A063776 -id:A063776 - cp's OEIS frontend

A053633 Triangular array T(n,k) giving coefficients in expansion of Product_{j=1..n} (1+x^j) mod x^(n+1)-1.

1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 4, 3, 3, 3, 3, 6, 5, 5, 6, 5, 5, 10, 9, 9, 9, 9, 9, 9, 16, 16, 16, 16, 16, 16, 16, 16, 30, 28, 28, 29, 28, 28, 29, 28, 28, 52, 51, 51, 51, 51, 52, 51, 51, 51, 51, 94, 93, 93, 93, 93, 93, 93, 93, 93, 93, 93, 172, 170, 170, 172, 170, 170, 172

Offset: 0

N. J. A. Sloane, Mar 22 2000

T(n,k) = number of binary vectors (x_1,...,x_n) satisfying Sum_{i=1..n} i*x_i = k (mod n+1) = size of Varshamov-Tenengolts code VT_k(n).

			Triangle begins:
  k  0    1    2    3    4    5    6    7    8    9
n
0    1;
1    1,   1;
2    2,   1,   1;
3    2,   2,   2,   2;
4    4,   3,   3,   3,   3;
5    6,   5,   5,   6,   5,   5;
6   10,   9,   9,   9,   9,   9,   9;
7   16,  16,  16,  16,  16,  16,  16,  16;
8   30,  28,  28,  29,  28,  28,  29,  28,  28;
9   52,  51,  51,  51,  51,  52,  51,  51,  51,  51;
    ...
[Edited by _Seiichi Manyama_, Mar 11 2018]

B. D. Ginsburg, On a number theory function applicable in coding theory, Problemy Kibernetiki, No. 19 (1967), pp. 249-252.

Seiichi Manyama, Rows n = 0..139, flattened
F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence [pdf, ps].
N. J. A. Sloane, On single-deletion-correcting codes
Index entries for sequences related to subset sums modulo m
Index entries for sequences related to Gijswijt's sequence

Cf. A053632, A063776, A300328, A300628. Leading coefficients give A000016, next column gives A000048.

with(numtheory): A053633 := proc(n,k) local t1,d; t1 := 0; for d from 1 to n do if n mod d = 0 and d mod 2 = 1 then t1 := t1+(1/(2*n))*2^(n/d)*phi(d)*mobius(d/gcd(d,k))/phi(d/gcd(d,k)); fi; od; t1; end;

Flatten[ Table[ CoefficientList[ PolynomialMod[ Product[1+x^j, {j,1,n}], x^(n+1)-1], x], {n,0,11}]] (* Jean-François Alcover, May 04 2011 *)

The Maple code gives an explicit formula.

A053634 a(n) = Sum_{ d divides n } phi(d)*2^(n/d)/(2n).

Original entry on oeis.org

2, 3, 4, 7, 10, 18, 30, 54, 94, 176, 316, 591, 1096, 2058, 3856, 7301, 13798, 26244, 49940, 95373, 182362, 349626, 671092, 1290714, 2485534, 4793790, 9256396, 17896284, 34636834, 67109898, 130150588, 252647064, 490853416, 954440950

Offset: 3

N. J. A. Sloane, Mar 23 2000

nonn

Offset is 3 because a(2) is 3/2, not an integer. - Michel Marcus, Sep 11 2013

G. C. Greubel, Table of n, a(n) for n = 3..3000
James East, Ron Niles, Integer polygons of given perimeter, arXiv:1710.11245 [math.CO], 2017.

Cf. A000016, A000031, A053635, A063776.

a[n_] := DivisorSum[n, EulerPhi[#]*2^(n/#)&]/(2n); Table[a[n], {n, 3, 36}] (* Jean-François Alcover, Dec 07 2015 *)

a(n) = sumdiv (n, d, eulerphi(d)*2^(n/d)/(2*n));  \\ Michel Marcus, Sep 11 2013

a(n) = A000031(n)/2.

A053636 a(n) = Sum_{odd d|n} phi(d)*2^(n/d).

Original entry on oeis.org

0, 2, 4, 12, 16, 40, 72, 140, 256, 540, 1040, 2068, 4128, 8216, 16408, 32880, 65536, 131104, 262296, 524324, 1048640, 2097480, 4194344, 8388652, 16777728, 33554600, 67108912, 134218836, 268435552, 536870968, 1073744160, 2147483708

Offset: 0

N. J. A. Sloane, Mar 23 2000

nonn

			2*x + 4*x^2 + 12*x^3 + 16*x^4 + 40*x^5 + 72*x^6 + 140*x^7 + 256*x^8 + 540*x^9 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..3321

Cf. A000010, A001227, A053635, A063776, A182469.

a053636 0 = 0
a053636 n = sum $ zipWith (*) (map a000010 ods) (map ((2 ^) . (div n)) ods)
            where ods = a182469_row n
-- Reinhard Zumkeller, Sep 13 2013

a[ n_] := If[ n < 1, 0, Sum[ Mod[ d, 2] EulerPhi[ d] 2^(n / d), {d, Divisors[ n]}]] (* Michael Somos, May 09 2013 *)

{a(n) = if( n<1, 0, sumdiv( n, d, (d % 2) * eulerphi(d) * 2^(n / d)))} /* Michael Somos, May 09 2013 */

from sympy import totient, divisors
def A053636(n): return (sum(totient(d)<>(~n&n-1).bit_length(),generator=True))<<1) # Chai Wah Wu, Feb 21 2023

a(n) = n * A063776(n).
a(n) = Sum_{k=1..A001227(n)} A000010(A182469(n,k)) * 2^(n/A182469(n, A001227(n)+1-k)). - Reinhard Zumkeller, Sep 13 2013
G.f.: Sum_{m >= 0} phi(2*m + 1)*2*x^(2*m + 1)/(1 - 2*x^(2*m + 1)). - Petros Hadjicostas, Jul 20 2019

A267632 Triangle T(n, k) read by rows: the k-th column of the n-th row lists the number of ways to select k distinct numbers (k >= 1) from [1..n] so that their sum is divisible by n.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 2, 2, 1, 1, 1, 2, 4, 3, 1, 0, 1, 3, 5, 5, 3, 1, 1, 1, 3, 7, 9, 7, 3, 1, 0, 1, 4, 10, 14, 14, 10, 4, 1, 1, 1, 4, 12, 22, 26, 20, 12, 5, 1, 0, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 5, 19, 42, 66, 76, 66, 43, 19, 5, 1, 0

Offset: 1

Dimitri Papadopoulos, Jan 18 2016

Row less the last element is palindrome for n=odd or n=power of 2 where n is the row number (observation-conjecture).
From Petros Hadjicostas, Jul 13 2019: (Start)
By reading carefully the proof of Lemma 5.1 (pp. 65-66) in Barnes (1959), we see that he actually proved a general result (even though he does not state it in the lemma).
According to the definition of this sequence, for 1 <= k <= n, T(n, k) is the number of unordered sets b_1, b_2, ..., b_k of k distinct integers from 1..n such that b_1 + b_2 + ... + b_k = 0 (mod n). The proof of Lemma 5.1 in Barnes (1959) implies that T(n, k) = (1/n) * Sum_{s | gcd(n, k)} (-1)^(k - (k/s)) * phi(s) * binomial(n/s, k/s) for 1 <= k <= n.
For fixed k >= 1, the g.f. of the column (T(n, k): n >= 1) (with T(n, k) = 0 for 1 <= n < k) is (x^k/k) * Sum_{s|k} phi(s) * (-1)^(k - (k/s)) / (1 - x^s)^(k/s), which generalizes Herbert Kociemba's formula from A032801.
Barnes' (1959) formula is a special case of Theorem 4 (p. 66) in Ramanathan (1944). If R(n, k, v) is the number of unordered sets b_1, b_2, ..., b_k of k distinct integers from 1..n such that b_1 + b_2 + ... + b_k = v (mod n), then he proved that R(n, k, v) = (1/n) * Sum_{s | gcd(n,k)} (-1)^(k - (k/s)) * binomial(n/s, k/s) * C_s(v), where C_s(v) = A054535(s, v) =  Sum_{d | gcd(s,v)} d * Moebius(s/d) is Ramanujan's sum (even though it was first discovered around 1900 by the Austrian mathematician R. D. von Sterneck).
Because C_s(v = 0) = phi(s), we get Barnes' (implicit) result; i.e., R(n, k, v=0) = T(n, k) for 1 <= k <= n.
For k=2, we have R(n, k=2, v=0) = T(n, k=2) = A004526(n-1) for n >= 1. For k=3, we have R(n, k=3, v=0) = T(n, k=3) = A058212(n) for n >= 1. For k=4, we have R(n, k=4, v=0) = A032801(n) for n >= 1. For k=5, we have R(n, k=5, v=0) = T(n, k=5) = A008646(n-5) for n >= 5.
The reason we have T(2*m+1, k) = A037306(2*m+1, k) = A047996(2*m+1, k) for m >= 0 and k >= 1 is the following. When n = 2*m + 1, all divisors s of gcd(n, k) are odd. In such a case, k - (k/s) is even for all k >= 1, and thus (-1)^(k - (k/s)) = 1, and thus T(n = 2*m+1, k) = (1/n) * Sum_{s | gcd(n, k)} phi(s) * binomial(n/s, k/s) = A037306(2*m+1, k) = A047996(2*m+1, k).
By summing the products of the g.f. of column k times y^k from k = 1 to k = infinity, we get the bivariate g.f. for the array: Sum_{n, k >= 1} T(n, k)*x^n*y^k = Sum_{s >= 1} (phi(s)/s) * log((1 - x^s + (-x*y)^s)/(1 - x^s)) = -x/(1 - x) - Sum_{s >= 1} (phi(s)/s) * log(1 - x^s + (-x*y)^s).
Letting y = 1 in the above bivariate g.f., we get the g.f. of the sequence (Sum_{1 <= k <= n} T(n, k): n >= 1) is -x/(1 - x) - Sum_{s >= 1} (phi(s)/s) * log(1 - x^s + (-x)^s) = -x/(1 - x) + Sum_{m >= 0} (phi(2*m + 1)/(2*m + 1)) * log(1 - 2*x^(2*m+1)), which is the g.f. of sequence A082550. Hence, sequence A082550 consists of the row sums.
There is another important interpretation of this array T(n, k) which is related to some of the references for sequence A047996, but because the discussion is too lengthy, we omit the details.
(End)

			For n = 5, there is one way to pick one number (5), two ways to pick two numbers (1+4, 2+3), two ways to pick 3 numbers (1+4+5, 2+3+5), one way to pick 4 numbers (1+2+3+4), and one way to pick 5 numbers (1+2+3+4+5) so that their sum is divisible by 5. Therefore, T(5,1) = 1, T(5,2) = 2, T(5,3) = 2, T(5,4) = 1, and T(5,5) = 1.
Table for T(n,k) begins as follows:
n\k 1 2   3    4    5    6    7    8    9   10
1   1
2   1 0
3   1 1   1
4   1 1   1    0
5   1 2   2    1    1
6   1 2   4    3    1    0
7   1 3   5    5    3    1    1
8   1 3   7    9    7    3    1    0
9   1 4  10   14   14   10    4    1    1
10  1 4  12   22   26   20   12    5    1    0
...

Alois P. Heinz, Rows n = 1..150, flattened
Eric Stephen Barnes, The construction of perfect and extreme forms I, Acta Arith., 5 (1959); see pp. 65-66.
Michiel Kosters, The subset sum problem for finite abelian groups, J. Combin. Theory Ser. A 120 (2013), 527-530.
Jiyou Li and Daqing Wan, Counting subset sums of finite abelian groups, J. Combin. Theory Ser. A 119 (2012), 170-182; see pp. 171-172.
K. G. Ramanathan, Some applications of Ramanujan's trigonometrical sum C_m(n), Proc. Indian Acad. Sci., Sect. A 20 (1944), 62-69; see p. 66.
R. D. von Sterneck, Ein Analogon zur additiven Zahlentheorie, Sitzungsber. Akad. Wiss. Sapientiae Math.-Naturwiss. Kl. 111 (1902), 1567-1601 (Abt. IIa).
R. D. von Sterneck, Über ein Analogon zur additiven Zahlentheorie, Jahresbericht der Deutschen Mathematiker-Vereinigung 12 (1903), 110-113.

Cf. A004526, A032801, A037306, A047996, A054532, A054533, A054534, A054535, A058212, A063776, A082550 (row sums), A309122.

A267632 := proc(n,k)
    local a,msel,p;
    a := 0 ;
    for msel in combinat[choose](n,k) do
        if modp(add(p,p=msel),n) = 0 then
            a := a+1 ;
        end if;
    end do:
    a;
end proc: # R. J. Mathar, May 15 2016
# second Maple program:
b:= proc(n, m, s) option remember; expand(`if`(n=0,
      `if`(s=0, 1, 0), b(n-1, m, s)+x*b(n-1, m, irem(s+n, m))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2, 0)):
seq(T(n), n=1..14);  # Alois P. Heinz, Aug 27 2018

f[k_, n_] :=  Length[Select[Select[Subsets[Range[n]], Length[#] == k &], IntegerQ[Total[#]/n] &]];MatrixForm[Table[{n, Table[f[k, n], {k, n}]}, {n, 10}]] (* Dimitri Papadopoulos, Jan 18 2016 *)

T(2n+1, k) = A037306(2n+1, k) = A047996(2n+1, k).
From Petros Hadjicostas, Jul 13 2019: (Start)
T(n, k) = (1/n) * Sum_{s | gcd(n, k)} (-1)^(k - (k/s)) * phi(s) * binomial(n/s, k/s) for 1 <= k <= n.
G.f. for column k >= 1: (x^k/k) * Sum_{s|k} phi(s) * (-1)^(k - (k/s)) / (1 - x^s)^(k/s).
Bivariate g.f.: Sum_{n, k >= 1} T(n, k)*x^n*y^k = -x/(1 - x) - Sum_{s >= 1} (phi(s)/s) * log(1 - x^s + (-x*y)^s).
(End)
Sum_{k=1..n} k * T(n,k) = A309122(n). - Alois P. Heinz, Jul 13 2019

A064355 Number of subsets of {1,2,..n} that sum to 1 mod n.

Original entry on oeis.org

2, 2, 2, 4, 6, 10, 18, 32, 56, 102, 186, 340, 630, 1170, 2182, 4096, 7710, 14560, 27594, 52428, 99858, 190650, 364722, 699040, 1342176, 2581110, 4971008, 9586980, 18512790, 35791358, 69273666, 134217728, 260300986, 505290270, 981706806, 1908874240, 3714566310

Offset: 1

Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Sep 25 2001

nonn

			a(7) = 18 because there are 18 subsets of {1,2,3,4,5,6,7} which sum to 1 mod 7:{1}, {1,7}, {2,6}, {3,5}, {1,2,5}, {1,3,4}, {2,6,7}, {3,5,7}, {4,5,6}, {1,2,5,7}, {1,3,4,7}, {1,3,5,6}, {2,3,4,6}, {4,5,6,7}, {1,2,3,4,5}, {1,3,5,6,7}, {2,3,4,6,7}, {1,2,3,4,5,7}.

Index entries for sequences related to subset sums modulo m.

Cf. A000048, A008683, A063776.

a[n_] := Block[{d = Select[Divisors@n, OddQ@ # &]}, Plus @@ (2^(n/d)*MoebiusMu@d)/n]; Array[a, 35] (* Robert G. Wilson v, Feb 20 2006 *)

a(n) = sumdiv(n, d, (d % 2) * 2^(n/d) * moebius(d)) / n; \\ Amiram Eldar, Jun 05 2025

a(n) = 2*A000048(n).

a(n) = (1/n) * sum_{d divides n and d is odd} 2^(n/d) * mu(d); (mu(d) is the Moebius function, sequence A008683).

More terms from Vladeta Jovovic, Sep 27 2001

A309128 (1/n) times the sum of the elements of all subsets of [n] whose sum is divisible by n.

Original entry on oeis.org

1, 1, 4, 4, 12, 20, 40, 70, 150, 284, 564, 1116, 2212, 4392, 8768, 17404, 34704, 69214, 137980, 275264, 549340, 1096244, 2188344, 4369196, 8724196, 17422500, 34797476, 69505628, 138845940, 277383904, 554189344, 1107296248, 2212559996, 4421289872, 8835361488

Offset: 1

Alois P. Heinz, Jul 13 2019

nonn

			The subsets of [5] whose sum is divisible by 5 are: {}, {5}, {1,4}, {2,3}, {1,4,5}, {2,3,5}, {1,2,3,4}, {1,2,3,4,5}.  The sum of their elements is 0 + 5 + 5 + 5 + 10 + 10 + 10 + 15 = 60.  So a(5) = 60/5 = 12.

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Cf. A000010, A001792 (the same for all subsets), A053636, A063776, A309122, A309280.

b:= proc(n, m, s) option remember; `if`(n=0, [`if`(s=0, 1, 0), 0],
      b(n-1, m, s) +(g-> g+[0, g[1]*n])(b(n-1, m, irem(s+n, m))))
    end:
a:= proc(n) option remember; forget(b); b(n$2, 0)[2]/n end:
seq(a(n), n=1..40);

b[n_, m_, s_] := b[n, m, s] = If[n == 0, {If[s == 0, 1, 0], 0},
     b[n-1, m, s] + Function[g, g+{0, g[[1]] n}][b[n-1, m, Mod[s+n, m]]]];
a[n_] := b[n, n, 0][[2]]/n;
Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Mar 19 2022, after Alois P. Heinz *)

Conjecture: a(n) = (n + 1) * A063776(n)/4 - (phi(n)/2) * (1 + (-1)^n)/2 = ((n + 1)/(4*n)) * A053636(n) - (phi(n)/2) * (1 + (-1)^n)/2. - Petros Hadjicostas, Jul 20 2019

a(n) = A309280(n,n). - Alois P. Heinz, Jul 21 2019

A300218 Number of solutions to 1 +- 3 +- 5 +- ... +- (2*n-1) == 0 mod n.

Original entry on oeis.org

1, 2, 2, 4, 4, 12, 10, 36, 30, 104, 94, 344, 316, 1172, 1096, 4132, 3856, 14572, 13798, 52432, 49940, 190652, 182362, 699416, 671092, 2581112, 2485534, 9586984, 9256396, 35791472, 34636834, 134221860, 130150588, 505290272, 490853416, 1908874584, 1857283156

Offset: 1

Seiichi Manyama, Feb 28 2018

nonn

			Solutions for n = 7:
----------------------------
1 +3 +5 +7 +9 +11 +13 =  49.
1 +3 +5 -7 +9 +11 +13 =  35.
1 +3 -5 +7 -9 +11 +13 =  21.
1 +3 -5 -7 -9 +11 +13 =   7.
1 -3 +5 +7 +9 -11 +13 =  21.
1 -3 +5 -7 +9 -11 +13 =   7.
1 -3 -5 +7 +9 +11 -13 =   7.
1 -3 -5 +7 -9 -11 +13 =  -7.
1 -3 -5 -7 +9 +11 -13 =  -7.
1 -3 -5 -7 -9 -11 +13 = -21.

Alois P. Heinz, Table of n, a(n) for n = 1..1000 (first 200 terms from Seiichi Manyama)

Cf. A063776, A156700, A300190.

b:= proc(n, i, m) option remember; `if`(i<1, `if`(n=0, 1, 0),
      add(b(irem(n+j, m), i-2, m), j=[i, m-i]))
    end:
a:= n-> b(n-1, 2*n-3, n):
seq(a(n), n=1..40);  # Alois P. Heinz, Mar 01 2018

Table[With[{s = Range[1, (2 n - 1), 2]}, Count[Map[Total[s #] &, Take[Tuples[{-1, 1}, Length@ s], -2^(n - 1)]], ?(Divisible[#, n] &)]], {n, 22}] (* _Michael De Vlieger, Mar 01 2018 *)

A054539 A000016 / 2.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 15, 26, 47, 86, 158, 293, 548, 1024, 1928, 3643, 6899, 13108, 24970, 47663, 91181, 174768, 335546, 645278, 1242767, 2396746, 4628198, 8947868, 17318417, 33554432, 65075294, 126322568, 245426708

Offset: 3

N. J. A. Sloane, Apr 10 2000

nonn

A063776(n) / 4, n>2.

A334125 Number of subsets of {1, 3, ..., 2n-1} which sum to 0 modulo 2n-1.

Original entry on oeis.org

2, 2, 2, 2, 4, 6, 10, 18, 30, 54, 98, 178, 328, 608, 1130, 2114, 3974, 7490, 14170, 26890, 51150, 97542, 186420, 356962, 684784, 1315870, 2532410, 4880646, 9418806, 18199014, 35204650, 68174116, 132152842, 256415958, 497967282, 967879954, 1882725390, 3665038872

Offset: 1

Jinyuan Wang, Apr 30 2020

nonn

			a(5) = 4 because there are 4 subsets of {1, 3, 5, 7, 9} which sum to 0 modulo 9: {}, {9}, {1, 3, 5}, {1, 3, 5, 9}.

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A063776, A178738.

f:= proc(n) local V, k;
  V:= Vector(2*n-1);
  V[2*n-1]:= 1;
  for k from 1 to 2*n-1 by 2 do
    V:= V + V[[$(k+1)..(2*n-1),$1..k]]
  od;
  V[2*n-1]
end proc:
map(f, [$1..40]); # Robert Israel, May 12 2020

a(n) = {my(v=Vec(prod(i=1, n, x^(2*i-1)+1))); sum(i=0, n^2\(2*n-1), v[n^2+1-i*(2*n-1)]); }

If 2*k - 1 is a prime, then a(k) = (2^k - 2*(-1)^floor(k/2))/(2*k - 1).

Conjecture: a(n) = 2*abs(A178738(n)).

A369878 Array read by antidiagonals: T(n,k) is the number of length n necklaces using at most k colors in which the convex hull of a set of beads of any color A cannot be transformed by rotation into the convex hull of a set of beads of any other color B (n >= 1, k >= 1).

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 3, 4, 1, 5, 4, 9, 4, 1, 6, 5, 16, 9, 8, 1, 7, 6, 25, 16, 27, 12, 1, 8, 7, 36, 25, 64, 93, 20, 1, 9, 8, 49, 36, 125, 304, 243, 32, 1, 10, 9, 64, 49, 216, 705, 928, 699, 60, 1, 11, 10, 81, 64, 343, 1356, 2405, 4384, 2019, 104, 1

Offset: 1

Maxim Karimov and Vladislav Sulima, Feb 03 2024

It is assumed that all beads lie on a circle and that the distance between any two adjacent beads is the same.

			n\k| 1  2    3     4     5      6      7      8       9 ...
---+---------------------------------------------------
 1 | 1  2    3     4     5      6      7      8       9 ... A000027
 2 | 1  2    3     4     5      6      7      8       9 ... A000027
 3 | 1  4    9    16    25     36     49     64      81 ... A000290
 4 | 1  4    9    16    25     36     49     64      81 ... A000290
 5 | 1  8   27    64   125    216    343    512     729 ... A000578
 6 | 1 12   93   304   705   1356   2317   3648    5409
 7 | 1 20  243   928  2405   5076   9415  15968   25353
 8 | 1 32  699  4384 15245  39216  84007 159104  275769
 9 | 1 60 2019 18256 72765 202236 457135 902784 1620441
...

Cf. A000012, A000027, A000290, A000578, A063776.

T(n,2) = A063776(n).

cp's OEIS Frontend

A053633 Triangular array T(n,k) giving coefficients in expansion of Product_{j=1..n} (1+x^j) mod x^(n+1)-1.

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A053634 a(n) = Sum_{ d divides n } phi(d)*2^(n/d)/(2n).

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A053636 a(n) = Sum_{odd d|n} phi(d)*2^(n/d).

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A267632 Triangle T(n, k) read by rows: the k-th column of the n-th row lists the number of ways to select k distinct numbers (k >= 1) from [1..n] so that their sum is divisible by n.

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A064355 Number of subsets of {1,2,..n} that sum to 1 mod n.

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A309128 (1/n) times the sum of the elements of all subsets of [n] whose sum is divisible by n.

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A300218 Number of solutions to 1 +- 3 +- 5 +- ... +- (2*n-1) == 0 mod n.

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A334125 Number of subsets of {1, 3, ..., 2n-1} which sum to 0 modulo 2n-1.