A026741 -id:A026741 - cp's OEIS frontend

A103628 Total sum of parts of multiplicity 1 in all partitions of n.

0, 1, 2, 6, 10, 21, 33, 59, 89, 145, 212, 325, 463, 680, 948, 1348, 1845, 2558, 3446, 4681, 6219, 8306, 10901, 14352, 18632, 24230, 31151, 40077, 51074, 65088, 82290, 103986, 130517, 163679, 204078, 254174, 314975, 389839, 480369, 591133, 724600, 886965

Offset: 0

Vladeta Jovovic, Mar 25 2005

Total number of parts of multiplicity 1 in all partitions of n is A024786(n+1).

Equals A000041 convolved with A026741. - Gary W. Adamson, Jun 11 2009

			Partitions of 4 are [1,1,1,1], [1,1,2], [2,2], [1,3], [4] and a(4) = 0 + 2 + 0 + (1+3) + 4 = 10.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A026741. - Gary W. Adamson, Jun 11 2009

Column k=1 of A222730. - Alois P. Heinz, Mar 03 2013

gf:=x*(1+x+x^2)/(1-x^2)^2/product((1-x^k), k=1..500): s:=series(gf, x, 100): for n from 0 to 60 do printf(`%d,`,coeff(s, x, n)) od: # James Sellers, Apr 22 2005
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, [1, 0],
      `if`(i<1, [0, 0], add((l->`if`(j=1, [l[1],
       l[2]+l[1]*i], l))(b(n-i*j, i-1)), j=0..n/i)))
    end:
a:= n-> b(n, n)[2]:
seq(a(n), n=0..50);  # Alois P. Heinz, Feb 03 2013

b[n_, p_] := b[n, p] = If[n == 0 && p == 0, {1, 0}, If[p == 0, Array[0&, n+2], Sum[Function[l, ReplacePart[l, m+2 -> p*l[[1]] + l[[m+2]]]][Join[b[n-p*m, p-1], Array[0&, p*m]]], {m, 0, n/p}]]]; a[n_] := b[n, n][[3]]; a[0] = 0; Table[a[n], {n, 0, 50}]  (* Jean-François Alcover, Jan 24 2014, after Alois P. Heinz *)

G.f.: x*(1+x+x^2)/(1-x^2)^2 /Product_{k>0}(1-x^k).
a(n) = A066186(n) - A194544(n). - Omar E. Pol, Nov 20 2011
a(n) = 3*A014153(n)/4 - 3*A000070(n)/4 - A270143(n+1)/4 + A087787(n)/4. - Vaclav Kotesovec, Nov 05 2016
a(n) ~ 3^(3/2) * exp(Pi*sqrt(2*n/3)) / (8*Pi^2) * (1 - Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Nov 05 2016

More terms from James Sellers, Apr 22 2005

A241269 Denominator of c(n) = (n^2+n+2)/((n+1)(n+2)(n+3)).

Original entry on oeis.org

3, 6, 15, 60, 105, 21, 126, 360, 495, 330, 429, 1092, 1365, 420, 1020, 2448, 2907, 1710, 1995, 4620, 5313, 759, 3450, 7800, 8775, 4914, 5481, 12180, 13485, 3720, 8184, 17952, 19635, 10710, 11655, 25308, 27417, 3705, 15990, 34440, 37023, 19866, 21285, 45540

Offset: 0

Paul Curtz, Apr 18 2014

All terms are multiples of 3.
Difference table of c(n):
   1/3,     1/6,    2/15,    7/60,    2/21,...
  -1/6,    -1/30,  -1/60,   -1/84,   -1/105,...
   2/15,    1/60,   1/210,   1/420,   1/630,...
  -7/60,   -1/84,  -1/420,  -1/1260, -1/2520,... .
This is an autosequence of the second kind; the inverse binomial transform is the signed sequence. The main diagonal is the first upper diagonal multiplied by 2.
Denominators of the main diagonal: A051133(n+1).
Denominators of the first upper diagonal; A000911(n).
c(n) is a companion to A026741(n)/A045896(n).
Based on the Akiyama-Tanigawa transform applied to 1/(n+1) which yields the Bernoulli numbers A164555(n)/A027642(n).
Are the numerators of the main diagonal (-1)^n? If yes, what is the value of 1/3 - 1/30 + 1/210,... or 1 - 1/10 + 1/70 - 1/420, ... , from A002802(n)?
Is a(n+40) - a(n) divisible by 10?
No: a(5) = 21 but a(45) = 12972. # Robert Israel, Jul 17 2023
Are the common divisors to A014206(n) and A007531(n+3) of period 16: repeat 2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2?
Reduce c(n) = f(n) = b(n)/a(n) = 1/3, 1/6, 2/15, 7/60, 11/105, 2/21, 11/126, 29/360, ... .
Consider the successively interleaved autosequences (also called eigensequences) of the second kind and of the first kind
  1,    1/2,  1/3,    1/4,     1/5,     1/6,   ...
  0,    1/6,  1/6,    3/20,    2/15,    5/42,  ...
  1/3,  1/6,  2/15,   7/60,   11/105,   2/21,  ...
  0,    1/10, 1/10,  13/140,   3/35,    5/63,  ...
  1/5,  1/10, 3/35,  11/140,  23/315,  43/630, ...
  0,    1/14, 1/14,  17/252,   4/63,   ...
This array is Au1(m,n). Au1(0,0)=1, Au1(0,1)=1/2.
Au1(m+1,n) = 2*Au1(m,n+1) - Au1(m,n).
First row: see A003506, Leibniz's Harmonic Triangle.
Second row: A026741/A045896.
a(n) is the denominator of the third row f(n).
The first column is 1, 0, 1/3, 0, 1/5, 0, 1/7, 0, ... . Numerators: A093178(n+1). This incites, considering tan(1), to introduce before the first row
Ta0(n) = 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, ... .

Robert Israel, Table of n, a(n) for n = 0..10000

seq(denom((n^2+n+2)/((n+1)*(n+2)*(n+3))),n=0..1000);

Denominator[Table[(n^2+n+2)/Times@@(n+{1,2,3}),{n,0,50}]] (* Harvey P. Dale, Mar 27 2015 *)

for(n=0, 100, print1(denominator((n^2+n+2)/((n+1)*(n+2)*(n+3))), ", ")) \\ Colin Barker, Apr 18 2014

c(n) = A014206(n)/A007531(n+3).
The sum of the difference table main diagonal is 1/3 - 1/30 + 1/210 - ... = 10*A086466-4 = 4*(sqrt(5)*log(phi)-1) = 0.3040894... - Jean-François Alcover, Apr 22 2014
a(n) = (n+1)*(n+2)*(n+3)/gcd(4*n - 4, n^2 + n + 2), where gcd(4*n - 4, n^2 + n + 2) is periodic with period 16. - Robert Israel, Jul 17 2023

More terms from Colin Barker, Apr 18 2014

A334006 Triangle read by rows: T(n,k) = (the number of nonnegative bases m < n such that m^k == m (mod n))/(the number of nonnegative bases m < n such that -m^k == m (mod n)) for nonnegative k < n, n >= 1.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 2, 1, 3, 1, 5, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 7, 1, 3, 1, 3, 1, 1, 4, 1, 5, 1, 5, 1, 5, 1, 9, 1, 3, 1, 3, 1, 7, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 11, 1, 3, 1, 3, 1, 3, 1, 3, 1, 1, 6, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 13, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 1, 7, 1, 3, 1, 3

Offset: 1

Juri-Stepan Gerasimov, Apr 12 2020

If the sum of proper divisors of q in row q <= q, then q are 1, 2, 3, 4, 5, 8, 16, 17, 32, 64, 128, 256, 257, ...(union of Fermat primes and powers of 2).

			Triangle T(n,k) begins:
  n\k| 0   1  2  3  4   5  6  7  8   9 10 11 12  13 14 15 16
  ---+------------------------------------------------------
   1 | 1;
   2 | 1,  1;
   3 | 1,  3, 1;
   4 | 1,  2, 1, 3;
   5 | 1,  5, 1, 1, 1;
   6 | 1,  3, 1, 3, 1,  3;
   7 | 1,  7, 1, 3, 1,  3, 1;
   8 | 1,  4, 1, 5, 1,  5, 1, 5;
   9 | 1,  9, 1, 3, 1,  3, 1, 7, 1;
  10 | 1,  5, 1, 1, 1,  5, 1, 1, 1,  5;
  11 | 1, 11, 1, 3, 1,  3, 1, 3, 1,  3, 1;
  12 | 1,  6, 1, 9, 1,  9, 1, 9, 1,  9, 1, 9;
  13 | 1, 13, 1, 1, 1,  5, 1, 1, 1,  5, 1, 1, 1;
  14 | 1,  7, 1, 3, 1,  3, 1, 7, 1,  3, 1, 3, 1, 7;
  15 | 1, 15, 1, 3, 1, 15, 1, 3, 1, 15, 1, 3, 1, 15, 1;
  16 | 1,  8, 1, 5, 1,  9, 1, 5, 1,  9, 1, 5, 1,  9, 1, 5;
  17 | 1, 17, 1, 1, 1,  1, 1, 1, 1,  1, 1, 1, 1,  1, 1, 1, 1;
  ...
For (n, k) = (7, 3), there are three nonnegative values of m < n such that m^3 == m (mod 7) (namely 0, 1, and 6) and one nonnegative value of m < n such that -m^3 == m (mod 7) (namely 0), so T(7,3) = 3/1 = 3.

Cf. A000012, A000079, A002322, A010684, A019434, A023506, A026741, A182816, A333570, A337454, A337632, A337633, A337820, A337910.

[[#[m: m in [0..n-1] | m^k mod n eq m]/#[m: m in [0..n-1] | -m^k mod n eq m]: k in [0..n-1]]: n in [1..17]];

T(n, k) = sum(m=0, n-1, Mod(m, n)^k == m)/sum(m=0, n-1, -Mod(m, n)^k == m);
matrix(7, 7, n, k, k--; if (k>=n, 0, T(n,k))) \\ to see the triangle \\ Michel Marcus, Apr 17 2020

Name corrected by Peter Kagey, Sep 12 2020

A013942 Triangle of numbers T(n,k) = floor(2n/k), k=1..2n, read by rows.

Original entry on oeis.org

2, 1, 4, 2, 1, 1, 6, 3, 2, 1, 1, 1, 8, 4, 2, 2, 1, 1, 1, 1, 10, 5, 3, 2, 2, 1, 1, 1, 1, 1, 12, 6, 4, 3, 2, 2, 1, 1, 1, 1, 1, 1, 14, 7, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 16, 8, 5, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 18, 9, 6, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 20, 10, 6, 5, 4, 3

Offset: 1

Clark Kimberling

a(n) is also the leading term in period of continued fraction for n-th nonsquare.

Row A026741(n) contains n and all rows with a smaller row number do not contain n. - Reinhard Zumkeller, Jun 04 2013

			First four rows:
  2 1
  4 2 1 1
  6 3 2 1 1 1
  8 4 2 2 1 1 1 1
  ...

Reinhard Zumkeller, Rows n = 1..100 of table, flattened

Cf. A010766.

Cf. A005843 (row lengths and left edge), A062550 (row sums).

a013942 n k = a013942_tabf !! (n-1) !! (k-1)
a013942_row n = map (div (n * 2)) [1 .. 2 * n]
a013942_tabf = map a013942_row [1 ..]
-- Reinhard Zumkeller, Jun 04 2013

f[n_,h_]:=FractionalPart[(n^2+h)^(1/2)];
g[n_,h_]:=Floor[1/f[n,h]];
TableForm[Table[g[n,h],{n,1,13},{h,1,2n}]]

T(n, k) = 2*n\k;
tabf(nn) = for (n=1, nn, for (k=1, 2*n, print1(T(n,k), ", ")); print()); \\ Michel Marcus, Sep 30 2016

T(n,k) = floor(2n/k), k=1,...,2n.

T(n,k) = [1/{sqrt(k+n^2)}], k=1,2,...,2n, {}=fractional part, []=floor.

Keyword tabl replaced by tabf and missing a(90)=1 inserted by Reinhard Zumkeller, Jun 04 2013

A098832 Square array read by antidiagonals: even-numbered rows of the table are of the form n(n+m) and odd-numbered rows are of the form n(n+m)/2.

Original entry on oeis.org

1, 3, 3, 6, 8, 2, 10, 15, 5, 5, 15, 24, 9, 12, 3, 21, 35, 14, 21, 7, 7, 28, 48, 20, 32, 12, 16, 4, 36, 63, 27, 45, 18, 27, 9, 9, 45, 80, 35, 60, 25, 40, 15, 20, 5, 55, 99, 44, 77, 33, 55, 22, 33, 11, 11, 66, 120, 54, 96, 42, 72, 30, 48, 18, 24, 6, 78, 143, 65, 117, 52, 91, 39, 65, 26, 39, 13, 13

Offset: 1

Eugene McDonnell (eemcd(AT)mac.com), Nov 02 2004

The rows of this table and that in A098737 are related. Given a function f = n/( 1 + (1+n) mod(2) ), row n of A098737 can be derived from row n of T by multiplying the latter by f(n); row n of T can be derived from row n of A098737 by dividing the latter by f(n).

			Array begins as:
  1,  3,  6, 10, 15, 21,  28,  36,  45 ... A000217;
  3,  8, 15, 24, 35, 48,  63,  80,  99 ... A005563;
  2,  5,  9, 14, 20, 27,  35,  44,  54 ... A000096;
  5, 12, 21, 32, 45, 60,  77,  96, 117 ... A028347;
  3,  7, 12, 18, 25, 33,  42,  52,  63 ... A027379;
  7, 16, 27, 40, 55, 72,  91, 112, 135 ... A028560;
  4,  9, 15, 22, 30, 39,  49,  60,  72 ... A055999;
  9, 20, 33, 48, 65, 84, 105, 128, 153 ... A028566;
  5, 11, 18, 26, 35, 45,  56,  68,  81 ... A056000;
Antidiagonals begin as:
   1;
   3,  3;
   6,  8,  2;
  10, 15,  5,  5;
  15, 24,  9, 12,  3;
  21, 35, 14, 21,  7,  7;
  28, 48, 20, 32, 12, 16,  4;
  36, 63, 27, 45, 18, 27,  9,  9;
  45, 80, 35, 60, 25, 40, 15, 20,  5;
  55, 99, 44, 77, 33, 55, 22, 33, 11, 11;

G. C. Greubel, Antidiagonals n = 1..50, flattened

Row m of array: A000217 (m=1), A005563 (m=2), A000096 (m=3), A028347 (m=4), A027379 (m=5), A028560 (m=6), A055999 (m=7), A028566 (m=8), A056000 (m=9), A098603 (m=10), A056115 (m=11), A098847 (m=12), A056119 (m=13), A098848 (m=14), A056121 (m=15), A098849 (m=16), A056126 (m=17), A098850 (m=18), A051942 (m=19).
Column m of array: A026741 (m=1), A022998 (m=2), A165351 (m=3).
Cf. A098737, A168509, A181900.

A098832:= func< n,k | (1/4)*(3+(-1)^k)*(n+1)*(n-k+1) >;
[A098832(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Jul 31 2022

A098832[n_, k_]:= (1/4)*(3+(-1)^k)*(n+1)*(n-k+1);
Table[A098832[n,k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Jul 31 2022 *)

def A098832(n,k): return (1/4)*(3+(-1)^k)*(n+1)*(n-k+1)
flatten([[A098832(n,k) for k in (1..n)] for n in (1..15)]) # G. C. Greubel, Jul 31 2022

Item m of row n of T is given (in infix form) by: n T m = n * (n + m) / (1 + m (mod 2)). E.g. Item 4 of row 3 of T: 3 T 4 = 14.
From G. C. Greubel, Jul 31 2022: (Start)
A(n, k) = (1/4)*(3 + (-1)^n)*k*(k+n) (array).
T(n, k) = (1/4)*(3 + (-1)^k)*(n+1)*(n-k+1) (antidiagonal triangle).
Sum_{k=1..n} T(n, k) = (1/8)*(n+1)*( (3*n-1)*(n+1) + (1+(-1)^n)/2 ).
T(2*n-1, n) = A181900(n).
T(2*n+1, n) = 2*A168509(n+1). (End)

Missing terms added by G. C. Greubel, Jul 31 2022

A160467 a(n) = 1 if n is odd; otherwise, a(n) = 2^(k-1) where 2^k is the largest power of 2 that divides n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 16, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 32, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 16

Offset: 1

Johannes W. Meijer, May 24 2009, Jun 28 2011

Fifth factor of the row sums A160466 of the Eta triangle A160464.
From Peter Luschny, May 31 2009: (Start)
Let odd(n) be the characteristic function of the odd numbers (A000035) and sigma(n) the number of 1's in binary expansion of n (A000120). Then a(n) = 2^(sigma(n-1) - sigma(n) + odd(n)).
Let B_{n} be the Bernoulli number. Then this sequence is also
a(n) = denominator(4*(4^n-1)*B_{2*n}/n). (End)

Antti Karttunen, Table of n, a(n) for n = 1..16384

Cf. A000265, A001620, A007814, A026741, A140670, A160464, A220466, A006519, A000010.

nmax:=96: p:= floor(log[2](nmax)): for n from 1 to nmax do a(n):=1 end do: for q from 1 to p do for n from 1 to nmax do if n mod 2^q = 0 then a(n):= 2^(q-1) end if: end do: end do: seq(a(n), n=1..nmax);
From Peter Luschny, May 31 2009: (Start)
a := proc(n) local sigma; sigma := proc(n) local i; add(i,i=convert(n,base,2)) end; 2^(sigma(n-1)-sigma(n)+`if`(type(n,odd),1,0)) end: seq(a(n), n=1..96);
a := proc(n) denom(4*(4^n-1)*bernoulli(2*n)/n) end: seq(a(n), n=1..96); (End)

a[n_] := If[OddQ[n], 1, 2^(IntegerExponent[n, 2] - 1)]; Array[a, 100] (* Amiram Eldar, Jul 02 2020 *)

A160467(n) = 2^max(valuation(n,2)-1,0); \\ Antti Karttunen, Nov 18 2017, after Max Alekseyev's Feb 09 2011 formula.

def A160467(n): return max(1,(n&-n)>>1) # Chai Wah Wu, Jul 08 2022

a(n) = A026741(n)/A000265(n). - Paul Curtz, Apr 18 2010
a(n) = 2^max(A007814(n) - 1, 0). - Max Alekseyev, Feb 09 2011
a((2*n-1)*2^p) = A011782(p), p >= 0 and n >= 1. - Johannes W. Meijer, Jan 25 2013
a(n) = (1 + A140670(n))/2. - Antti Karttunen, Nov 18 2017
From Amiram Eldar, Dec 31 2022: (Start)
Dirichlet g.f.: zeta(s)*(2^s-2+1/2^s)/(2^s-2).
Sum_{k=1..n} a(k) ~ (1/(4*log(2)))*n*log(n) + (5/8 + (gamma-1)/(4*log(2)))*n, where gamma is Euler's constant (A001620). (End)
a(n) = A006519(n)/gcd(n,2). - Ridouane Oudra, Feb 08 2025
a(n) = A000010(A006519(n)). - Ridouane Oudra, Jul 27 2025

Keyword mult added by Max Alekseyev, Feb 09 2011

Name changed by Antti Karttunen, Nov 18 2017

A165351 Numerator of 3*n/2.

Original entry on oeis.org

0, 3, 3, 9, 6, 15, 9, 21, 12, 27, 15, 33, 18, 39, 21, 45, 24, 51, 27, 57, 30, 63, 33, 69, 36, 75, 39, 81, 42, 87, 45, 93, 48, 99, 51, 105, 54, 111, 57, 117, 60, 123, 63, 129, 66, 135, 69, 141, 72, 147, 75, 153, 78, 159, 81, 165, 84, 171, 87, 177, 90, 183, 93, 189, 96, 195

Offset: 0

Paul Curtz, Sep 16 2009

First trisection of A026741. The other trisections are A165355 and A165367.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A000034 (denominator).

Cf. A008585, A016945, A026741, A098832, A165355, A165367.

[Numerator(3*n/2): n in [0..100]]; // Vincenzo Librandi, Mar 03 2014

A165351:=n->numer(3*n/2); seq(A165351(k), k=0..100); # Wesley Ivan Hurt, Oct 11 2013

Table[Numerator[3n/2], {n, 0, 100}] (* Wesley Ivan Hurt, Oct 11 2013 *)
CoefficientList[Series[3*x*(1+x+x^2)/(1-x^2)^2, {x, 0, 70}], x] (* Vincenzo Librandi, Mar 03 2014 *)
LinearRecurrence[{0,2,0,-1},{0,3,3,9},70] (* Harvey P. Dale, Jun 20 2021 *)

[3*n*(3-(-1)^n)/4 for n in (0..100)] # G. C. Greubel, Jul 31 2022

a(n) = A026741(3*n) = 3*A026741(n).
a(2n) = A008585(n).
a(2n+1) = A016945(n).
G.f.: 3*x*(1+x+x^2)/((1-x)^2 * (1+x)^2).
a(n) = numerator(3n/2). - Wesley Ivan Hurt, Oct 11 2013
a(n) = 3*n / (1 + ((n+1) mod 2)). - Wesley Ivan Hurt, Feb 25 2014
From G. C. Greubel, Jul 31 2022: (Start)
a(n) = 3*n*(3 - (-1)^n)/4.
E.g.f.: (3*x/2)*( 2*cosh(x) + sinh(x) ). (End)

Edited and extended by R. J. Mathar, Sep 26 2009

New name from Wesley Ivan Hurt, Oct 13 2013

A227042 Triangle of denominators of harmonic mean of n and m, 1 <= m <= n.

Original entry on oeis.org

1, 3, 1, 2, 5, 1, 5, 3, 7, 1, 3, 7, 4, 9, 1, 7, 1, 1, 5, 11, 1, 4, 9, 5, 11, 6, 13, 1, 9, 5, 11, 3, 13, 7, 15, 1, 5, 11, 2, 13, 7, 5, 8, 17, 1, 11, 3, 13, 7, 3, 2, 17, 9, 19, 1, 6, 13, 7, 15, 8, 17, 9, 19, 10, 21, 1

Offset: 1

Wolfdieter Lang, Jul 01 2013

See the comments under A227041. a(n,m) gives the denominator of H(n,m) = 2*n*m/(n+m) in lowest terms.

			The triangle of denominators of H(n,m), called a(n,m) begins:
n\m  1   2   3   4   5    6    7    8    9   10  11 ...
1:   1
2:   3   1
3:   2   5   1
4:   5   3   7   1
5:   3   7   4   9   1
6:   7   1   1   5  11    1
7:   4   9   5  11   6   13    1
8;   9   5  11   3  13    7   15    1
9:   5  11   2  13   7    5    8   17    1
10: 11   3  13   7   3    2   17    9   19    1
11:  6  13   7  15   8   17    9   19   10   21   1
...
For the triangle of the rationals H(n,m) see the example section of A227041.
H(4,2) = denominator(16/6) = denominator(8/3) = 3 = 6/gcd(6,8) = 6/2.

Eric Weisstein's World of Mathematics, Harmonic Mean.

Cf. A227041, A026741 (column m=1), A000265 (m=2), A106619 (m=3), A227140(n+8) (m=4), A227108 (m=5), A221918/A221919.

a(n,m) = denominator(2*n*m/(n+m)), 1 <= m <= n.

a(n,m) = (n+m)/gcd(2*n*m, n+m) = (n+m)/gcd(n+m, 2*m^2), 1 <= m <= n.

A317311 Multiples of 11 and odd numbers interleaved.

Original entry on oeis.org

0, 1, 11, 3, 22, 5, 33, 7, 44, 9, 55, 11, 66, 13, 77, 15, 88, 17, 99, 19, 110, 21, 121, 23, 132, 25, 143, 27, 154, 29, 165, 31, 176, 33, 187, 35, 198, 37, 209, 39, 220, 41, 231, 43, 242, 45, 253, 47, 264, 49, 275, 51, 286, 53, 297, 55, 308, 57, 319, 59, 330, 61, 341, 63, 352, 65, 363, 67, 374, 69

Offset: 0

Omar E. Pol, Jul 25 2018

Partial sums give the generalized 15-gonal numbers (A277082).

a(n) is also the length of the n-th line segment of the rectangular spiral wh0se vertices are the generalized 15-gonal numbers.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A008593 and A005408 interleaved.
Column 11 of A195151.
Sequences whose partial sums give the generalized k-gonal numbers: A026741 (k=5), A001477 (k=6), zero together with A080512 (k=7), A022998 (k=8), A195140 (k=9), zero together with A165998 (k=10), A195159 (k=11), A195161 (k=12), A195312 (k=13), A195817 (k=14).
Cf. A277082.

{0}~Join~Riffle[2 Range@ # - 1, 11 Range@ #] &@ 35 (* or *)
CoefficientList[Series[x (1 + 11 x + x^2)/((1 - x)^2*(1 + x)^2), {x, 0, 69}], x] (* Michael De Vlieger, Jul 26 2018 *)
LinearRecurrence[{0,2,0,-1},{0,1,11,3},90] (* Harvey P. Dale, Aug 28 2022 *)

concat(0, Vec(x*(1 + 11*x + x^2) / ((1 - x)^2*(1 + x)^2) + O(x^40))) \\ Colin Barker, Jul 26 2018

a(2n) = 11*n, a(2n+1) = 2*n + 1.
From Colin Barker, Jul 26 2018: (Start)
G.f.: x*(1 + 11*x + x^2) / ((1 - x)^2*(1 + x)^2).
a(n) = 2*a(n-2) - a(n-4) for n>3. (End)
Multiplicative with a(2^e) = 11*2^(e-1), and a(p^e) = p^e for an odd prime p. - Amiram Eldar, Oct 14 2023
Dirichlet g.f.: zeta(s-1) * (1 + 9/2^s). - Amiram Eldar, Oct 25 2023
a(n) = (13 + 9*(-1)^n)*n/4. - Aaron J Grech, Aug 20 2024

A328203 Expansion of Sum_{k>=1} k * x^k / (1 - x^(2*k))^2.

Original entry on oeis.org

1, 2, 5, 4, 8, 10, 11, 8, 20, 16, 17, 20, 20, 22, 42, 16, 26, 40, 29, 32, 58, 34, 35, 40, 53, 40, 74, 44, 44, 84, 47, 32, 90, 52, 94, 80, 56, 58, 106, 64, 62, 116, 65, 68, 174, 70, 71, 80, 102, 106, 138, 80, 80, 148, 146, 88, 154, 88, 89, 168, 92, 94, 241, 64, 172

Offset: 1

Ilya Gutkovskiy, Oct 07 2019

nonn

Antti Karttunen, Table of n, a(n) for n = 1..20000
Jon Maiga, Computer-generated formulas for A328203, Sequence Machine.

Cf. A000005, A000079 (fixed points), A000203, A001227, A002131, A003602, A006519, A006918, A026741, A038040, A109168, A113415, A140472, A152211, A193356, A245579, A349353 (Dirichlet inverse), A349354.

Cf. also A347957.

a:=[]; for k in [1..65] do if IsOdd(k) then a[k]:=(k * #Divisors(k) + DivisorSigma(1,k)) / 2; else a[k]:=(k * (#Divisors(k) - #Divisors(k div 2)) + DivisorSigma(1,k) - DivisorSigma(1,k div 2)) / 2;  end if; end for; a; // Marius A. Burtea, Oct 07 2019

nmax = 65; CoefficientList[Series[Sum[k x^k/(1 - x^(2 k))^2, {k, 1, nmax}], {x, 0, nmax}], x] // Rest
a[n_] := DivisorSum[n, (n Mod[#, 2] + Boole[OddQ[n/#]] #)/2 &]; Table[a[n], {n, 1, 65}]

A328203(n) = if(n%2,(1/2)*(sigma(n)+(n*numdiv(n))),2*A328203(n/2)); \\ Antti Karttunen, Nov 13 2021

a(n) = (n * d(n) + sigma(n)) / 2 if n odd, (n * (d(n) - d(n/2)) + sigma(n) - sigma(n/2)) / 2 if n even.
a(n) = (n * A001227(n) + A002131(n)) / 2.
a(2*n) = 2 * a(n).
From Antti Karttunen, Nov 13 2021: (Start)
The following two convolutions were found by Jon Maiga's Sequence Machine search algorithm. Both are easy to prove:
a(n) = Sum_{d|n} A003602(d) * A026741(n/d).
a(n) = Sum_{d|n} A109168(d) * A193356(n/d), where A109168(d) = A140472(d) = (d+A006519(d))/2.
(End)

cp's OEIS Frontend

A103628 Total sum of parts of multiplicity 1 in all partitions of n.

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A241269 Denominator of c(n) = (n^2+n+2)/((n+1)*(n+2)*(n+3)).

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A334006 Triangle read by rows: T(n,k) = (the number of nonnegative bases m < n such that m^k == m (mod n))/(the number of nonnegative bases m < n such that -m^k == m (mod n)) for nonnegative k < n, n >= 1.

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A013942 Triangle of numbers T(n,k) = floor(2n/k), k=1..2n, read by rows.

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A098832 Square array read by antidiagonals: even-numbered rows of the table are of the form n*(n+m) and odd-numbered rows are of the form n*(n+m)/2.

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A160467 a(n) = 1 if n is odd; otherwise, a(n) = 2^(k-1) where 2^k is the largest power of 2 that divides n.

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A165351 Numerator of 3*n/2.

A241269 Denominator of c(n) = (n^2+n+2)/((n+1)(n+2)(n+3)).

A098832 Square array read by antidiagonals: even-numbered rows of the table are of the form n(n+m) and odd-numbered rows are of the form n(n+m)/2.