A133910 -id:A133910 - cp's OEIS frontend

A134332 Integer part of the arithmetic mean of the prime factors (counted with multiplicity) of the period numbers defined by A133900.

1, 2, 3, 2, 5, 2, 7, 2, 3, 3, 11, 2, 13, 3, 3, 2, 17, 2, 19, 2, 4, 4, 23, 2, 5, 5, 3, 3, 29, 3, 31, 2, 5, 5, 5, 2, 37, 6, 6, 2, 41, 3, 43, 4, 3, 7, 47, 2, 7, 3, 7, 4, 53, 2, 7, 3, 8, 8, 59, 2, 61, 9, 4, 2, 8, 3, 67, 5, 9, 3, 71, 2, 73, 9, 4, 5, 8, 4, 79, 2, 3, 9, 83, 3, 9, 11, 10, 3, 89, 2, 9, 6, 11, 12

Offset: 1

Hieronymus Fischer, Oct 23 2007

nonn

			a(6)=2, since floor(A134331(6)/A133911(6))=floor(12/5)=2.
a(7)=7, since floor(A134331(7)/A133911(7))=floor(14/2)=7.

Cf. A000040, A100118, A046363, A133620, A133621, A133623, A133630, A133635.

Cf. A133872, A133880, A133890, A133900, A133910, A133911, A134331.

a(n)=floor(A134331(n)/A133911(n)) for n>1, defining a(1):=1.

a(n)=n, if n is a prime or 1.

A078177 Composite numbers with an integer arithmetic mean of all prime factors.

Original entry on oeis.org

4, 8, 9, 15, 16, 20, 21, 25, 27, 32, 33, 35, 39, 42, 44, 49, 50, 51, 55, 57, 60, 64, 65, 68, 69, 77, 78, 81, 85, 87, 91, 92, 93, 95, 105, 110, 111, 112, 114, 115, 116, 119, 121, 123, 125, 128, 129, 133, 140, 141, 143, 145, 155, 156, 159, 161, 164, 169, 170, 177, 180

Offset: 1

Reinhard Zumkeller, Nov 20 2002

nonn

That is, composite numbers such that the arithmetic mean of their prime factors (counted with multiplicity) is an integer.

			60 = 2*2*3*5: (2+2+3+5)/4 = 3, therefore 60 is a term.

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

Cf. A078175, A001222, A100118, A046363, A133620, A133621.

Cf. A133880, A133890, A133900, A133910, A133911, A046346, A134331, A134332, A134333, A134334, A134344, A134376.

Select[Range[200], CompositeQ[#] && IntegerQ[Mean[Flatten[Table[#[[1]], #[[2]]]& /@ FactorInteger[#]]]]&] (* Jean-François Alcover, Aug 03 2018 *)

lista(nn) = {forcomposite(n=1, nn, my(f = factor(n)); if (! (sum(k=1, #f~, f[k,1]*f[k,2]) % vecsum(f[,2])), print1(n, ", ")););} \\ Michel Marcus, Feb 22 2016

A001414(a(n)) == 0 modulo A001222(a(n)).

Edited by N. J. A. Sloane, May 30 2008 at the suggestion of R. J. Mathar

A133873 n modulo 3 repeated 3 times.

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Oct 10 2007

Periodic with length 3^2=9.

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

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

Cf. A133883, A133880, A133890, A133900, A133910.

G.f.: (1 + 2*x^3)*(1 - x^3)/((1 - x)*(1 - x^9)).
a(n) = (1 + floor(n/3)) mod 3.
a(n) = A010872(A002264(n+3)).
a(n) = 1+floor(n/3)-3*floor((n+3)/9).
a(n) = (((n+3) mod 9)-(n mod 3))/3.
a(n) = ((n+3-(n mod 3))/3) mod 3.
a(n) = binomial(n+3,n) mod 3 = binomial(n+3,3) mod 3.

A133874 n modulo 4 repeated 4 times.

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Oct 10 2007

nonn

Periodic with length 4^2 = 16.

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

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

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

Flatten[Table[Table[Mod[n,4],{4}],{n,30}]] (* Harvey P. Dale, Dec 22 2013 *)

def A133874(n): return 1+(n>>2)&3 # Chai Wah Wu, Jan 18 2023

a(n) = (1 + floor(n/4)) mod 4.
a(n) = A010873(A002265(n+4)).
a(n) = 1 + floor(n/4) - 4*floor((n+4)/16).
a(n) = (((n+4) mod 16) - (n mod 4))/4.
a(n) = ((n + 4 - (n mod 4))/4) mod 4.
G.f. g(x) = (1 + x + x^2 + x^3 + 2x^4 + 2x^5 + 2x^6 + 2x^7 + 3x^8 + 3x^9 + 3x^10 + 3x^11)/(1-x^16).
G.f. g(x) = ((1-x^4)*(1+2x^4+3x^8))/((1-x)*(1-x^16)).
G.f. g(x) = (3x^16-4x^12+1)/((1-x)*(1-x^4)*(1-x^16)).
G.f. g(x) = (1+2x^4+3x^8)/((1-x)*(1+x^4)*(1+x^8)).

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

A133906 Least number m such that binomial(n+m, m) mod m = 1.

Original entry on oeis.org

2, 3, 5, 2, 2, 7, 9, 2, 2, 3, 3, 2, 2, 17, 17, 2, 2, 3, 3, 2, 2, 23, 25, 2, 2, 4, 3, 2, 2, 31, 37, 2, 2, 8, 8, 2, 2, 3, 41, 2, 2, 4, 4, 2, 2, 3, 3, 2, 2, 5, 5, 2, 2, 3, 3, 2, 2, 4, 4, 2, 2, 67, 3, 2, 2, 44, 44, 2, 2, 16, 16, 2, 2, 3, 4, 2, 2, 5, 5, 2, 2, 3, 3, 2, 2, 89, 9, 2, 2, 3, 3, 2, 2, 97, 97, 2, 2, 7

Offset: 1

Hieronymus Fischer, Oct 20 2007

nonn

			a(1)=2, since binomial(1+2, 2) mod 2 = 3 mod 2 = 1 and 2 is the minimal number with this property.
a(7)=9 because of binomial(7+9, 9) = 11440 = 1271*9 + 1, but binomial(7+k, k) mod k <> 1 for all numbers < 9.

Seiichi Manyama, Table of n, a(n) for n = 1..1000

Cf. A000040, A133620, A133621, A133623, A133630, A133635.

Cf. A133872, A133880, A133890, A133900, A133907, A133910.

Table[Block[{m = 1}, While[Mod[Binomial[n + m, m], m] != 1, m++]; m], {n, 98}] (* Michael De Vlieger, Jul 30 2018 *)

a(n) = {my(m = 1, ok = 0); until (ok, if (binomial(n+m, m) % m == 1, ok = 1, m++);); return (m);} \\ Michel Marcus, Jul 15 2013

A133907 Least prime number p such that binomial(n+p, p) mod p = 1.

Original entry on oeis.org

2, 3, 5, 2, 2, 7, 11, 2, 2, 3, 3, 2, 2, 17, 17, 2, 2, 3, 3, 2, 2, 23, 29, 2, 2, 5, 3, 2, 2, 31, 37, 2, 2, 37, 37, 2, 2, 3, 41, 2, 2, 43, 47, 2, 2, 3, 3, 2, 2, 5, 5, 2, 2, 3, 3, 2, 2, 59, 61, 2, 2, 67, 3, 2, 2, 67, 71, 2, 2, 71, 73, 2, 2, 3, 5, 2, 2, 5, 5, 2, 2, 3, 3, 2, 2, 89, 89, 2, 2, 3, 3, 2, 2, 97

Offset: 1

Hieronymus Fischer, Oct 20 2007

nonn

Also the least prime number p such that p divides floor(n/p) or p > n.
a(n) = 2 if and only if n is in A042948. - Robert Israel, May 11 2017
Conjecture: a(n) is the smallest prime p such that Sum_{k=1..n} k^(p-1) == n (mod p). Thus a(n) >= A317358(n). - Thomas Ordowski, Jul 29 2018

			a(2)=3, since binomial(2+3,3) mod 3 = 10 mod 3 = 1 and 3 is the minimal prime number with this property.
a(7)=11 because of binomial(7+11, 11) = 31824 = 2893*11 + 1, but binomial(7+k, k) mod k <> 1 for all primes < 11.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000040, A042948, A133620, A133621, A133623, A133630, A133635.

Cf. A133872, A133880, A133890, A133900, A133906, A133910, A317358.

f:= proc(n) local m;
  m:= 2:
  while floor(n/m) mod m <> 0 do m:= nextprime(m) od:
  m
end proc:
map(f, [$1..100]); # Robert Israel, May 11 2017

a[n_] := Module[{p}, For[p = 2, True, p = NextPrime[p], If[Mod[Binomial[n+p, p], p] == 1, Return[p]]]];
Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Feb 05 2023 *)

a(n) = my(p=2); while (binomial(n+p, p) % p != 1, p = nextprime(p+1)); p; \\ Michel Marcus, Dec 17 2022

from sympy import nextprime, ff
def A133907(n):
    p, m = 2, (n+2)*(n+1)>>1
    while m%p != 1:
        q = nextprime(p)
        m = m*ff(n+q,q-p)//ff(q,q-p)
        p = q
    return p # Chai Wah Wu, Feb 22 2023

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

A133876 n modulo 6 repeated 6 times.

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Oct 10 2007

nonn

Periodic with length 6^2=36.

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

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

Cf. A133886, A133880, A133890, A133900, A133910.

Table[PadRight[{},6,Mod[n,6]],{n,20}]//Flatten (* Harvey P. Dale, Nov 15 2023 *)

a(n)=(1+floor(n/6)) mod 6.
a(n)=1+floor(n/6)-6*floor((n+6)/36).
a(n)=(((n+6) mod 36)-(n mod 6))/6.
a(n)=((n+6-(n mod 6))/6) mod 6.
G.f. g(x)=(1-x^6)(1+2x^5+3x^12+4x^18+5x^24)/((1-x)(1-x^36)).
G.f. g(x)=(5x^36-6x^30+1)/((1-x)(1-x^6)(1-x^36)).

A133877 n modulo 7 repeated 7 times.

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Oct 10 2007

nonn

Periodic with length 7^2=49.

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

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

Cf. A133887, A133880, A133890, A133900, A133910.

a(n)=(1+floor(n/7)) mod 7.
a(n)=1+floor(n/7)-7*floor((n+7)/49).
a(n)=(((n+7) mod 49)-(n mod 7))/7.
a(n)=((n+7-(n mod 7))/7) mod 7.
a(n)=binomial(n+7,n) mod 7 =binomial(n+7,7) mod 7.
G.f. g(x)=(1-x^7)(1+2x^7+3x^14+4x^21+5x^28+6x^35)/((1-x)(1-x^49)).
G.f. g(x)=(6x^49-7x^42+1)/((1-x)(1-x^7)(1-x^49)).

cp's OEIS Frontend

A134332 Integer part of the arithmetic mean of the prime factors (counted with multiplicity) of the period numbers defined by A133900.

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A078177 Composite numbers with an integer arithmetic mean of all prime factors.

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A133873 n modulo 3 repeated 3 times.

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A133874 n modulo 4 repeated 4 times.

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

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Magma

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A133906 Least number m such that binomial(n+m, m) mod m = 1.

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A133907 Least prime number p such that binomial(n+p, p) mod p = 1.

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Maple

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

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