A048158 -id:A048158 - cp's OEIS frontend

A372523 Triangle read by rows: T(n, k) is equal to n/k if k | n, else to the concatenation of A003988(n, k) = floor(n/k) and A051127(k, n) = n mod k.

1, 2, 1, 3, 11, 1, 4, 2, 11, 1, 5, 21, 12, 11, 1, 6, 3, 2, 12, 11, 1, 7, 31, 21, 13, 12, 11, 1, 8, 4, 22, 2, 13, 12, 11, 1, 9, 41, 3, 21, 14, 13, 12, 11, 1, 10, 5, 31, 22, 2, 14, 13, 12, 11, 1, 11, 51, 32, 23, 21, 15, 14, 13, 12, 11, 1, 12, 6, 4, 3, 22, 2, 15, 14, 13, 12, 11, 1

Offset: 1

Stefano Spezia, May 04 2024

			The triangle begins:
  1;
  2,  1;
  3, 11,  1;
  4,  2, 11,  1;
  5, 21, 12, 11,  1;
  6,  3,  2, 12, 11,  1;
  7, 31, 21, 13, 12, 11, 1;
  ...

Stefano Spezia, First 150 rows of the triangle, flattened

Cf. A003988, A004216, A051127, A055642.
Cf. A000012 (right diagonal), A000027 (1st column).
Cf. A048158, A051126, A051127, A234575.

T[n_,k_]:=If[Divisible[n,k],n/k,FromDigits[Join[IntegerDigits[Floor[n/k]],IntegerDigits[Mod[n,k]]]]]; Table[T[n,k],{n,12},{k,n}]//Flatten (* or *)
T[n_,k_]:=Floor[n/k]10^IntegerLength[Mod[n,k]]+Mod[n,k]; Table[T[n,k],{n,12},{k,n}]//Flatten (* or *)
T[n_, k_]:=SeriesCoefficient[x^k(1+Sum[(i + 10^(1+Floor[Log10[Mod[n,k]]]))*x^i, {i, k-1}] - Sum[i*x^(k+i), {i, k-1}])/(1-x^k)^2, {x, 0, n}]; Table[T[n, k], {n, 12}, {k, n}]//Flatten

T(n, k) = floor(n/k)*10^(1+floor(log10(n mod k))) + (n mod k) if n is not divisible by k.
T(n, n) = 1.
T(n, 1) = n.
T(n, k) = 2*T(n-k, k) - T(n-2*k, k) for n >= 3*k.
T(n, k) = [x^n] x^k*(1 + (Sum_{i=1..k-1} (i + 10^(1+floor(log10(n mod k))))*x^i) - (Sum_{i=1..k-1} i*x^(k+i)))/(1 - x^k)^2.

A378275 Numbers m which satisfy the equation: (m - floor((m - k)/k)) mod k = 1 (1 <= k <= m) only for k = 2 and m - 1.

Original entry on oeis.org

3, 4, 7, 11, 19, 23, 59, 83, 167, 227, 491, 659, 839, 983, 1019, 1091, 1319, 1459, 1523, 1847, 2179, 2503, 2963, 3719, 3767, 4519, 4871, 4919, 5059, 6563, 9239, 9419, 10883, 12107, 12539, 14891, 15383, 20071, 20747, 23819, 25219, 26759, 33851, 35591, 37379, 45191

Offset: 1

Lechoslaw Ratajczak, Nov 21 2024

nonn

Every term greater than 4 has the form 4*t + 3.
Let b(z) be the number of elements of this sequence <= z:
-------------
   z  | b(z)
-------------
 10^2 |   8
 10^3 |  14
 10^4 |  32
 10^5 |  55
 10^6 | 125
 10^7 | 347
 10^8 | 950
-------------
Every term greater than 4 is prime.

			Let T(i,j) be the triangle read by rows: T(i,j) = (i - floor((i - j)/j)) mod j for 1 <= j <= i. The triangle begins:
 i\j | 1 2 3 4 5 6 7 8 9 ...
-----+------------------
   1 | 0
   2 | 0 0
   3 | 0 1 0
   4 | 0 1 1 0
   5 | 0 0 2 1 0
   6 | 0 0 2 2 1 0
   7 | 0 1 0 3 2 1 0
   8 | 0 1 1 3 3 2 1 0
   9 | 0 0 1 0 4 3 2 1 0
 ...
The j-th column has period j^2, r-th element of this period has the form (r - 1 - floor((r - 1)/j)) mod j (1 <= r <= j^2). The period of j-th column consists of the sequence (0,1,2,...,j-1) and its consecutive j-1 right rotations (moving rightmost element to the left end).
7 is in this sequence because the only k's satisfying the equation (7 - floor((7 - k)/k)) mod k = 1 are 2 and (7-1).

Jinyuan Wang, Table of n, a(n) for n = 1..3000

Cf. A000040, A048158, A051731, A375007, A375595.

(f(i, j):=mod((i-floor((i-j)/j)), j),
(n:3, for t:7 thru 100000 step 4 do
(for k:3 while f(t, k)#1 and k

is(m) = if(m%4==3, for(k=3, m\2, if((m-m\k)%k==0, return(0))); 1, m==4); \\ Jinyuan Wang, Jan 14 2025

A049765 Triangular array T, read by rows: T(n,k) = (k mod n) + (n mod k), for k = 1..n and n >= 1.

Original entry on oeis.org

0, 1, 0, 1, 3, 0, 1, 2, 4, 0, 1, 3, 5, 5, 0, 1, 2, 3, 6, 6, 0, 1, 3, 4, 7, 7, 7, 0, 1, 2, 5, 4, 8, 8, 8, 0, 1, 3, 3, 5, 9, 9, 9, 9, 0, 1, 2, 4, 6, 5, 10, 10, 10, 10, 0, 1, 3, 5, 7, 6, 11, 11, 11, 11, 11, 0, 1, 2, 3, 4, 7, 6, 12, 12, 12, 12, 12, 0

Offset: 1

Clark Kimberling

			Triangle T(n,k) (with rows n >= 1 and columns k >= 1) begins as follows:
  0;
  1, 0;
  1, 3, 0;
  1, 2, 4, 0;
  1, 3, 5, 5, 0;
  1, 2, 3, 6, 6,  0;
  1, 3, 4, 7, 7,  7,  0;
  1, 2, 5, 4, 8,  8,  8,  0;
  1, 3, 3, 5, 9,  9,  9,  9,  0;
  1, 2, 4, 6, 5, 10, 10, 10, 10, 0;
  ...

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Row sums are in A049766.

Cf. A048158, A049767, A049768.

Flat(List([1..15], n-> List([1..n], k-> (k mod n) + (n mod k) ))); # G. C. Greubel, Dec 13 2019

[[(k mod n) + (n mod k): k in [1..n]]: n in [1..15]]; // G. C. Greubel, Dec 13 2019

seq(seq( `mod`(k, n) + `mod`(n, k), k = 1..n), n = 1..15); # G. C. Greubel, Dec 13 2019

Table[Mod[k,n] + Mod[n,k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Dec 13 2019 *)

T(n,k) = k%n + n%k;
for(n=1,15, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 13 2019

[[(k%n) + (n%k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Dec 13 2019

A309176 a(n) = n^2 * (n + 1)/2 - Sum_{k=1..n} sigma_2(k).

Original entry on oeis.org

0, 0, 2, 3, 12, 13, 33, 40, 66, 81, 135, 135, 212, 249, 319, 354, 489, 511, 681, 725, 876, 981, 1233, 1235, 1509, 1660, 1920, 2032, 2437, 2472, 2936, 3091, 3488, 3755, 4275, 4290, 4955, 5292, 5854, 6024, 6843, 6968, 7870, 8190, 8839, 9340, 10420, 10442, 11568, 12038, 13014, 13474, 14851, 15098, 16436

Offset: 1

Ilya Gutkovskiy, Jul 15 2019

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000326, A001157, A004125, A048158, A051126, A154585, A256532.

Cf. A002411, A064602.

Table[n^2 (n + 1)/2 - Sum[DivisorSigma[2, k], {k, 1, n}], {n, 1, 55}]
nmax = 55; CoefficientList[Series[x (1 + 2 x)/(1 - x)^4 - 1/(1 - x) Sum[k^2 x^k/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x] // Rest
Table[Sum[Mod[n, k] k, {k, 1, n}], {n, 1, 55}]

a(n) = n^2*(n+1)/2 - sum(k=1, n, sigma(k, 2)); \\ Michel Marcus, Sep 18 2021

from math import isqrt
def A309176(n): return (n**2*(n+1)>>1)+((s:=isqrt(n))**2*(s+1)*(2*s+1)-sum((q:=n//k)*(6*k**2+q*(2*q+3)+1) for k in range(1,s+1)))//6 # Chai Wah Wu, Oct 21 2023

G.f.: x * (1 + 2*x)/(1 - x)^4 - (1/(1 - x)) * Sum_{k>=1} k^2 * x^k/(1 - x^k).
a(n) = Sum_{k=1..n} (n mod k) * k.
a(n) = A002411(n) - A064602(n).

A324472 a(n) = 1000 mod n.

Original entry on oeis.org

0, 0, 1, 0, 0, 4, 6, 0, 1, 0, 10, 4, 12, 6, 10, 8, 14, 10, 12, 0, 13, 10, 11, 16, 0, 12, 1, 20, 14, 10, 8, 8, 10, 14, 20, 28, 1, 12, 25, 0, 16, 34, 11, 32, 10, 34, 13, 40, 20, 0, 31, 12, 46, 28, 10, 48, 31, 14, 56, 40, 24, 8, 55, 40, 25, 10, 62, 48, 34, 20, 6, 64, 51, 38, 25, 12, 76

Offset: 1

N. J. A. Sloane, Mar 07 2019, following a suggestion from Charles Kusniec

			a(98) = 1000 mod 98 = 20 as 1000 = 98 * 10 + 20. - _David A. Corneth_, Mar 07 2019

David A. Corneth, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A018766, A018767, A048158, A051127, A090976, A324471.

Mod[1000,Range[100]] (* Paolo Xausa, Nov 14 2023 *)

a(n) = 1000 % n \\ David A. Corneth, Mar 07 2019

a(n) = 1000 for n >= 1001.
a(n) = 0 <=> n in { A018767 }.
a(n) = 1 <=> n > 1 and n in { A018766 }.

A369311 Square array A(n, k), n >= 0, k > 0, read by upwards antidiagonals: P(A(n, k)) is the remainder of the polynomial long division of P(n) by P(k) (where P(m) denotes the polynomial over GF(2) whose coefficients are encoded in the binary expansion of the nonnegative integer m).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 1, 3, 2, 1, 0, 0, 0, 0, 0, 3, 2, 1, 0, 0, 1, 0, 1, 1, 3, 2, 1, 0, 0, 0, 1, 2, 0, 2, 3, 2, 1, 0, 0, 1, 1, 3, 3, 3, 3, 3, 2, 1, 0, 0, 0, 0, 0, 2, 0, 2, 4, 3, 2, 1, 0, 0, 1, 0, 1, 2, 1, 1, 5, 4, 3, 2, 1, 0

Offset: 0

Rémy Sigrist, Jan 19 2024

For any number m >= 0 with binary expansion Sum_{k >= 0} b_k * 2^k, P(m) = Sum_{k >= 0} b_k * X^k.

			Array A(n, k) begins:
  n\k | 1  2  3  4  5  6  7  8  9  10  11  12
  ----+--------------------------------------
    0 | 0  0  0  0  0  0  0  0  0   0   0   0
    1 | 0  1  1  1  1  1  1  1  1   1   1   1
    2 | 0  0  1  2  2  2  2  2  2   2   2   2
    3 | 0  1  0  3  3  3  3  3  3   3   3   3
    4 | 0  0  1  0  1  2  3  4  4   4   4   4
    5 | 0  1  0  1  0  3  2  5  5   5   5   5
    6 | 0  0  0  2  3  0  1  6  6   6   6   6
    7 | 0  1  1  3  2  1  0  7  7   7   7   7
    8 | 0  0  1  0  2  2  1  0  1   2   3   4
    9 | 0  1  0  1  3  3  0  1  0   3   2   5
   10 | 0  0  0  2  0  0  3  2  3   0   1   6
   11 | 0  1  1  3  1  1  2  3  2   1   0   7
   12 | 0  0  0  0  3  0  2  4  5   6   7   0

Rémy Sigrist, Colored scatterplot of (n, k) for n, k < 2^10 (where the color is function of A(n, k))
Index entries for sequences operating on GF(2)[X]-polynomials

See A369312 for the corresponding quotients.

Cf. A048158, A048720.

A(n, k) = { fromdigits(lift(Vec( (Mod(1, 2) * Pol(binary(n))) % (Mod(1, 2) * Pol(binary(k))))), 2) }

A372727 Triangle read by rows: T(n, k) = n if k = 0, otherwise n - k*floor(n/k). The binary modulo operation.

Original entry on oeis.org

0, 1, 0, 2, 0, 0, 3, 0, 1, 0, 4, 0, 0, 1, 0, 5, 0, 1, 2, 1, 0, 6, 0, 0, 0, 2, 1, 0, 7, 0, 1, 1, 3, 2, 1, 0, 8, 0, 0, 2, 0, 3, 2, 1, 0, 9, 0, 1, 0, 1, 4, 3, 2, 1, 0, 10, 0, 0, 1, 2, 0, 4, 3, 2, 1, 0, 11, 0, 1, 2, 3, 1, 5, 4, 3, 2, 1, 0, 12, 0, 0, 0, 0, 2, 0, 5, 4, 3, 2, 1, 0

Offset: 0

Peter Luschny, May 13 2024

The binary operation 'mod' as defined here is discussed in 'Concrete Mathematics' by Graham et. al. on p. 82 and the connection with the congruence relation '(mod)' on p. 123. See also Bach & Shallit, p. 21, and Apostol, p. 14.

This definition is implemented in Sage, but not in Python. For example, Sage answers 0.mod(0) = 0, whereas in Python 0 % 0 leads to a 'ZeroDivisionError'. What is often misunderstood is that the operation 'mod' gives answers to divisibility, not to division. Apostol shows that n|0 (every integer divides zero), but 0|n implies n = 0 (zero divides only zero), and thus confirms the result given by Sage.

			Triangle begins:
  [ 0]  0;
  [ 1]  1, 0;
  [ 2]  2, 0, 0;
  [ 3]  3, 0, 1, 0;
  [ 4]  4, 0, 0, 1, 0;
  [ 5]  5, 0, 1, 2, 1, 0;
  [ 6]  6, 0, 0, 0, 2, 1, 0;
  [ 7]  7, 0, 1, 1, 3, 2, 1, 0;
  [ 8]  8, 0, 0, 2, 0, 3, 2, 1, 0;
  [ 9]  9, 0, 1, 0, 1, 4, 3, 2, 1, 0;
  [10] 10, 0, 0, 1, 2, 0, 4, 3, 2, 1, 0;
  [11] 11, 0, 1, 2, 3, 1, 5, 4, 3, 2, 1, 0;
.
The triangle shows the modulo operation in the range 0 <= k <= n. Test your
computer implementation in the range R X R where R = [-6, ..., 0, ..., 6].
According to Graham et al. it should look like this:
   0, -1, -2,  0,  0, 0, -6, 0, 0, 0, 2, 4, 0
  -5,  0, -1, -2, -1, 0, -5, 0, 1, 1, 3, 0, 1
  -4, -4,  0, -1,  0, 0, -4, 0, 0, 2, 0, 1, 2
  -3, -3, -3,  0, -1, 0, -3, 0, 1, 0, 1, 2, 3
  -2, -2, -2, -2,  0, 0, -2, 0, 0, 1, 2, 3, 4
  -1, -1, -1, -1, -1, 0, -1, 0, 1, 2, 3, 4, 5
   0,  0,  0,  0,  0, 0,  0, 0, 0, 0, 0, 0, 0
  -5, -4, -3, -2, -1, 0,  1, 0, 1, 1, 1, 1, 1
  -4, -3, -2, -1,  0, 0,  2, 0, 0, 2, 2, 2, 2
  -3, -2, -1,  0, -1, 0,  3, 0, 1, 0, 3, 3, 3
  -2, -1,  0, -2,  0, 0,  4, 0, 0, 1, 0, 4, 4
  -1,  0, -3, -1, -1, 0,  5, 0, 1, 2, 1, 0, 5
   0, -4, -2,  0,  0, 0,  6, 0, 0, 0, 2, 1, 0

Tom Apostol, Introduction to analytic number theory, 1976, Springer, page 14.
Eric Bach and Jeffrey Shallit, Algorithmic Number Theory, 1997, p. 21.
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994, 34th printing 2022, p. 81f.

Cf. A111490 (row sums).

Cf. A048158.

MOD := (n, k) -> ifelse(k = 0, n, n - k * iquo(n, k)):
seq( seq(MOD(n, k), k = 0..n), n = 0..12);

def T(n, k): return n if k == 0 else n - k * (n // k)
for n in range(12): print([T(n, k) for k in range(n + 1)])

def A372727_T(n, k): return n % k if k else n # Chai Wah Wu, May 14 2024

def T(n, k): return n.mod(k)
for n in srange(12): print([T(n, k) for k in range(n + 1)])

A380153 Numbers m for which the sum of all values of k satisfying the equation: (m - floor((m - k)/k)) mod k = 0 (1 <= k <= m) equals 2*m.

Original entry on oeis.org

39, 4395, 29055, 57931, 81115, 152571, 164955, 410731, 664747, 877435, 2080875, 2521087, 2539515

Offset: 1

Lechoslaw Ratajczak, Jan 13 2025

a(14) > 3*10^6 (if it exists). Is there any even term?

			Let T(i,j) be the triangle read by rows: T(i,j) = (i - floor((i - j)/j)) mod j for 1 <= j <= i. The triangle begins:
 i\j | 1 2 3 4 5 6 7 8 9 10 11 ...
-----+------------------------
   1 | 0
   2 | 0 0
   3 | 0 1 0
   4 | 0 1 1 0
   5 | 0 0 2 1 0
   6 | 0 0 2 2 1 0
   7 | 0 1 0 3 2 1 0
   8 | 0 1 1 3 3 2 1 0
   9 | 0 0 1 0 4 3 2 1 0
  10 | 0 0 2 1 4 4 3 2 1  0
  11 | 0 1 0 2 0 5 4 3 2  1  0
  ...
The j-th column has period j^2, r-th element of this period has the form (r - 1 - floor((r - 1)/j)) mod j (1 <= r <= j^2). The period of j-th column consists of the sequence (0,1,2,...,j-1) and its consecutive j-1 right rotations (moving rightmost element to the left end).
39 is in this sequence because the only k's <= 39 satisfying the equation (39 - floor((39 - k)/k)) mod k = 0 are: 1, 3, 7, 9, 19, 39, hence: 1+3+7+9+19+39 = 78 = 2*39.

Cf. A048158, A375595, A378275.

(f(i, j):=mod(i-floor((i-j)/j), j),
(n:0, for m:2 thru 5000 do
(s:0, for k:1 thru floor(m/2) do
(if f(m, k)=0 then
(s:s+k)), if s=m then
(n:n+1, print(n , "" , m)))));

a(9)-a(13) from Jinyuan Wang, Jan 14 2025

A283190 a(n) is the number of different values n mod k for 1 <= k <= floor(n/2).

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 2, 2, 2, 3, 4, 2, 3, 3, 3, 4, 5, 4, 5, 4, 4, 5, 6, 5, 6, 7, 7, 7, 8, 6, 7, 7, 7, 8, 9, 8, 9, 9, 9, 10, 11, 9, 10, 9, 9, 10, 11, 10, 11, 12, 12, 12, 13, 12, 13, 13, 13, 14, 15, 13, 14, 14, 14, 15, 16, 15, 16, 15, 15, 16, 17, 16, 17, 17, 17, 17, 18, 17

Offset: 1

Thomas Kerscher, Mar 02 2017

a(n) is the number of distinct terms in the first half of the n-th row of the A048158 triangle. - Michel Marcus, Mar 04 2017
a(n)/n appears to converge to a constant, approximately 0.2296. Can this be proved, and does the constant have a closed form? - Robert Israel, Mar 13 2017
The constant that a(n)/n approaches is Sum {p prime} 1/(p^2+p)* Product {q prime < p} (q-1)/q. - Michael R Peake, Mar 16 2017

			a(7) = 2 because 7=0 (mod 1), 7=1 (mod 2), 7=1 (mod 3), two different results.

Robert Israel, Table of n, a(n) for n = 1..10000
Omkar Baraskar and Ingrid Vukusic, Bounds for sets of remainders, arXiv:2508.20853 [math.NT], 2025.
Michael R Peake, Explanation of limiting value of a(n)/n

Cf. A048158.

N:= 100: # to get a(1)..a(N)
V:= Vector(N,1):
V[1]:= 0:
for m from 2 to N-1 do
  k:= m/min(numtheory:-factorset(m));
  ns:= [seq(n,n=m+1..min(N,m+k-1))];
  V[ns]:= map(`+`,V[ns],1);
od:
convert(V,list); # Robert Israel, Mar 13 2017

Table[Length@ Union@ Map[Mod[n, #] &, Range@ Floor[n/2]], {n, 78}] (* Michael De Vlieger, Mar 03 2017 *)

a(n) = #vecsort(vector(n\2, k, n % k),,8); \\ Michel Marcus, Mar 02 2017

A324471 a(n) = 10 mod n.

Original entry on oeis.org

0, 0, 1, 2, 0, 4, 3, 2, 1, 0, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10

Offset: 1

N. J. A. Sloane, Mar 07 2019, following a suggestion from Charles Kusniec

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

Cf. A010692, A010879, A048158, A051127, A090976, A324472.

Mod[10,Range[100]] (* Paolo Xausa, Nov 14 2023 *)

From Elmo R. Oliveira, Aug 03 2024: (Start)
G.f.: x^3*(1 + x - 2*x^2 + 4*x^3 - x^4 - x^5 - x^6 - x^7 + 10*x^8)/(1 - x).
a(n) = 10 for n > 10. (End)

cp's OEIS Frontend

A372523 Triangle read by rows: T(n, k) is equal to n/k if k | n, else to the concatenation of A003988(n, k) = floor(n/k) and A051127(k, n) = n mod k.

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A378275 Numbers m which satisfy the equation: (m - floor((m - k)/k)) mod k = 1 (1 <= k <= m) only for k = 2 and m - 1.

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A049765 Triangular array T, read by rows: T(n,k) = (k mod n) + (n mod k), for k = 1..n and n >= 1.

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GAP

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A309176 a(n) = n^2 * (n + 1)/2 - Sum_{k=1..n} sigma_2(k).

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A324472 a(n) = 1000 mod n.

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A369311 Square array A(n, k), n >= 0, k > 0, read by upwards antidiagonals: P(A(n, k)) is the remainder of the polynomial long division of P(n) by P(k) (where P(m) denotes the polynomial over GF(2) whose coefficients are encoded in the binary expansion of the nonnegative integer m).

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A372727 Triangle read by rows: T(n, k) = n if k = 0, otherwise n - k*floor(n/k). The binary modulo operation.

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