A011848 -id:A011848 - cp's OEIS frontend

A054040 a(n) terms of series {1/sqrt(j)} are >= n.

1, 3, 5, 7, 10, 14, 18, 22, 27, 33, 39, 45, 52, 60, 68, 76, 85, 95, 105, 115, 126, 138, 150, 162, 175, 189, 202, 217, 232, 247, 263, 280, 297, 314, 332, 351, 370, 389, 409, 430, 451, 472, 494, 517, 540, 563, 587, 612, 637, 662, 688, 715, 741, 769, 797, 825

Offset: 1

Asher Auel, Apr 13 2000

nonn

In many cases the first differences have the form {2k, 2k, 2k, 2k+1} (A004524). In such cases the second differences are {0, 0, 1, 1}. See A082915 for the exceptions. In as many as these, the first differences have the form {2k-1, 2k-1, 2k-1, 2k}. - Robert G. Wilson v, Apr 18 2003 [Corrected by Carmine Suriano, Nov 08 2013]
a(100)=2574, a(1000)=250731 & a(10000)=25007302 which differs from Sum{i=4..104}A004524(i)=2625, Sum{i=4..1004}A004524(i)=251250 & Sum{i=4..10004}A004524(i)=25012500. - Robert G. Wilson v, Apr 18 2003
A054040(n) <= A011848(n+2), A054040(10000)=25007302 and A011848(n+2)=25007500. - Robert G. Wilson v, Apr 18 2003

			Let b(k) = 1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k):
.k.......1....2.....3.....4.....5.....6.....7
-------------------------------------------------
b(k)...1.00..1.71..2.28..2.78..3.23..3.64..4.01
For A019529 we have:
n=0: smallest k is a(0) = 1 since 1.00 > 0
n=1: smallest k is a(1) = 2 since 1.71 > 1
n=2: smallest k is a(2) = 3 since 2.28 > 2
n=3: smallest k is a(3) = 5 since 3.23 > 3
n=4: smallest k is a(4) = 7 since 4.01 > 4
For this sequence we have:
n=1: smallest k is a(1) = 1 since 1.00 >= 1
n=2: smallest k is a(2) = 3 since 2.28 >= 2
n=3: smallest k is a(3) = 5 since 3.23 >= 3
n=4: smallest k is a(4) = 7 since 4.01 >= 4

Sela Fried, On the partial sums of the Zeta function Sum_{n>=1} 1/n^s for 0 < s < 1, 2024.
Hugo Pfoertner, Plot of a(n)-(1/4)*(n^2-2*zeta(1/2)*n), n=1..1000.

Cf. A002387, A004080, A059750, A082915, A054044, A011848, A082915.

See A019529 for a different version.

f[n_] := Block[{k = 0, s = 0}, While[s < n, k++; s = N[s + 1/Sqrt[k], 50]]; k]; Table[f[n], {n, 1, 60}]

a(n)=if(n<0,0,t=1;z=1;while(zBenoit Cloitre, Sep 23 2012

Let f(n) = (1/4)*(n^2-2*zeta(1/2)*n) then we have a(n) = f(n) + O(1). More precisely we claim that for n >= 2 we have a(n) = floor(f(n)+c) where c > Max{a(n)-f(n) : n>=1} = a(153) - f(153) = 1.032880076066608813953... and we believe we can take c = 1.033. - Benoit Cloitre, Sep 23 2012

Definition and offset modified by N. J. A. Sloane, Sep 01 2009

A281026 a(n) = floor(3n(n+1)/4).

Original entry on oeis.org

0, 1, 4, 9, 15, 22, 31, 42, 54, 67, 82, 99, 117, 136, 157, 180, 204, 229, 256, 285, 315, 346, 379, 414, 450, 487, 526, 567, 609, 652, 697, 744, 792, 841, 892, 945, 999, 1054, 1111, 1170, 1230, 1291, 1354, 1419, 1485, 1552, 1621, 1692, 1764, 1837, 1912, 1989, 2067, 2146

Offset: 0

Bruno Berselli, Jan 13 2017

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of the initial terms.
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Subsequence of A214068.
Partial sums of A047273.
Cf. A011865, A045943, A274757 (subsequence).
Cf. sequences with formula floor(k*n*(n+1)/4): A011848 (k=1), A000217 (k=2), this sequence (k=3), A002378 (k=4).
Cf. sequences with formula floor(k*n*(n+1)/(k+1)): A000217 (k=1), A143978 (k=2), this sequence (k=3), A281151 (k=4), A194275 (k=5).

Magma
```
[3*n*(n+1) div 4: n in [0..60]];
```

A281026:=n->floor(3*n*(n+1)/4): seq(A281026(n), n=0..100); # Wesley Ivan Hurt, Jan 13 2017

Table[Floor[3 n (n + 1)/4], {n, 0, 60}]
LinearRecurrence[{3,-4,4,-3,1},{0,1,4,9,15},60] (* Harvey P. Dale, Jun 04 2023 *)

makelist(floor(3*n*(n+1)/4), n, 0, 60);

PARI
```
vector(60, n, n--; floor(3*n*(n+1)/4))
```
Python
```
[int(3*n*(n+1)/4) for n in range(60)]
```

[floor(3*n*(n+1)/4) for n in range(60)]

O.g.f.: x*(1 + x + x^2)/((1 + x^2)*(1 - x)^3).
E.g.f.: -(1 - 6*x - 3*x^2)*exp(x)/4 - (1 + i)*(i - exp(2*i*x))*exp(-i*x)/8, where i=sqrt(-1).
a(n) = a(-n-1) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) = a(n-4) + 6*n - 9.
a(n) = 3*n*(n+1)/4 + (i^(n*(n+1)) - 1)/4. Therefore:
a(4*k+r) = 12*k^2 + 3*(2*r+1)*k + r^2, where 0 <= r <= 3.
a(n) = n^2 - floor((n-1)*(n-2)/4).
a(n) = A011865(3*n+2).

A288852 Number T(n,k) of matchings of size k in the n X n X n triangular grid; triangle T(n,k), n>=0, 0<=k<=floor(n*(n+1)/4), read by rows.

Original entry on oeis.org

1, 1, 1, 3, 1, 9, 15, 2, 1, 18, 99, 193, 108, 6, 1, 30, 333, 1734, 4416, 5193, 2331, 240, 1, 45, 825, 8027, 45261, 151707, 298357, 327237, 180234, 40464, 2238, 1, 63, 1710, 26335, 255123, 1629474, 6995539, 20211423, 38743020, 47768064, 35913207, 15071019

Offset: 0

Alois P. Heinz, Jun 18 2017

The n X n X n triangular grid has n rows with i vertices in row i. Each vertex is connected to the neighbors in the same row and up to two vertices in each of the neighboring rows. The graph has A000217(n) vertices and 3*A000217(n-1) edges altogether.

			Triangle T(n,k) begins:
  1;
  1;
  1,  3;
  1,  9,  15,    2;
  1, 18,  99,  193,   108,      6;
  1, 30, 333, 1734,  4416,   5193,   2331,    240;
  1, 45, 825, 8027, 45261, 151707, 298357, 327237, 180234, 40464, 2238;

Alois P. Heinz, Rows n = 0..17, flattened
Eric Weisstein's World of Mathematics, Matching-Generating Polynomial
Wikipedia, Matching (graph theory)
Wikipedia, Triangular grid graph

Columns k=0-1 give: A000012, A045943(n-1) for n>0.
Row sums give A269869.
Last elements of rows give A271610.
Cf. A000217, A011848, A271612, A271614, A271616.

b:= proc(l) option remember;  local n, k; n:= nops(l);
      if n=0 then 1
    elif min(l)>0 then b(subsop(-1=NULL, map(h-> h-1, l)))
    else for k to n while l[k]>0 do od; b(subsop(k=1, l))+
         expand(x*(`if`(k1 and l[k-1]=1, b(subsop(k=1, k-1=2, l)), 0)))
      fi
    end:
T:= n-> (p-> seq(coeff(p,x,i), i=0..degree(p)))(b([0$n])):
seq(T(n), n=0..10);

b[l_] := b[l] = Module[{n = Length[l], k}, Which[n == 0, 1, Min[l] > 0, b[ReplacePart[l - 1, -1 -> Nothing]], True, For[k = 1, k <= n && l[[k]] > 0, k++]; b[ReplacePart[l, k -> 1]] + x*Expand[If[k < n, b[ReplacePart[l, k -> 2]], 0] + If[k < n && l[[k + 1]] == 0, b[ReplacePart[l, {k -> 1, k + 1 -> 1}]], 0] + If[k > 1 && l[[k - 1]] == 1, b[ReplacePart[l, {k -> 1, k - 1 -> 2}]], 0]]]];
T[n_] := Table[Coefficient[#, x, i], {i, 0, Exponent[#, x]}]&[b[Table[0, n] ]];
Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, May 24 2018, translated from Maple *)

T(n,floor(n*(n+1)/4)) = A271610(n).
Sum_{i=0..1} T(n,floor(n*(n+1)/4)-i) = A271612(n).
Sum_{i=0..2} T(n,floor(n*(n+1)/4)-i) = A271614(n).
Sum_{i=0..3} T(n,floor(n*(n+1)/4)-i) = A271616(n).

A166454 Triangle read by rows: T(n, k) = (1/2)*(A007318(n,k) - A047999(n,k)).

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 2, 5, 5, 2, 3, 7, 10, 7, 3, 3, 10, 17, 17, 10, 3, 4, 14, 28, 35, 28, 14, 4, 4, 18, 42, 63, 63, 42, 18, 4, 5, 22, 60, 105, 126, 105, 60, 22, 5, 5, 27, 82, 165, 231, 231, 165, 82, 27, 5, 6, 33, 110, 247, 396, 462, 396, 247, 110, 33, 6

Offset: 2

Gary W. Adamson, Oct 14 2009

Row sums = A120739: (1, 2, 7, 14, 30, 60, 127, 254, ...).

			First few rows of the triangle:
  1;
  1,   1;
  2,   3,   2;
  2,   5,   5,   2;
  3,   7,  10,   7,   3;
  3,  10,  17,  17,  10,   3;
  4,  14,  28,  35,  28,  14,   4;
  4,  18,  42,  63,  63,  42,  18,   4;
  5,  22,  60, 105, 126, 105,  60,  22,   5;
  5,  27,  82, 165, 231, 231, 165,  82,  27,   5;
  6,  33, 110, 247, 396, 462, 396, 247, 110,  33,   6;
  ...

Reinhard Zumkeller, Rows n = 2..125 of triangle, flattened

Cf. A047999, A120739.

Cf. A007318, A011848, A001700 (central terms).

Flat(List([2..12],n->List([1..n-1],m->Int(Binomial(n,m)/2)))); # Muniru A Asiru, Apr 14 2019

Following Bagula's formula
a166454 n k = a166454_tabl !! (n-2) !! (k-1)
a166454_row n = a166454_tabl !! (n-2)
a166454_tabl = map (map (flip div 2) . init . tail) $ drop 2 a007318_tabl
-- Reinhard Zumkeller, Mar 04 2015

[[Floor(Binomial(n,k)/2): k in [1..n-1]]: n in [2..12]]; // G. C. Greubel, Apr 16 2019

seq(seq(floor(binomial(n,m)/2),m=1..n-1),n=2..12); # Muniru A Asiru, Apr 14 2019

T[n_, m_] = Floor[Binomial[n, m]/2]; Table[T[n, m], {n, 2, 12}, {m, 1, n-1}]//Flatten (* Roger L. Bagula, Mar 07 2010*)

{T(n,k) = binomial(n,k)\2 };
for(n=2,12, for(k=1,n-1, print1(T(n,k), ", "))) \\ G. C. Greubel, Apr 16 2019

[[floor(binomial(n,k)/2) for k in (1..n-1)] for n in (2..12)] # G. C. Greubel, Apr 16 2019

T(n, k) = (1/2)*(A007318(n,k) - A047999(n,k)), nonzero terms.

T(n, m) = floor(binomial(n, m)/2). - Roger L. Bagula, Mar 07 2010

A350393 Smallest degree of x with the largest coefficient in Product_{k=1..n} (1 + x^k).

Original entry on oeis.org

0, 0, 0, 3, 3, 5, 9, 12, 18, 21, 27, 33, 39, 45, 52, 60, 68, 76, 85, 95, 105, 115, 126, 138, 150, 162, 175, 189, 203, 217, 232, 248, 264, 280, 297, 315, 333, 351, 370, 390, 410, 430, 451, 473, 495, 517, 540, 564, 588, 612, 637, 663, 689, 715, 742, 770, 798, 826, 855, 885, 915, 945, 976

Offset: 0

Max Alekseyev, Dec 28 2021

nonn

Apparently, a(n) = A011848(n+1) for n >= 10. - Hugo Pfoertner, Dec 30 2021

Cf. A025591 (largest coefficient), A350394 (largest degree of x), A350395, A350396.

Cf. A011848.

{ A350393(n) = my(v,t,x='x); v = Vecrev(prod(k=1,n,1+x^k)); vecmax(v,&t); t-1; }

A118404 Triangle T, read by rows, where all columns of T are different and yet all columns of the matrix square T^2 (A118407) are equal; also equals the matrix inverse of triangle A118400.

Original entry on oeis.org

1, 1, -1, -1, 0, 1, -1, 1, -1, -1, 1, 0, 0, 2, 1, 1, -1, 0, -2, -3, -1, -1, 0, 1, 2, 5, 4, 1, -1, 1, -1, -3, -7, -9, -5, -1, 1, 0, 0, 4, 10, 16, 14, 6, 1, 1, -1, 0, -4, -14, -26, -30, -20, -7, -1, -1, 0, 1, 4, 18, 40, 56, 50, 27, 8, 1, -1, 1, -1, -5, -22, -58, -96, -106, -77, -35, -9, -1, 1, 0, 0, 6, 27, 80, 154, 202, 183, 112, 44, 10, 1, 1, -1, 0, -6, -33, -107, -234, -356, -385, -295, -156, -54, -11, -1, -1, 0, 1, 6, 39, 140, 341, 590, 741, 680, 451, 210, 65, 12, 1, -1, 1, -1, -7, -45, -179, -481, -931, -1331, -1421, -1131, -661, -275, -77, -13, -1, 1, 0, 0, 8, 52, 224, 660, 1412, 2262, 2752, 2552, 1792, 936, 352, 90, 14, 1

Offset: 0

Paul D. Hanna, Apr 27 2006

Appears to coincide with triangle (5.2) in Lee-Oh (2016), although there is no obvious connection! - N. J. A. Sloane, Dec 07 2016

			Triangle begins:
1;
1,-1;
-1, 0, 1;
-1, 1,-1,-1;
1, 0, 0, 2, 1;
1,-1, 0,-2,-3,-1;
-1, 0, 1, 2, 5, 4, 1;
-1, 1,-1,-3,-7,-9,-5,-1;
1, 0, 0, 4, 10, 16, 14, 6, 1;
1,-1, 0,-4,-14,-26,-30,-20,-7,-1;
-1, 0, 1, 4, 18, 40, 56, 50, 27, 8, 1;
-1, 1,-1,-5,-22,-58,-96,-106,-77,-35,-9,-1;
1, 0, 0, 6, 27, 80, 154, 202, 183, 112, 44, 10, 1;
1, -1, 0, -6, -33, -107, -234, -356, -385, -295, -156, -54, -11, -1;
-1, 0, 1, 6, 39, 140, 341, 590, 741, 680, 451, 210, 65, 12, 1;
-1, 1, -1, -7, -45, -179, -481, -931, -1331, -1421, -1131, -661, -275, -77, -13, -1;
1, 0, 0, 8, 52, 224, 660, 1412, 2262, 2752, 2552, 1792, 936, 352, 90, 14, 1;
1, -1, 0, -8, -60, -276, -884, -2072, -3674, -5014, -5304, -4344, -2728, -1288, -442, -104, -15, -1;
-1, 0, 1, 8, 68, 336, 1160, 2956, 5746, 8688, 10318, 9648, 7072, 4016, 1730, 546, 119, 16, 1; ...
The matrix square is A118407:
1;
0, 1;
-2, 0, 1;
2,-2, 0, 1;
0, 2,-2, 0, 1;
-2, 0, 2,-2, 0, 1;
4,-2, 0, 2,-2, 0, 1;
-6, 4,-2, 0, 2,-2, 0, 1;
4,-6, 4,-2, 0, 2,-2, 0, 1;
6, 4,-6, 4,-2, 0, 2,-2, 0, 1; ...
in which all columns are equal.

Kyu-Hwan Lee, Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.

Cf. A118405 (row sums), A118406 (unsigned row sums), A118407 (matrix square), A118400 (matrix inverse).

Columns or diagonals (modulo offsets): A219977, A011848, A212342, A007598, A005581, A007910.

T[n_, k_] := SeriesCoefficient[(-1)^k/((1+x^2)(1+x)^(k-1)), {x, 0, n-k}];
Table[T[n, k], {n, 0, 16}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2018 *)

{T(n,k)=polcoeff(polcoeff((1+x)^2/(1+x^2)/(1+x+x*y +x*O(x^n)),n,x)+y*O(y^k),k,y)}
for(n=0, 16, for(k=0, n, print1(T(n, k), ", ")); print(""))

G.f.: A(x,y) = (1+x)^2 / ( (1+x^2) * (1+x + x*y) ).

G.f. of column k: (-1)^k / ( (1+x^2) * (1+x)^(k-1) ) for k>=0.

A231754 Products of distinct primes congruent to 1 modulo 4 (A002144).

Original entry on oeis.org

1, 5, 13, 17, 29, 37, 41, 53, 61, 65, 73, 85, 89, 97, 101, 109, 113, 137, 145, 149, 157, 173, 181, 185, 193, 197, 205, 221, 229, 233, 241, 257, 265, 269, 277, 281, 293, 305, 313, 317, 337, 349, 353, 365, 373, 377, 389, 397, 401, 409, 421, 433, 445, 449

Offset: 1

Michel Marcus, Nov 13 2013

Contains A002144 as a subsequence, and is a subsequence of A016813 and of A005117.

Also, these numbers satisfy A231589(n) = floor(n*(n-1)/4) (A011848).

			65 = 5*13 is in the sequence since both 5 and 13 are congruent to 1 modulo 4.

R. J. Mathar, Table of n, a(n) for n = 1..4999
Rafael Jakimczuk, Generalizations of Mertens's Formula and k-Free and s-Full Numbers with Prime Divisors in Arithmetic Progression, ResearchGate, 2024.
Shailesh A. Shirali, A family portrait of primes-a case study in discrimination, Math. Mag., Vol. 70, No. 4 (Oct., 1997), pp. 263-272.

Intersection of A005117 and A004613.
Cf. A002144, A011848, A016813, A167181, A231589.
Cf. A088539, A175647.

isA231754 := proc(n)
    local d;
    for d in ifactors(n)[2] do
        if op(2,d) > 1 then
            return false;
        elif modp(op(1,d),4) <> 1 then
            return false;
        end if;
    end do:
    true ;
end proc:
for n from 1 to 500 do
    if isA231754(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Mar 16 2016

Select[Range[500], # == 1 || AllTrue[FactorInteger[#], Last[#1] == 1 && Mod[First[#1], 4] == 1 &] &] (* Amiram Eldar, Mar 08 2024 *)

isok(n) = if (! issquarefree(n), return (0)); if (n > 1, f = factor(n); for (i=1, #f~, if (f[i, 1] % 4 != 1, return (0)))); 1

The number of terms that do not exceed x is ~ c * x / sqrt(log(x)), where c = A088539 * sqrt(A175647) / Pi = 0.3097281805... (Jakimczuk, 2024, Theorem 3.10, p. 26). - Amiram Eldar, Mar 08 2024

A160438 Number of partitions of n*(n+1)/2 with at most four parts that can be obtained from grouping (with parentheses) a permutation of the sum 1+2+...+n.

Original entry on oeis.org

1, 1, 2, 5, 13, 35, 93, 215, 437, 815, 1436, 2413, 3886, 6041, 9125, 13436, 19323, 27221, 37670, 51293, 68797, 91025, 118982, 153797, 196721, 249206, 312935, 389761, 481709, 591080, 720485, 872763, 1050980, 1258565, 1499351, 1777462

Offset: 0

Hagen von Eitzen, May 13 2009

nonn

a(n) is the number of integer quadruples (x,y,z,w) with x >= y >= z >= w >= 0 and x+y+z+w = n*(n+1)/2 such that the set {1,2,...,n} can be partitioned into four (possibly empty) subsets with respective element sums x, y, z, w.

			For n = 3 the a(3) = 5 solutions are 6 = (1+2+3), 5+1 = (2+3)+(1), 4+2 = (1+3)+(2), 3+3 = (3)+(1+2), 3+2+1 = (3)+(2)+(1). Note that 3+3 = (1+2)+(3) is the same as (3)+(1+2) as both are 3+3.
For n = 6 the partition 10+4+4+3 is *not* among the a(6) = 93 solutions because 4 can only come from grouping either (4) or (1+3), hence both groupings would have to occur; but (1+3) conflicts with both possible groupings (3) and (1+2) which could produce 3.

H. v. Eitzen, Table of n, a(n) for n=0..5262 (i.e. a(n) less than 2^64)
"AI", (sci.math thread that inspired investigating the sequence)
H. v. Eitzen, How to Build Triangles from Integers

If n >= 8 then a(n) = A001400(n*(n+1)/2) - 2*A011848(n+1) - 5.

A160637 Hankel transform of A114464(n+1).

Original entry on oeis.org

1, 1, -2, -8, -32, -128, 1024, 16384, 262144, 4194304, -134217728, -8589934592, -549755813888, -35184372088832, 4503599627370496, 1152921504606846976, 295147905179352825856, 75557863725914323419136, -38685626227668133590597632, -39614081257132168796771975168

Offset: 0

Paul Barry, May 21 2009

Hankel transform of A114464(n) is A160636.

This is a generalized Somos-4 sequence. - Michael Somos, Mar 14 2020

G. C. Greubel, Table of n, a(n) for n = 0..114

Cf. A011848, A114464, A160636.

[(-2)^Floor(Binomial(n+1,2)/2): n in [0..50]]; // G. C. Greubel, May 03 2018

Table[(-2)^Floor[Binomial[n + 1, 2]/2], {n, 0, 50}] (* G. C. Greubel, May 03 2018 *)
a[ n_] := (-2)^Quotient[n (n + 1), 4]; (* Michael Somos, Mar 14 2020 *)

for(n=0, 50, print1((-2)^floor(binomial(n+1,2)/2), ", ")) \\ G. C. Greubel, May 03 2018

a(n) = (-2)^floor(C(n+1,2)/2) = (-2)^A011848(n+1).

0 = a(n)*a(n+4) - 2*a(n+1)*a(n+3) + 4*a(n+2)^2 = a(n)*a(n+5) - 4*a(n+1)*a(n+4) for all n in Z. - Michael Somos, Mar 14 2020

A120739 a(n) = Sum_{k=0..n} floor(C(n,k)/2).

Original entry on oeis.org

0, 0, 1, 2, 7, 14, 30, 60, 127, 254, 510, 1020, 2046, 4092, 8188, 16376, 32767, 65534, 131070, 262140, 524286, 1048572, 2097148, 4194296, 8388606, 16777212, 33554428, 67108856, 134217724, 268435448, 536870904, 1073741808, 2147483647

Offset: 0

Paul Barry, Jun 29 2006

Nonzero terms = row sums of triangle A166454. - Gary W. Adamson, Oct 14 2009

Charles R Greathouse IV, Table of n, a(n) for n = 0..3322
Pantelimon Stanica, Tsutomu Sasao, Jon T. Butler, Distance Duality on Some Classes of Boolean Functions, Journal of Combinatorial Mathematics and Combinatorial Computing (to appear), 2017. [Theorem 9.]

Cf. A001316, A011848, A166454.

a120739 n = if n < 2 then 0 else sum $ a166454_row n
-- Reinhard Zumkeller, Mar 04 2015

[(&+[Floor(Binomial(n,k)/2): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Apr 18 2019

Table[Sum[Floor[Binomial[n,k]/2], {k,0,n}], {n,0,40}] (* G. C. Greubel, Apr 18 2019 *)

PARI
```
a(n)=(2^n-2^norml2(binary(n)))/2
```

{a(n) = sum(k=0,n, binomial(n,k)\2)}; \\ G. C. Greubel, Apr 18 2019

[sum(floor(binomial(n,k)/2) for k in (0..n)) for n in (0..40)] # G. C. Greubel, Apr 18 2019

a(n) = (2^n - A001316(n))/2.

cp's OEIS Frontend

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