A133874 -id:A133874 - cp's OEIS frontend

A133634 Nonprime numbers k such that binomial(k+p,k) mod k = 1, where p=4.

10, 25, 26, 34, 35, 49, 50, 55, 58, 65, 74, 77, 82, 85, 91, 95, 98, 106, 115, 119, 121, 122, 125, 130, 133, 143, 145, 146, 154, 155, 161, 169, 170, 175, 178, 185, 187, 194, 202, 203, 205, 209, 215, 217, 218, 221, 226, 235, 242, 245, 247, 250, 253, 259, 265, 266

Offset: 1

Hieronymus Fischer, Sep 30 2007

nonn

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.

Cf. A133874, A133884, A133880, A133890, A133900, A133910.

Select[Range[300],!PrimeQ[#]&&Mod[Binomial[#+4,#],#]==1&] (* Harvey P. Dale, Oct 09 2011 *)

A133884 a(n) = binomial(n+4,n) mod 4.

Original entry on oeis.org

1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3

Offset: 0

Hieronymus Fischer, Oct 10 2007

Periodic with length 4^2=16.

			For n=2, binomial(6,2) = 6*5/2 = 15, which is 3 (mod 4) so a(2) = 3. - _Michael B. Porter_, Jul 19 2016

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

Cf. A000040, A133630, A038509.
Cf. A133620, A133621, A133622, A133623, A133624, A133625
Cf. A133634, A133635, A133636.
Cf. A133874, A133880, A133890, A133900, A133910.

[Binomial(n+4,n) mod 4: n in [0..100]]; // Vincenzo Librandi, Jul 15 2016

Table[Mod[Binomial[n + 4, 4], 4], {n, 0, 100}] (* Vincenzo Librandi, Jul 15 2016 *)

a(n) = binomial(n+4,4) mod 4.
G.f.: (1 + x + 3*x^2 + 3*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + 2*x^7 + 3*x^8 + 3*x^9 + x^10 + x^11)/(1 - x^16) = (1 + 2*x^2 + 2*x^6 + x^8)/((1 - x)*(1 + x^4)*(1 + x^8)).
a(n) = A000505(n+5) mod 4. - John M. Campbell, Jul 14 2016
a(n) = A000506(n+6) mod 4. - John M. Campbell, Jul 15 2016

G.f. corrected by Bruno Berselli, Jul 19 2016

A133624 Binomial(n+p, n) mod n, where p=4.

Original entry on oeis.org

0, 1, 2, 2, 1, 0, 1, 7, 4, 1, 1, 8, 1, 8, 6, 13, 1, 7, 1, 6, 8, 12, 1, 3, 1, 1, 10, 8, 1, 26, 1, 25, 12, 1, 1, 22, 1, 20, 14, 31, 1, 15, 1, 12, 16, 24, 1, 5, 1, 1, 18, 14, 1, 46, 1, 43, 20, 1, 1, 36, 1, 32, 22, 49, 1, 23, 1, 18, 24, 36, 1, 7, 1, 1, 26, 20, 1, 66, 1, 61, 28, 1, 1, 50, 1, 44, 30

Offset: 1

Hieronymus Fischer, Sep 30 2007

nonn

Let d(m)...d(2)d(1)d(0) be the base-n representation of n+p. The relation a(n)=d(1) holds, if n is a prime index. For this reason there are infinitely many terms which are equal to 1.

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

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636, A133874, A133884, A133880, A133890, A133900, A133910, A362686, A362687, A362688, A362689.

[Binomial(n+4,4) mod n: n in [1..100]]; // Vincenzo Librandi, Apr 27 2014

Table[Mod[Binomial[n+4,n],n],{n,90}] (* Harvey P. Dale, Apr 26 2014 *)

a(n) = binomial(n+4,4) mod n.
a(n)=1 if n is a prime > 4, since binomial(n+4,n) == (1+floor(4/n))(mod n), provided n is a prime.
From Chai Wah Wu, May 26 2016: (Start)
a(n) = (n^4 + 10*n^3 + 11*n^2 + 2*n + 24)/24 mod n.
For n > 6:
if n mod 24 == 0, then a(n) = n/12 + 1.
if n mod 24 is in {1, 2, 5, 7, 10, 11, 13, 17, 19, 23}, then a(n) = 1.
if n mod 24 is in {3, 9, 15, 18, 21}, then a(n) = n/3 + 1.
if n mod 24 is in {4, 20}, then a(n) = n/4 + 1.
if n mod 24 == 6, then a(n) = 5*n/6 + 1.
if n mod 24 is in {8, 16}, then a(n) = 3*n/4 + 1.
if n mod 24 == 12, then a(n) = 7*n/12 + 1.
if n mod 24 is in {14, 22}, then a(n) = n/2 + 1.
(End)
For n > 54, a(n) = 2*a(n-24) - a(n-48). - Ray Chandler, Apr 23 2023

A133894 Numbers m such that binomial(m+4,m) mod 4 = 0.

Original entry on oeis.org

12, 13, 14, 15, 28, 29, 30, 31, 44, 45, 46, 47, 60, 61, 62, 63, 76, 77, 78, 79, 92, 93, 94, 95, 108, 109, 110, 111, 124, 125, 126, 127, 140, 141, 142, 143, 156, 157, 158, 159, 172, 173, 174, 175, 188, 189, 190, 191, 204, 205, 206, 207, 220, 221, 222, 223, 236, 237

Offset: 1

Hieronymus Fischer, Oct 20 2007

Also numbers m such that floor(1+(m/4)) mod 4 = 0.
Partial sums of the sequence 12,1,1,1,13,1,1,1,13, ... which has period 4.
Numbers congruent to {12,13,14,15} mod 16. Numbers n such that n xor 12 = n - 12. [Brad Clardy, May 06 2013]

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

Cf. A000040, A133620, A133621, A133623, A133630, A133635, A133874, A133884, A133890, A133900, A133910.

a(n)=(n+3)\4*16-[1,4,3,2][n%4+1] \\ Charles R Greathouse IV, May 06 2013

a(n) = 4*n + 12 - 3*(n mod 4).
G.f.: 12/(1-x)+x(1+x+x^2+13x^3)/((1-x^4)(1-x)) = (12+x+x^2+x^3+x^4)/((1-x^4)(1-x)) = (12-11x-x^5)/((1-x^4)(1-x)^2).
a(n) = 4*n+3*((1-i)*i^n+(1+i)*(-i)^n+(-1)^n+5)/2, where i=sqrt(-1). - Bruno Berselli, Apr 08 2011

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A133634 Nonprime numbers k such that binomial(k+p,k) mod k = 1, where p=4.

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A133884 a(n) = binomial(n+4,n) mod 4.

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A133624 Binomial(n+p, n) mod n, where p=4.

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A133894 Numbers m such that binomial(m+4,m) mod 4 = 0.

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