A053204 -id:A053204 - cp's OEIS frontend

A053200 Binomial coefficients C(n,k) reduced modulo n, read by rows; T(0,0)=0 by convention.

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

Offset: 0

Asher Auel, Dec 12 1999

Pascal's triangle read by rows, where row n is read mod n.
A number n is a prime if and only if (1+x)^n == 1+x^n (mod n), i.e., if and only if the n-th row is 1,0,0,...,0,1. This result underlies the proof of Agrawal, Kayal and Saxena that there is a polynomial-time algorithm for primality testing. - N. J. A. Sloane, Feb 20 2004
A020475(n) = number of zeros in n-th row, for n > 0. - Reinhard Zumkeller, Jan 01 2013

			Row 4 = 1 mod 4, 4 mod 4, 6 mod 4, 4 mod 4, 1 mod 4 = 1, 0, 2, 0, 1.
Triangle begins:
  0;
  0,0;
  1,0,1;
  1,0,0,1;
  1,0,2,0,1;
  1,0,0,0,0,1;
  1,0,3,2,3,0,1;
  1,0,0,0,0,0,0,1;
  1,0,4,0,6,0,4,0,1;
  1,0,0,3,0,0,3,0,0,1;
  1,0,5,0,0,2,0,0,5,0,1;
  1,0,0,0,0,0,0,0,0,0,0,1;
  1,0,6,4,3,0,0,0,3,4,6,0,1;
  1,0,0,0,0,0,0,0,0,0,0,0,0,1;

T. D. Noe, Rows n = 0..100 of triangle, flattened
M. Agrawal, N. Kayal & N. Saxena, PRIMES is in P, Annals of Maths., 160:2 (2004), pp. 781-793. [alternate link]
Index entries for triangles and arrays related to Pascal's triangle

Row sums give A053204. Cf. A053201, A053202, A053203, A007318 (Pascal's triangle).
Cf. also A092241.
Cf. A053214 (central terms, apart from initial 1).

a053200 n k = a053200_tabl !! n !! k
a053200_row n = a053200_tabl !! n
a053200_tabl = [0] : zipWith (map . flip mod) [1..] (tail a007318_tabl)
-- Reinhard Zumkeller, Jul 10 2015, Jan 01 2013

f := n -> seriestolist( series( expand( (1+x)^n ) mod n, x, n+1)); # N. J. A. Sloane

Flatten[Join[{0},Table[Mod[Binomial[n,Range[0,n]],n],{n,20}]]] (* Harvey P. Dale, Apr 29 2013 *)

T(n,k)=if(n, binomial(n,k)%n, 0) \\ Charles R Greathouse IV, Feb 07 2017

Corrected by T. D. Noe, Feb 08 2008

Edited by N. J. A. Sloane, Aug 29 2008 at the suggestion of R. J. Mathar

A053205 Row sums of A053201.

Original entry on oeis.org

0, 0, 2, 0, 8, 0, 14, 6, 12, 0, 26, 0, 44, 36, 62, 0, 44, 0, 74, 48, 68, 0, 134, 30, 80, 78, 154, 0, 242, 0, 254, 72, 36, 86, 170, 0, 116, 84, 254, 0, 440, 0, 322, 330, 324, 0, 590, 126, 272, 414, 430, 0, 674, 96, 646, 120, 408, 0, 794, 0, 932, 384, 958, 290, 524, 0

Offset: 2

Asher Auel, Dec 12 1999

			a(6) = 0 + 3 + 2 + 3 + 0 = 8.

Reinhard Zumkeller, Table of n, a(n) for n = 2..1000

Cf. A053201, A053202, A053204.

a053205 = sum . a053201_row  -- Reinhard Zumkeller, Jan 24 2014

a(n) = A053204(n) - 2.

a(prime(n)) = 0.

Corrected and extended by James Sellers, Dec 13 1999

A090249 a(n) = 28a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 28.

Original entry on oeis.org

2, 28, 782, 21868, 611522, 17100748, 478209422, 13372763068, 373959156482, 10457483618428, 292435582159502, 8177738816847628, 228684251289574082, 6394981297291226668, 178830792072864772622, 5000867196742922406748

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 24 2004

a(n+1)/a(n) converges to (14+sqrt(195)) =27.96424004... Lim a(n)/a(n+1) as n approaches infinity = 0.03575995... = 1/(14+sqrt(195)) = (14-sqrt(195)). Lim a(n+1)/a(n) as n approaches infinity = 27.96424004... = (14+sqrt(195)) = 1/(14-sqrt(195)). Lim a(n)/a(n+1) = 28 - Lim a(n+1)/a(n).

			a(4) = 611522 = 28a(3) - a(2) = 28*21868 - 782 =(14+sqrt(195))^4 + (14-sqrt(195))^4 =611521.999998364 + 0.000001635 =611522.

Indranil Ghosh, Table of n, a(n) for n = 0..689
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (28, -1).

Cf. A053204, A063872.

a[0] = 2; a[1] = 28; a[n_] := 28a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{28,-1},{2,28},20] (* or *) CoefficientList[ Series[ (2-28x)/(x^2-28x+1),{x,0,20}],x] (* Harvey P. Dale, Jun 25 2011 *)

[lucas_number2(n,28,1) for n in range(0,16)] # Zerinvary Lajos, Jun 27 2008

a(n) = 28a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 28. a(n) = (14+sqrt(195))^n + (14-sqrt(195))^n. (a(n))^2 =a(2n)+2.

G.f.: (2-28*x)/(1-28*x+x^2). - Philippe Deléham, Nov 02 2008

More terms from Robert G. Wilson v, Jan 30 2004

A053206 Row sums of A053203.

Original entry on oeis.org

2, 0, 6, 6, 2, 0, 14, 0, 30, 36, 46, 0, 26, 0, 54, 48, 46, 0, 110, 30, 54, 78, 126, 0, 212, 0, 222, 72, 2, 86, 134, 0, 78, 84, 214, 0, 398, 0, 278, 330, 278, 0, 542, 126, 222, 414, 378, 0, 620, 96, 590, 120, 350, 0, 734, 0, 870, 384, 894, 290, 458, 0, 150, 558, 742, 0, 1142

Offset: 6

Asher Auel, Dec 12 1999

			a(6) = 3 + 0 + 0 + 3 = 6.

Reinhard Zumkeller, Table of n, a(n) for n = 6..1000

Cf. A053201, A053202, A053204.

a053206 = sum . a053203_row  -- Reinhard Zumkeller, Jan 24 2014

a(n) = A053204(n) - n*((n+1) mod 2) - 2.

a(prime(n)) = 0, where prime(n) is the n-th prime.

Corrected and extended by James Sellers, Dec 13 1999

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A053200 Binomial coefficients C(n,k) reduced modulo n, read by rows; T(0,0)=0 by convention.

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A053205 Row sums of A053201.

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A090249 a(n) = 28a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 28.

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A053206 Row sums of A053203.

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