 0; gmax(n) = min(n\2, G);
Q = matrix(N + 1, N + 1);
Qget(n, g) = { if (g < 0 || g > n/2, 0, Q[n+1, g+1]) };
Qset(n, g, v) = { Q[n+1, g+1] = v };
Quadric({x=1}) = {
  Qset(0, 0, x);
  for (n = 1, length(Q)-1, for (g = 0, gmax(n),
    my(t1 = (1+x)*(2*n-1)/3 * Qget(n-1, g),
       t2 = (2*n-3)*(2*n-2)*(2*n-1)/12 * Qget(n-2, g-1),
       t3 = 1/2 * sum(k = 1, n-1, sum(i = 0, g,
       (2*k-1) * (2*(n-k)-1) * Qget(k-1, i) * Qget(n-k-1, g-i))));
    Qset(n, g, (t1 + t2 + t3) * 6/(n+1))));
};
Quadric('x);
concat(apply(p->Vecrev(p/'x), vector(N+1 - 2*G, n, Qget(n-1 + 2*G, G))))

A045621 a(n) = 2^n - binomial(n, floor(n/2)).

Original entry on oeis.org

0, 1, 2, 5, 10, 22, 44, 93, 186, 386, 772, 1586, 3172, 6476, 12952, 26333, 52666, 106762, 213524, 431910, 863820, 1744436, 3488872, 7036530, 14073060, 28354132, 56708264, 114159428, 228318856, 459312152, 918624304, 1846943453, 3693886906

Offset: 0

David M Bloom, Brooklyn College

nonn

p(n) = a(n)/2^n is the probability that a majority of heads had occurred at some point after n flips of a fair coin. For example, after 3 flips of a coin, the probability is 5/8 that a majority of heads had occurred at some point. (First flip is heads, p=1/2, or sequence THH, p=1/8.) - Brian Galebach, May 14 2001
Hankel transform is (-1)^n*n. - Paul Barry, Jan 11 2007
Hankel transform of a(n+1) is A127630. - Paul Barry, Sep 01 2009
a(n) is the number of n-step walks on the number line that are positive at some point along the walk. - Benjamin Phillabaum, Mar 06 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..247
Kairi Kangro, Mozhgan Pourmoradnasseri, Dirk Oliver Theis, Short note on the number of 1-ascents in dispersed dyck paths, arXiv:1603.01422 [math.CO], 2016.
S. Mason and J. Parsley, A geometric and combinatorial view of weighted voting, arXiv preprint arXiv:1109.1082 [math.CO], 2011.

Cf. A000108, A001405, A000346, A054336, A068551.

List([0..35], n-> 2^n - Binomial(n, Int(n/2)) ); # G. C. Greubel, Jan 13 2020

[2^n - Binomial(n, Floor(n/2)): n in [0..35]]; // Bruno Berselli, Mar 08 2011

seq( 2^n -binomial(n,floor(n/2)), n=0..35); # G. C. Greubel, Jan 13 2020

Table[2^n - Binomial[n, Floor[n/2]], {n, 0, 35}] (* Roger L. Bagula, Aug 26 2006 *)

{a(n)=if(n<0, 0, 2^n -binomial(n, n\2))} /* Michael Somos, Oct 31 2006 */

[2^n -binomial(n,floor(n/2)) for n in (0..35)] # G. C. Greubel, Jan 13 2020

a(n) = 2^n - A001405(n).
a(2*k) = 2*a(2*k-1), a(2*k+1) = 2*a(2*k) + Catalan(k).
a(n+1) = b(0)*b(n)+b(1)*b(n-1)+...+b(n)*b(0), b(k)=C(k, [ k/2 ]).
G.f.: c(x^2)*x/(1-2*x) where c(x) = g.f. for Catalan numbers A000108.
a(n) = A054336(n, 1) (second column of triangle).
E.g.f.: exp(2*x) - I_0(2*x) - I_1(2*x) where I_n(x) is n-th modified Bessel function as a function of x. - Benjamin Phillabaum, Mar 06 2011
a(2*n+1) = A000346(n); a(2*n) = A068551(n). - Emeric Deutsch, Nov 16 2003
a(n) = Sum_{k=0..n-1} binomial(n, floor(k/2)). - Paul Barry, Aug 05 2004
a(n+1) = 2*a(n) + Catalan(n/2)*(1+(-1)^n)/2. - Paul Barry, Aug 05 2004
a(n+1) = Sum_{k=0..floor(n/2)} 2^(n-2*k)*A000108(k). - Paul Barry, Sep 01 2009
(n+1)*a(n) +2*(-n-1)*a(n-1) +4*(-n+2)*a(n-2) +8*(n-2)*a(n-3) = 0. - R. J. Mathar, Dec 02 2012

Edited by N. J. A. Sloane, Oct 08 2006

Adjustments to formulas (correcting offsets) from Michael Somos, Oct 31 2006

A345908 Traces of the matrices (A345197) counting integer compositions by length and alternating sum.

Original entry on oeis.org

1, 1, 0, 1, 3, 3, 6, 15, 24, 43, 92, 171, 315, 629, 1218, 2313, 4523, 8835, 17076, 33299, 65169

Offset: 0

Gus Wiseman, Jul 26 2021

The matrices (A345197) count the integer compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2. Here, the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. So a(n) is the number of compositions of n of length (n + s)/2, where s is the alternating sum of the composition.

			The a(0) = 1 through a(7) = 15 compositions of n = 0..7 of length (n + s)/2 where s = alternating sum (empty column indicated by dot):
  ()  (1)  .  (2,1)  (2,2)    (2,3)    (2,4)      (2,5)
                     (1,1,2)  (1,2,2)  (1,3,2)    (1,4,2)
                     (2,1,1)  (2,2,1)  (2,3,1)    (2,4,1)
                                       (1,1,3,1)  (1,1,3,2)
                                       (2,1,2,1)  (1,2,3,1)
                                       (3,1,1,1)  (2,1,2,2)
                                                  (2,2,2,1)
                                                  (3,1,1,2)
                                                  (3,2,1,1)
                                                  (1,1,1,1,3)
                                                  (1,1,2,1,2)
                                                  (1,1,3,1,1)
                                                  (2,1,1,1,2)
                                                  (2,1,2,1,1)
                                                  (3,1,1,1,1)

Traces of the matrices given by A345197.
Diagonals and antidiagonals of the same matrices are A346632 and A345907.
Row sums of A346632.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Other diagonals are A008277 of A318393 and A055884 of A320808.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A001405, A007318, A008549, A025047, A114121, A163493.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==(n+ats[#])/2&]],{n,0,15}]

A348614 Numbers k such that the k-th composition in standard order has sum equal to twice its alternating sum.

Original entry on oeis.org

0, 9, 11, 14, 130, 133, 135, 138, 141, 143, 148, 153, 155, 158, 168, 177, 179, 182, 188, 208, 225, 227, 230, 236, 248, 2052, 2057, 2059, 2062, 2066, 2069, 2071, 2074, 2077, 2079, 2084, 2089, 2091, 2094, 2098, 2101, 2103, 2106, 2109, 2111, 2120, 2129, 2131

Offset: 1

Gus Wiseman, Oct 29 2021

nonn

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			The terms together with their binary indices begin:
    0: ()
    9: (3,1)
   11: (2,1,1)
   14: (1,1,2)
  130: (6,2)
  133: (5,2,1)
  135: (5,1,1,1)
  138: (4,2,2)
  141: (4,1,2,1)
  143: (4,1,1,1,1)
  148: (3,2,3)
  153: (3,1,3,1)
  155: (3,1,2,1,1)
  158: (3,1,1,1,2)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The unordered case (partitions) is counted by A000712, reverse A006330.
These compositions are counted by A262977.
Except for 0, a subset of A345917 (which is itself a subset of A345913).
A000346 = even-length compositions with alt sum != 0, complement A001700.
A011782 counts compositions.
A025047 counts wiggly compositions, ranked by A345167.
A034871 counts compositions of 2n with alternating sum 2k.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A116406 counts compositions with alternating sum >=0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
A345197 counts compositions by length and alternating sum.
Cf. A000984, A002458, A004331, A005810, A224274.
Cf. A008549, A013777, A027306, A058622, A088218, A114121, A120452, A126869, A163493, A294175, A344604.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,1000],Total[stc[#]]==2*ats[stc[#]]&]

A054143 Triangular array T given by T(n,k) = Sum_{0 <= j <= i-n+k, n-k <= i <= n} C(i,j) for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 4, 7, 1, 5, 11, 15, 1, 6, 16, 26, 31, 1, 7, 22, 42, 57, 63, 1, 8, 29, 64, 99, 120, 127, 1, 9, 37, 93, 163, 219, 247, 255, 1, 10, 46, 130, 256, 382, 466, 502, 511, 1, 11, 56, 176, 386, 638, 848, 968, 1013, 1023, 1, 12, 67, 232, 562, 1024, 1486, 1816, 1981, 2036, 2047

Offset: 0

Clark Kimberling, Mar 18 2000

Row sums given by A001787.
T(n, n) = -1 + 2^(n+1).
T(2*n, n) = 4^n.
T(2*n+1, n) = A000346(n).
T(2*n-1, n) = A032443(n).
A054143 is the fission of the polynomial sequence ((x+1)^n) by the polynomial sequence (q(n,x)) given by q(n,x) = x^n + x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,  3;
  1,  4,  7;
  1,  5, 11, 15;
  1,  6, 16, 26, 31;
  1,  7, 22, 42, 57, 63;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A000346, A001787, A032443.

Diagonal sums give A005672. - Paul Barry, Feb 07 2003

Flat(List([0..12], n-> List([0..n], k-> Sum([n-k..n], i-> Sum([0..i-n+k], j-> Binomial(i,j) ))))); # G. C. Greubel, Aug 01 2019

T:= func< n,k | (&+[ (&+[ Binomial(i,j): j in [0..i-n+k]]): i in [n-k..n]]) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019

A054143_row := proc(n) add(add(binomial(n,n-i)*x^(k+1),i=0..k),k=0..n-1); coeffs(sort(%)) end; seq(print(A054143_row(n)),n=1..6); # Peter Luschny, Sep 29 2011

(* First program *)
z=10;
p[n_,x_]:=(x+1)^n;
q[0,x_]:=1;q[n_,x_]:=x*q[n-1,x]+1;
p1[n_,k_]:=Coefficient[p[n,x],x^k];p1[n_,0]:=p[n,x]/.x->0;
d[n_,x_]:=Sum[p1[n,k]*q[n-1-k,x],{k,0,n-1}]
h[n_]:=CoefficientList[d[n,x],{x}]
TableForm[Table[Reverse[h[n]],{n,0,z}]]
Flatten[Table[Reverse[h[n]],{n,-1,z}]] (* A054143 *)
TableForm[Table[h[n],{n,0,z}]]
Flatten[Table[h[n],{n,-1,z}]] (* A104709 *)
(* Second program *)
Table[Sum[Binomial[i, j], {i, n-k, n}, {j,0,i-n+k}], {n,0,12}, {k,0,n}]// Flatten (* G. C. Greubel, Aug 01 2019 *)

T(n,k) = sum(i=n-k,n, sum(j=0,i-n+k, binomial(i,j)));
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Aug 01 2019

def T(n, k): return sum(sum( binomial(i,j) for j in (0..i-n+k)) for i in (n-k..n))
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

T(n,k) = Sum_{0 <= j <= i-n+k, n-k <= i <= n} binomial(i,j).
T(n,k) = T(n-1,k) + 3*T(n-1,k-1) - 2*T(n-2,k-1) - 2*T(n-2,k-2), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 30 2013
From Petros Hadjicostas, Jun 05 2020: (Start)
Bivariate o.g.f.: Sum_{n,k >= 0} T(n,k)*x^n*y^k = 1/(1 - x - 3*x*y + 2*x^2*y + 2*x^2*y^2) = 1/((1 - 2*x*y)*(1 - x*(y+1))).
n-th row o.g.f.: ((1 + y)^(n+1) - (2*y)^(n+1))/(1 - y). (End)

Name edited by Petros Hadjicostas, Jun 04 2020

A213118 T(n,k)=Number of binary arrays of length n+2*k-1 with fewer than k ones in any length 2k subsequence (=less than 50% duty cycle).

Original entry on oeis.org

1, 5, 1, 22, 7, 1, 93, 34, 10, 1, 386, 151, 54, 14, 1, 1586, 646, 252, 86, 19, 1, 6476, 2710, 1110, 424, 136, 26, 1, 26333, 11236, 4748, 1926, 714, 212, 36, 1, 106762, 46231, 19964, 8404, 3354, 1198, 324, 50, 1, 431910, 189214, 83024, 35836, 14946, 5842, 1996, 498

Offset: 1

R. H. Hardin Jun 05 2012

Table starts
.1..5..22...93...386..1586...6476...26333..106762...431910...1744436...7036530
.1..7..34..151...646..2710..11236...46231..189214...771442...3136156..12720982
.1.10..54..252..1110..4748..19964...83024..342678..1406748...5751636..23443240
.1.14..86..424..1926..8404..35836..150604..626726..2589844..10646676..43594464
.1.19.136..714..3354.14946..64664..274676.1152494..4793874..19813536..81495084
.1.26.212.1198..5842.26630.116992..502492.2126238..8903350..36998056.152862180
.1.36.324.1996.10154.47448.211888..920744.3930286.16570608..69240296.287379592
.1.50.498.3292.17578.84424.383728.1688200.7272622.30880672.129768616.541108840

			Some solutions for n=3 k=4
..0....0....1....0....1....1....1....0....0....1....1....0....0....0....1....0
..1....0....1....0....1....0....0....0....0....0....0....0....0....1....0....0
..1....0....0....1....1....1....0....0....1....1....1....0....0....0....0....0
..1....0....0....0....0....0....0....1....1....1....0....1....0....0....1....1
..0....0....0....0....0....1....1....0....0....0....1....0....0....0....0....0
..0....1....0....0....0....0....1....0....1....0....0....0....1....1....0....1
..0....1....0....1....0....0....0....1....0....0....0....1....0....1....0....0
..0....0....0....1....0....0....0....0....0....0....0....0....0....0....1....0
..0....0....0....0....1....0....1....0....0....0....0....1....1....0....1....1
..1....1....1....0....0....0....0....0....0....0....1....0....1....1....0....0

R. H. Hardin, Table of n, a(n) for n = 1..4938

Column 2 is A003269(n+7)
Column 3 is A133523(n+5)
Row 1 is A000346(n-1)

A029760 A sum with next-to-central binomial coefficients of even order, Catalan related.

Original entry on oeis.org

1, 8, 47, 244, 1186, 5536, 25147, 112028, 491870, 2135440, 9188406, 39249768, 166656772, 704069248, 2961699667, 12412521388, 51854046982, 216013684528, 897632738722, 3721813363288, 15401045060572, 63616796642368, 262357557683422, 1080387930269464

Offset: 0

Wolfdieter Lang

nonn

Proof by induction.
a(n) = total area below paths consisting of steps east (1,0) and north (0,1) from (0,0) to (n+2,n+2) that stay weakly below y=x. For example, the two paths with n=0 are
. _|.....|
|....._|
The first has area 1 below it, the second area 0 and so a(0)=1. - David Callan, Dec 09 2004
Convolution of A000346 with A001700. - Philippe Deléham, May 19 2009

Michael De Vlieger, Table of n, a(n) for n = 0..1657
Ran Pan, Jeffrey B. Remmel, Paired patterns in lattice paths, arXiv:1601.07988 [math.CO], 2016.
Sittipong Thamrongpairoj, Dowling Set Partitions, and Positional Marked Patterns, Ph. D. Dissertation, University of California-San Diego (2019).

Cf. A000108, A002457, A008549, A000984, A139262.

a[n_] := (n+3)^2 CatalanNumber[n+2]/2 - 2^(2n+3);
Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Sep 25 2018 *)

a(n) = 4^(n+1)*Sum_{k=1..n+1} binomial(2k, k-1)/4^k = ((n+3)^2)*C(n+2)/2-2^(2*n+3), C = Catalan. Also a(n+1)=4*a(n)+binomial(2(n+2), n+1).
G.f.: (d/dx)c(x)/(1-4*x), where c(x) = g.f. for Catalan numbers; convolution of A001791 and powers of 4. G.f. also c(x)^2/(1-4*x)^(3/2); convolution of Catalan numbers A000108 C(n), n >= 1, with A002457; convolution of A008549(n), n >= 1, with A000984 (central binomial coefficients).
a(n) = Sum_{k=0..n+1} A039598(n+1,k)*k^2. - Philippe Deléham, Dec 16 2007

cp's OEIS Frontend

A345916 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum <= 0.

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A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

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A002664 a(n) = 2^n - C(n,0)- ... - C(n,4).

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A269920 Triangle read by rows: T(n,f) is the number of rooted maps with n edges and f faces on an orientable surface of genus 0.

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A045621 a(n) = 2^n - binomial(n, floor(n/2)).

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A345908 Traces of the matrices (A345197) counting integer compositions by length and alternating sum.

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A348614 Numbers k such that the k-th composition in standard order has sum equal to twice its alternating sum.

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