A048152 -id:A048152 - cp's OEIS frontend

A225126 Central terms of the triangle in A048152.

0, 1, 4, 2, 7, 3, 10, 4, 13, 5, 16, 6, 19, 7, 22, 8, 25, 9, 28, 10, 31, 11, 34, 12, 37, 13, 40, 14, 43, 15, 46, 16, 49, 17, 52, 18, 55, 19, 58, 20, 61, 21, 64, 22, 67, 23, 70, 24, 73, 25, 76, 26, 79, 27, 82, 28, 85, 29, 88, 30, 91, 31, 94, 32, 97, 33, 100

Offset: 1

Reinhard Zumkeller, Apr 29 2013

a(n) = A123684(n) for n > 1;
a(n) = A048152(2*n-1,n), central terms;
also a(n) = A060036(2*n-1,n-1) for n > 1.
a(n+1)=the remainder when n^2 is divided by 2n+1. - J. M. Bergot, Jun 25 2013

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

Haskell
```
a225126 n = a048152 (2 * n - 1)  n
```

a(n)=n--^2%(2*n+1) \\ Charles R Greathouse IV, Jun 25 2013

a(1) = 0, a(2*n) = n and a(2*n+1) = 3*n+1.

a(n) = 2*a(n-2)-a(n-4) for n>5. G.f.: -x^2*(x^3-4*x-1) / ((x-1)^2*(x+1)^2). - Colin Barker, May 01 2013

A063987 Irregular triangle in which n-th row gives quadratic residues modulo the n-th prime.

Original entry on oeis.org

1, 1, 1, 4, 1, 2, 4, 1, 3, 4, 5, 9, 1, 3, 4, 9, 10, 12, 1, 2, 4, 8, 9, 13, 15, 16, 1, 4, 5, 6, 7, 9, 11, 16, 17, 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18, 1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28, 1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 18, 19, 20, 25, 28, 1, 3, 4, 7, 9, 10, 11, 12, 16, 21, 25

Offset: 1

Suggested by Gary W. Adamson, Sep 18 2001

For n >= 2, row lengths are (prime(n) - 1)/2. For example, since 17 is the 7th prime number, the length of row 7 is (17 - 1)/2 = 8. - Geoffrey Critzer, Apr 04 2015

			Modulo the 5th prime, 11, the (11 - 1)/2 = 5 quadratic residues are 1,3,4,5,9 and the 5 non-residues are 2, 6, 7, 8, 10.
The irregular triangle T(n, k) begins (p is prime(n)):
   n    p  \k 1 2 3 4  5  6  7  8  9 10 11 12 13 14
   1,   2:    1
   2,   3:    1
   3,   5:    1 4
   4,   7:    1 2 4
   5,  11:    1 3 4 5  9
   6:  13:    1 3 4 9 10 12
   7,  17:    1 2 4 8  9 13 15 16
   8,  19:    1 4 5 6  7  9 11 16 17
   9,  23:    1 2 3 4  6  8  9 12 13 16 18
  10,  29:    1 4 5 6  7  9 13 16 20 22 23 24 25 28
  ...  reformatted by _Wolfdieter Lang_, Mar 06 2016

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 82 at p. 202.

T. D. Noe, Rows n = 1..100 of triangle, flattened
C. F. Gauss, Vierter Abschnitt. Von den Congruenzen zweiten Grades. Quadratische Reste und Nichtreste. Art. 97, in "Untersuchungen über die höhere Arithmetik", Hrsg. H. Maser, Verlag von Julius Springer, Berlin, 1889.

Cf. A063988, A010379 (6th row), A010381 (7th row), A010385 (8th row), A010391 (9th row), A010392 (10th row), A278580 (row 23), A230077.
Cf. A076409 (row sums).
Cf. A046071 (for all n), A048152 (for all n, with 0's).

with(numtheory): for n from 1 to 20 do for j from 1 to ithprime(n)-1 do if legendre(j, ithprime(n)) = 1 then printf(`%d,`,j) fi; od: od:
# Alternative:
QR := (a, n) -> NumberTheory:-QuadraticResidue(a, n):
for n from 1 to 10 do p := ithprime(n):
print(select(a -> 1 = QR(a, p), [seq(1..p-1)])) od:  # Peter Luschny, Jun 02 2024

row[n_] := (p = Prime[n]; Select[ Range[p - 1], JacobiSymbol[#, p] == 1 &]); Flatten[ Table[ row[n], {n, 1, 12}]] (* Jean-François Alcover, Dec 21 2011 *)

residue(n,m)=local(r);r=0;for(i=0,floor(m/2),if(i^2%m==n,r=1));r
isA063987(n,m)=residue(n,prime(m)) /* Michael B. Porter, May 07 2010 */

row(n) = my(p=prime(n)); select(x->issquare(Mod(x,p)), [1..p-1]); \\ Michel Marcus, Nov 07 2020

from sympy import jacobi_symbol as J, prime
def a(n):
    p = prime(n)
    return [1] if n==1 else [i for i in range(1, p) if J(i, p)==1]
for n in range(1, 11): print(a(n)) # Indranil Ghosh, May 27 2017

for p in prime_range(30): print(quadratic_residues(p)[1:])
# Peter Luschny, Jun 02 2024

Edited by Wolfdieter Lang, Mar 06 2016

A048153 a(n) = Sum_{k=1..n} (k^2 mod n).

Original entry on oeis.org

0, 1, 2, 2, 10, 13, 14, 12, 24, 45, 44, 38, 78, 77, 70, 56, 136, 129, 152, 130, 182, 209, 184, 148, 250, 325, 288, 294, 406, 365, 372, 304, 484, 561, 490, 402, 666, 665, 572, 540, 820, 805, 860, 726, 840, 897, 846, 680, 980, 1125, 1156, 1170, 1378, 1305, 1210

Offset: 1

Clark Kimberling

nonn

See A048152 for the array T[n,k] = k^2 mod n.
Starting with a(2)=1 each 4th term is odd: a(n=2+4*k) = 1, 13, 45, 77, 129, 209, 325, 365, ... - Zak Seidov, Apr 22 2009
Positions of squares in A048153: 1, 2, 33, 51, 69, 105, 195, 250, 294, 1250, 4913, 9583, 13778, 48778, 65603, 83521.
Corresponding values of squares are: {0, 1, 22, 34, 46, 70, 130, 175, 203, 875, 3468, 6734, 9711, 34481, 46308, 58956}^2 = {0, 1, 484, 1156, 2116, 4900, 16900, 30625, 41209, 765625, 12027024, 45346756, 94303521, 1188939361, 2144430864, 3475809936}. - Zak Seidov, Nov 02 2011
For n > 1 also row sums of A060036. - Reinhard Zumkeller, Apr 29 2013
Conjecture: a(n) <= (n^2-1)/2. - Aspen A.M. Meissner, Mar 06 2025

			a(5) = 1^2 + 2^2 + (3^2 mod 5) + (4^2 mod 5) + (5^2 mod 5) = 1 + 4 + 4 + 1 + 0 = 10. (It is easily seen that the last term, n^2 mod n, is always zero and would not need to be included.) - _M. F. Hasler_, Oct 21 2013

Zak Seidov, Table of n, a(n) for n = 1..10000

Cf. A000330, A048152, A215573.

a048153 = sum . a048152_row  -- Reinhard Zumkeller, Apr 29 2013

Table[Sum[PowerMod[k,2,n], {k,n-1}], {n,1,10000}] (* Zak Seidov, Nov 02 2011 *)

a(n)=sum(k=1,n,k^2%n) \\ Charles R Greathouse IV, Oct 21 2013

def A048153(n): return sum(k**2%n for k in range(1,n)) # Chai Wah Wu, Jun 02 2024

a(n) == n*(n+1)*(2n+1)/6 (mod n). - Charles R Greathouse IV, Dec 28 2011
a(n) == n*(n-1)*(2n-1)/6 (mod n). - Chai Wah Wu, Jun 02 2024
a(n) mod n = A215573(n). - Alois P. Heinz, Jun 03 2024

Definition made more explicit by M. F. Hasler, Oct 21 2013

A133819 Triangle whose rows are sequences of increasing squares: 1; 1,4; 1,4,9; ... .

Original entry on oeis.org

1, 1, 4, 1, 4, 9, 1, 4, 9, 16, 1, 4, 9, 16, 25, 1, 4, 9, 16, 25, 36, 1, 4, 9, 16, 25, 36, 49, 1, 4, 9, 16, 25, 36, 49, 64, 1, 4, 9, 16, 25, 36, 49, 64, 81, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

Offset: 1

Peter Bala, Sep 25 2007

Reading the triangle by rows produces the sequence 1,1,4,1,4,9,1,4,9,16,..., analogous to A002260.

Sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. A133819 is reluctant sequence of A000290. - Boris Putievskiy, Jan 11 2013

			The triangle T(n, k) starts:
1;
1, 4;
1, 4, 9;
1, 4, 9, 16;
1, 4, 9, 16, 25;

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
M. de Frenicle, Methode pour trouver la solutions des problemes par les exclusions, in: Divers ouvrages de mathematiques et de physique par messieurs de l'academie royale des sciences, (1693) pp 1-44, page 11. - _Paul Curtz_, Aug 18 2008
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A215630, A215631, A000290, A002260.

Cf. A048152, A060036.

a133819 n k = a133819_tabl !! (n-1) !! (k-1)
a133819_row n = a133819_tabl !! (n-1)
a133819_tabl = map (`take` (tail a000290_list)) [1..]
-- Reinhard Zumkeller, Nov 11 2012

With[{sqs=Range[12]^2},Flatten[Table[Take[sqs,n],{n,12}]]] (* Harvey P. Dale, Sep 09 2012 *)

T(n, k) = k^2, n >= k >= 1. - Wolfdieter Lang, Dec 02 2014
O.g.f.: (1+qx)/((1-x)(1-qx)^3) = 1 + x(1 + 4q) + x^2(1 + 4q + 9q^2) + ... .
a(n) = A000290(m+1), where m = n-t(t+1)/2, t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Jan 11 2013

Offset changed by Reinhard Zumkeller, Nov 11 2012

A070442 a(n) = n^2 mod 20.

Original entry on oeis.org

0, 1, 4, 9, 16, 5, 16, 9, 4, 1, 0, 1, 4, 9, 16, 5, 16, 9, 4, 1, 0, 1, 4, 9, 16, 5, 16, 9, 4, 1, 0, 1, 4, 9, 16, 5, 16, 9, 4, 1, 0, 1, 4, 9, 16, 5, 16, 9, 4, 1, 0, 1, 4, 9, 16, 5, 16, 9, 4, 1, 0, 1, 4, 9, 16, 5, 16, 9, 4, 1, 0, 1, 4, 9, 16, 5, 16, 9, 4, 1, 0, 1, 4, 9, 16, 5, 16, 9, 4, 1, 0, 1, 4, 9, 16

Offset: 0

N. J. A. Sloane, May 12 2002

Also, n^6 mod 20.

Equivalently n^10 mod 20. - Zerinvary Lajos, Oct 31 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,1).

Cf. A000290, A008959, A010382, A070431, A070435, A070438, A070452, A159852.

Row 20 of A048152.

Table[Mod[n^2,20],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 23 2011 *)
LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 1},{0, 1, 4, 9, 16, 5, 16, 9, 4, 1},95] (* Ray Chandler, Aug 26 2015 *)
PowerMod[Range[0,100],2,20] (* or *) PadRight[{},120,{0,1,4,9,16,5,16,9,4,1}] (* Harvey P. Dale, Jan 06 2019 *)

a(n)=n^2%20 \\ Charles R Greathouse IV, Jun 11 2015

[power_mod(n,10,20) for n in range(0, 88)] # Zerinvary Lajos, Oct 31 2009

From Reinhard Zumkeller, Apr 24 2009: (Start)
a(m*n) = a(m)*a(n) mod 20.
a(5*n+k) = a(5*n-k) for k <= 5*n.
a(n+10) = a(n). (End)
G.f. -x*(1+4*x+9*x^2+16*x^3+5*x^4+16*x^5+9*x^6+4*x^7+x^8) / ( (x-1) *(1+x) *(x^4+x^3+x^2+x+1) *(x^4-x^3+x^2-x+1) ). - R. J. Mathar, Aug 27 2013

A070438 a(n) = n^2 mod 15.

Original entry on oeis.org

0, 1, 4, 9, 1, 10, 6, 4, 4, 6, 10, 1, 9, 4, 1, 0, 1, 4, 9, 1, 10, 6, 4, 4, 6, 10, 1, 9, 4, 1, 0, 1, 4, 9, 1, 10, 6, 4, 4, 6, 10, 1, 9, 4, 1, 0, 1, 4, 9, 1, 10, 6, 4, 4, 6, 10, 1, 9, 4, 1, 0, 1, 4, 9, 1, 10, 6, 4, 4, 6, 10, 1, 9, 4, 1, 0, 1, 4, 9, 1, 10, 6, 4, 4, 6, 10, 1, 9, 4, 1, 0, 1, 4, 9, 1, 10, 6

Offset: 0

N. J. A. Sloane, May 12 2002

Equivalently, n^6 mod 15. - Ray Chandler, Dec 27 2023

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

Cf. A000290, A008959, A010378, A070431, A070435, A070442, A070452, A159852.

Row 15 of A048152.

Table[Mod[n^2,15],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 21 2011 *)
LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1},{0, 1, 4, 9, 1, 10, 6, 4, 4, 6, 10, 1, 9, 4, 1},97] (* Ray Chandler, Aug 26 2015 *)

a(n)=n^2%15 \\ Charles R Greathouse IV, Sep 28 2015

[power_mod(n,2,15)for n in range(0, 97)] # Zerinvary Lajos, Nov 06 2009

From Reinhard Zumkeller, Apr 24 2009: (Start)
a(m*n) = a(m)*a(n) mod 15.
a(15*n+7+k) = a(15*n+8-k) for k <= 15*n+7.
a(15*n+k) = a(15*n-k) for k <= 15*n.
a(n+15) = a(n). (End)
From R. J. Mathar, Mar 14 2011: (Start)
a(n) = a(n-15).
G.f.: -x*(1+x) *(x^12+3*x^11+6*x^10-5*x^9+15*x^8-9*x^7+13*x^6-9*x^5+15*x^4-5*x^3+6*x^2+3*x+1) / ( (x-1) *(1+x^4+x^3+x^2+x) *(1+x+x^2) *(1-x+x^3-x^4+x^5-x^7+x^8) ). (End)
G.f.: (x^14 +4*x^13 +9*x^12 +x^11 +10*x^10 +6*x^9 +4*x^8 +4*x^7 +6*x^6 +10*x^5 +x^4 +9*x^3 +4*x^2 +x)/(-x^15 +1). - Colin Barker, Aug 14 2012

A049759 Triangular array T read by rows: T(n,k)=n^2 mod k, for k=1,2,...,n, n=1,2,...

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 4, 1, 0, 0, 0, 1, 0, 4, 4, 1, 0, 0, 1, 0, 1, 1, 3, 4, 1, 0, 0, 0, 1, 0, 0, 4, 2, 4, 1, 0, 0, 1, 1, 1, 1, 1, 2, 1, 4, 1, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 4, 1, 0, 0, 1, 1, 1, 4, 1, 1, 1, 7, 9, 4, 1, 0

Offset: 1

Clark Kimberling

			Triangle begins:
  0;
  0, 0;
  0, 1, 0;
  0, 0, 1, 0;
  0, 1, 1, 1, 0;
  0, 0, 0, 0, 1, 0;
  0, 1, 1, 1, 4, 1, 0;
  0, 0, 1, 0, 4, 4, 1, 0;
  0, 1, 0, 1, 1, 3, 4, 1, 0;
  0, 0, 1, 0, 0, 4, 2, 4, 1, 0;
  0, 1, 1, 1, 1, 1, 2, 1, 4, 1, 0;
  0, 0, 0, 0, 4, 0, 4, 0, 0, 4, 1, 0;
  0, 1, 1, 1, 4, 1, 1, 1, 7, 9, 4, 1, 0;

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

Cf. A048152.

Flat(List([1..15], n-> List([1..n], k-> PowerMod(n,2,k) ))); # G. C. Greubel, Dec 13 2019

[[Modexp(n,2,k): k in [1..n]]: n in [1..15]]; // G. C. Greubel, Dec 13 2019

seq(seq( `mod`(n^2, k), k = 1..n), n = 1..15); # G. C. Greubel, Dec 13 2019

Table[PowerMod[n,2,k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Dec 13 2019 *)

tabl(nn) = {for (n=1, nn, for (k=1, n, print1(n^2 % k, ", ");); print(););} \\ Michel Marcus, Mar 31 2014

[[power_mod(n,2,k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Dec 13 2019

A060036 Triangular array T read by rows: T(n,k) = k^2 mod n, for k = 1,2,...,n-1, n = 2,3,...

Original entry on oeis.org

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

Offset: 2

N. J. A. Sloane, Mar 17 2001

T(n,k) = A048152(n-1,k), 1 <= k < n; T(2*n-1,n-1) = A123684(n-1) = A225126(n-1). - Reinhard Zumkeller, Apr 29 2013

			The triangle T(n,k) begins:
n\k 1 2 3 4 5 6 7 8 9 10 11 ...
-------------------------------
2:  1
3:  1 1
4:  1 0 1
5:  1 4 4 1
6:  1 4 3 4 1
7:  1 4 2 2 4 1
6:  1 4 1 0 1 4 1
9:  1 4 0 7 7 0 4 1
10: 1 4 9 6 5 6 9 4 1
11: 1 4 9 5 3 3 5 9 4  1
12: 1 4 9 4 1 0 1 4 9  4  1
...  reformatted by - _Wolfdieter Lang_, Dec 17 2018

T. D. Noe, Rows n = 2..100 of triangle, flattened

Cf. A048152, A060037.

Cf. A048153 (row sums).

a060036 n k = a060036_tabl !! (n-2) !! (k-1)
a060036_row n = a060036_tabl !! (n-2)
a060036_tabl = map init $ tail a048152_tabl
-- Reinhard Zumkeller, Apr 29 2013

Flatten[Table[PowerMod[k,2,n],{n,2,20},{k,n-1}]] (* Harvey P. Dale, Feb 27 2012 *)

{ n=1; for (m=2, 46, for (k=1, m-1, write("b060036.txt", n++, " ", k^2 % m)); ) } \\ Harry J. Smith, Jul 01 2009

More terms from Larry Reeves (larryr(AT)acm.org), Mar 20 2001

A049767 Triangular array T, read by rows: T(n,k) = (k^2 mod n) + (n^2 mod k), for k = 1..n and n >= 1.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

			Triangle T(n,k) (with rows n >= 1 and columns k >= 1) begins as follows:
  0;
  1,  0;
  1,  2,  0;
  1,  0,  2,  0;
  1,  5,  5,  2,  0;
  1,  4,  3,  4,  2,  0;
  1,  5,  3,  3,  8,  2,  0;
  1,  4,  2,  0,  5,  8,  2,  0;
  1,  5,  0,  8,  8,  3,  8,  2,  0;
  1,  4, 10,  6,  5, 10, 11,  8,  2,  0;
  ...

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

Row sums are in A049768.

Cf. A048152, A049759.

Flat(List([1..15], n-> List([1..n], k-> PowerMod(k,2,n) + PowerMod(n,2,k) ))); # G. C. Greubel, Dec 13 2019

[[Modexp(k,2,n) + Modexp(n,2,k): k in [1..n]]: n in [1..15]]; // G. C. Greubel, Dec 13 2019

seq(seq( `mod`(k^2, n) + `mod`(n^2, k), k = 1..n), n = 1..15); # G. C. Greubel, Dec 13 2019

Table[PowerMod[k,2,n] + PowerMod[n,2,k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Dec 13 2019 *)

T(n,k) = lift(Mod(k,n)^2) + lift(Mod(n,k)^2);
for(n=1,15, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 13 2019

[[power_mod(k,2,n) + power_mod(n,2,k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Dec 13 2019

T(n, k) = A048152(n, k) + A049759(n, k). - Michel Marcus, Nov 21 2019

A373749 Triangle read by rows: T(n, k) = MOD(k^2, n) where MOD(a, n) = a if n = 0 and otherwise a - n*floor(a/n). (Quadratic residue.)

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 4, 4, 1, 0, 0, 1, 4, 3, 4, 1, 0, 0, 1, 4, 2, 2, 4, 1, 0, 0, 1, 4, 1, 0, 1, 4, 1, 0, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 0, 1, 4, 9, 4, 1, 0, 1, 4, 9, 4, 1, 0

Offset: 0

Peter Luschny, Jun 23 2024

The definition of the binary operation MOD in the name follows CMath (Graham et al.) and Bach & Shallit. This is important because some CAS unfortunately do not follow this definition and throw a 'division by zero' error if n = 0.

Row n reduced to a set is the set of the quadratic residues mod n.

			Triangle starts:
  [0] 0;
  [1] 0, 0;
  [2] 0, 1, 0;
  [3] 0, 1, 1, 0;
  [4] 0, 1, 0, 1, 0;
  [5] 0, 1, 4, 4, 1, 0;
  [6] 0, 1, 4, 3, 4, 1, 0;
  [7] 0, 1, 4, 2, 2, 4, 1, 0;
  [8] 0, 1, 4, 1, 0, 1, 4, 1, 0;
  [9] 0, 1, 4, 0, 7, 7, 0, 4, 1, 0;
 [10] 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0;

Eric Bach and Jeffrey Shallit, Algorithmic Number Theory, p. 21.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, 2nd ed., pp. 81f.

Variants: A048152, A096008.

Cf. A048153 (row sums), A373750 (middle terms).

Mod(n, k) = k == 0 ? n : mod(n, k)
T(n, k) = Mod(k^2, n)
for n in 0:10
    [T(n, k) for k in 0:n] |> println
end

REM := (n, k) -> ifelse(k = 0, n, irem(n, k)):
T := n -> local k; seq(REM(k^2, n), k = 0..n):
seq(T(n), n = 0..12);

MOD[n_, k_] := If[k == 0, n, Mod[n, k]];
Table[MOD[k^2, n], {n, 0, 10}, {k, 0, n}]

def A373749(n, k): return mod(k^2, n)
for n in range(11): print([A373749(n, k) for k in range(n + 1)])

cp's OEIS Frontend

A225126 Central terms of the triangle in A048152.

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A063987 Irregular triangle in which n-th row gives quadratic residues modulo the n-th prime.

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A048153 a(n) = Sum_{k=1..n} (k^2 mod n).

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A133819 Triangle whose rows are sequences of increasing squares: 1; 1,4; 1,4,9; ... .

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A070442 a(n) = n^2 mod 20.

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A070438 a(n) = n^2 mod 15.

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