A174709 -id:A174709 - cp's OEIS frontend

A277646 Triangle T(n,k) = floor(n^2/k) for 1 <= k <= n^2, read by rows.

1, 4, 2, 1, 1, 9, 4, 3, 2, 1, 1, 1, 1, 1, 16, 8, 5, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 25, 12, 8, 6, 5, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 36, 18, 12, 9, 7, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 49, 24, 16, 12, 9, 8, 7, 6

Offset: 1

Jason Kimberley, Nov 09 2016

			The first five rows of the triangle are:
1;
4, 2, 1, 1;
9, 4, 3, 2, 1, 1, 1, 1, 1;
16, 8, 5, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1;
25, 12, 8, 6, 5, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;

Jason Kimberley, Table of n, a(n) for n = 1..10416 (the first 31 rows of the triangle)

Cf. Related triangles: A010766, A277647, A277648.
Rows of this triangle (with infinite trailing zeros):
T(1,k) = A000007(k-1),
T(2,k) = A033324(k),
T(3,k) = A033329(k),
T(4,k) = A033336(k),
T(5,k) = A033345(k),
T(6,k) = A033356(k),
T(7,k) = A033369(k),
T(8,k) = A033384(k),
T(9,k) = A033401(k),
T(10,k) = A033420(k),
T(100,k) = A033422(k),
T(10^3,k) = A033426(k),
T(10^4,k) = A033424(k).
Columns of this triangle:
T(n,1) = A000290(n),
T(n,2) = A007590(n),
T(n,3) = A000212(n),
T(n,4) = A002620(n),
T(n,5) = A118015(n),
T(n,6) = A056827(n),
T(n,7) = A056834(n),
T(n,8) = A130519(n+1),
T(n,9) = A056838(n),
T(n,10)= A056865(n),
T(n,12)= A174709(n+2).

A277646:=func;
[A277646(n,k):k in[1..n^2],n in[1..7]];

Table[Floor[n^2/k], {n, 7}, {k, n^2}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

T(n,k) = A010766(n^2,k).

A134546 Triangle read by rows: T(n, k) = Sum_{j=0..n} floor(j / k).

Original entry on oeis.org

1, 3, 1, 6, 2, 1, 10, 4, 2, 1, 15, 6, 3, 2, 1, 21, 9, 5, 3, 2, 1, 28, 12, 7, 4, 3, 2, 1, 36, 16, 9, 6, 4, 3, 2, 1, 45, 20, 12, 8, 5, 4, 3, 2, 1, 55, 25, 15, 10, 7, 5, 4, 3, 2, 1, 66, 30, 18, 12, 9, 6, 5, 4, 3, 2, 1, 78, 36, 22, 15, 11, 8, 6, 5, 4, 3, 2, 1, 91, 42, 26, 18, 13, 10, 7, 6, 5, 4, 3, 2, 1

Offset: 1

Gary W. Adamson, Oct 31 2007

From Bob Selcoe, Aug 08 2016: (Start)
Columns are partial sums of k-repeating increasing positive integers:
  Column 1 is {1+2+3+4+5+...}         = A000217 (triangular numbers);
  Column 2 is {1+1+2+2+3+3+4+4+...}   = A002620 (quarter-squares);
  Column 3 is {1+1+1+2+2+2+3+3+3+...} = A130518.
  Columns k = 4..7 are A130519, A130520, A174709 and A174738, respectively.
T(n, k) is the number of positive multiples of k which can be expressed as i-j, {i=1..n; j=0..n-1}. So for example, T(5, 2) = 6 because there are 6 ways to express i-j {i<=5} as a multiple of 2: {5-3, 4-2, 3-1, 2-0, 5-1 and 4-0}. (End)
Conjecture: For T(n, k) n >= k^(3/2), there is at least one prime in the interval [T(n-1, k+1), T(n, k)]. - Bob Selcoe, Aug 21 2016
Theorem: For n >= 3*k, T(n, k) is composite. - Daniel Hoying, Jul 08 2020

			The triangle T(n, k) begins:
   n\k  1   2   3   4  5  6  7  8  9  10 ...
   1:   1
   2:   3   1
   3:   6   2   1
   4:  10   4   2   1
   5:  15   6   3   2  1
   6:  21   9   5   3  2  1
   7:  28  12   7   4  3  2  1
   8:  36  16   9   6  4  3  2  1
   9:  45  20  12   8  5  4  3  2  1
  10:  55  25  15  10  7  5  4  3  2   1
... Reformatted. - _Wolfdieter Lang_, Feb 04 2015
T(10,3) = 15: 3*floor(10/3)*floor(13/3)/2 - floor(10/3)*(3-1 - 13 mod 3) = 3*3*4/2 - 3*(3-1-1) = 18 - 3 = 15. - _Bob Selcoe_, Aug 21 2016

Michael De Vlieger, Table of n, a(n) for n = 1..11325 (1 <= n <= 150)

Cf. A078567 (row sums), A000217 (column 1).

Cf. A004736, A051731, A002620, A130518, A130519, A130520, A174709, A174738.

T := proc(n, k) option remember: `if`(n = k, 1, T(n-1, k) + iquo(n,k)) end:
seq(seq(T(n,k), k=1..n),n=1..16); # Peter Luschny, May 26 2020

nn = 12; f[w_] := Map[PadRight[#, nn] &, w]; MapIndexed[Take[#1, First@ #2] &, f@ Table[Reverse@ Range@ n, {n, nn}].f@ Table[Boole@ Divisible[n, #] & /@ Range@ n, {n, nn}]] // Flatten (* Michael De Vlieger, Aug 10 2016 *)

t(n, k) = if (k>n, 0, if (n==1, 1, t(n-1, k) + n\k));
tabl(nn) = {m = matrix(nn, nn, n , k, t(n,k)); for (n=1, nn, for (k=1, n, print1(m[n, k], ", ");); print(););} \\ Michel Marcus, Jan 18 2015

trg(nn) = {ma = matrix(nn, nn, n, k, if (k<=n, n-k+1, 0)); mb = matrix(nn, nn, n, k, if (k<=n, !(n%k), 0)); ma*mb;} \\ Michel Marcus, Jan 20 2015

Original definition: T = A004736 * A051731 as infinite lower triangular matrices.
In other words: T(n, k) = Sum_{m=k..n} A004736(n, m)*A051731(m, k).
T(n, k) = 0 if n < k, T(1, 1) = 1, and T(n, k) = T(n-1, k) + floor(n/k), for n >= 2. - Richard R. Forberg, Jan 17 2015
T(n, k) = k*floor(n/k)*floor((n+k)/k)/2 - floor(n/k)*(k-1-(n mod k)). - Bob Selcoe, Aug 21 2016
T(n, k) = k*A000217(b) + (b+1)*[(n +1)-(b + 1)*k] for 1 <= k <= floor[(n + 1) / 2] where b = floor[(n - k + 1) / k], T(n, k) = n-k+1 for floor[(n + 1) / 2] < k <= n and T(n, k) = 0 for k > n. - Henri Gonin, May 12 2020
T(n, k) = (-k/2)*floor(n/k)^2+(n-k/2+1)*floor(n/k). - Daniel Hoying, May 25 2020
From Daniel Hoying, Jul 06 2020: (Start)
T(m + 2*n - 1, m + n) = n for n > 0, m >= 0.
T(3*m + 3*ceiling((n-3)/6) + (n+1)/2, 2*m + 2*ceiling((n-3)/6) + 1) = n for n > 0, n odd, 0 <= m <= floor(n/3).
T(3*m + 3*ceiling(n/6) + n/2 - 1, 2*m + 2*ceiling(n/6)) = n for n > 0, n even, 0 <= m <= floor(n/3). (End)

Edited. Name clarified. Formulas rewritten. - Wolfdieter Lang, Feb 04 2015
Corrected and extended by Michael De Vlieger, Aug 10 2016
Edited and new name from Peter Luschny, Apr 02 2025

A221912 Partial sums of floor(n/12).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 125, 130, 135, 140, 145, 150, 155

Offset: 0

Philippe Deléham, Mar 27 2013

Apart from the initial zeros, the same as A008730.

			..0....0....0....0....0....0....0....0....0....0....0....0
..1....2....3....4....5....6....7....8....9...10...11...12
.14...16...18...20...22...24...26...28...30...32...34...36
.39...42...45...48...51...54...57...60...63...66...69...72
.76...80...84...88...92...96..100..104..108..112..116..120
125..130..135..140..145..150..155..160..165..170..175..180
186..192..198..204..210..216..222..228..234..240..246..252
259..266..273..280..287..294..301..308..315..322..329..336
344..352..360..368..376..384..392..400..408..416..424..432
441..450..459..468..477..486..495..504..513..522..531..540
...

Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,0,0,0,1,-2,1).

Cf. A000217, A002620, A130518, A130519, A130520, A174709, A174738, A118729, A218470, A131242, A218530.

Accumulate[Floor[Range[0,70]/12]] (* or *) LinearRecurrence[{2,-1,0,0,0,0,0,0,0,0,0,1,-2,1},{0,0,0,0,0,0,0,0,0,0,0,0,1,2},70] (* Harvey P. Dale, Mar 23 2015 *)

a(12n) = A051866(n).
a(12n+1) = A139267(n).
a(12n+2) = A094159(n).
a(12n+3) = A033579(n).
a(12n+4) = A049452(n).
a(12n+5) = A033581(n).
a(12n+6) = A049453(n).
a(12n+7) = A033580(n).
a(12n+8) = A195319(n).
a(12n+9) = A202804(n).
a(12n+10) = A211014(n).
a(12n+11) = A049598(n).
G.f.: x^12/((1-x)^2*(1-x^12)).
a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=0, a(5)=0, a(6)=0, a(7)=0, a(8)=0, a(9)=0, a(10)=0, a(11)=0, a(12)=1, a(13)=2, a(n)=2*a(n-1)- a(n-2)+ a(n-12)- 2*a(n-13)+ a(n-14). - Harvey P. Dale, Mar 23 2015

A269445 a(n) = Sum_{k=0..n} floor(k/13).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175

Offset: 0

Ilya Gutkovskiy, Feb 27 2016

Partial sums of A090620.

More generally, the ordinary generating function for the Sum_{k=0..n} floor(k/m) is x^m/((1 - x^m)*(1 - x)^2).

Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,0,0,0,0,1,-2,1).

Cf. A090620.

Cf. similar sequences of Sum_{k=0..n} floor(k/m): A002620 (m=2), A130518 (m=3), A130519 (m=4), A130520 (m=5), A174709 (m=6), A174738 (m=7), A118729 (m=8), A218470 (m=9), A131242 (m=10), A218530 (m=11), A221912 (m=12), this sequence (m=13).

Table[Sum[Floor[k/13], {k, 0, n}], {n, 0, 73}]
LinearRecurrence[{2, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -2, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2}, 74]

G.f.: x^13/((1 - x^13)*(1 - x)^2).

a(n) = 2*a(n-1) - a(n-2) + a(n-13) - 2*a(n-14) + a(n-15).

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A277646 Triangle T(n,k) = floor(n^2/k) for 1 <= k <= n^2, read by rows.

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A134546 Triangle read by rows: T(n, k) = Sum_{j=0..n} floor(j / k).

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A221912 Partial sums of floor(n/12).

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A269445 a(n) = Sum_{k=0..n} floor(k/13).

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