A133623 -id:A133623 - cp's OEIS frontend

A133910 Period numbers of A133900 divided by n^2.

1, 1, 1, 1, 1, 2, 1, 1, 1, 4, 1, 6, 1, 4, 3, 1, 1, 8, 1, 4, 3, 8, 1, 6, 1, 8, 1, 4, 1, 360, 1, 1, 9, 16, 5, 24, 1, 16, 9, 20, 1, 144, 1, 8, 15, 16, 1, 18, 1, 16, 9, 8, 1, 16, 5, 28, 9, 16, 1, 360, 1, 16, 21, 1, 5, 288, 1, 16, 9, 1120, 1, 24, 1, 32, 9, 16, 7, 288, 1, 20, 1, 64, 1, 6048, 5, 32, 27, 8

Offset: 1

Hieronymus Fischer, Oct 20 2007

nonn

			a(6)=2, since A133900(6)/6^2=72/36=2.
a(18)=8, since A133900(18)/18^2=2592/324=8.

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

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

a(n)=A133900(n)/n^2.
a(n)=1, iff n is a prime or a power of a prime (including n=1).
If a prime p is a factor of a(n), then p is also a factor of n.

A133622 a(n) = 1 if n is odd, a(n) = n/2+1 if n is even.

Original entry on oeis.org

1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 1, 12, 1, 13, 1, 14, 1, 15, 1, 16, 1, 17, 1, 18, 1, 19, 1, 20, 1, 21, 1, 22, 1, 23, 1, 24, 1, 25, 1, 26, 1, 27, 1, 28, 1, 29, 1, 30, 1, 31, 1, 32, 1, 33, 1, 34, 1, 35, 1, 36, 1, 37, 1, 38, 1, 39, 1, 40, 1, 41, 1, 42, 1, 43, 1, 44, 1

Offset: 1

Hieronymus Fischer, Sep 30 2007

a(n) is the count of terms a(n+1) present so far in the sequence, with a(n+1) included in the count; example: a(1) = 1 "says" that there is 1 term "2" so far in the sequence; a(2) = 2 "says" that there are 2 terms "1" so far in the sequence... etc. This comment was inspired by A039617. - Eric Angelini, Mar 03 2020

Index entries for linear recurrences with constant coefficients, signature (0, 2, 0, -1).

Cf. A133620, A133621, A133623, A133624, A133625, A133630, A038509, A133634-A133636, A133872, A133882, A133880, A133890, A133900, A133910, A165157 (partial sums).

Other related sequences: A000217, A007879, A057979.

import Data.List (transpose)
a133622 n = (1 - m) * n' + 1 where (n', m) = divMod n 2
a133622_list = concat $ transpose [[1, 1 ..], [2 ..]]
-- Reinhard Zumkeller, Feb 20 2015

seq([1,n][],n=2..100); # Robert Israel, May 27 2016

Riffle[Range[2,50],1,{1,-1,2}] (* Harvey P. Dale, Jan 19 2013 *)

a(n)=if(n%2,1,n/2+1) \\ Charles R Greathouse IV, Sep 02 2015

a(n)=1+(binomial(n+1,2)mod n)=1+(binomial(n+1,n-1)mod n).
a(n)=binomial(n+2,2) mod n = binomial(n+2,n) mod n for n>2.
a(n)=1+(1+(-1)^n)*n/4.
a(n)=1+(A000217(n) mod n).
a(n)=a(n-2)+1, if n is even, a(n)=a(n-2) if n is odd.
a(n)=a(n-2)+1-(n mod 2)=a(n-2)+(1+(-1)^n)/2 for n>2.
a(n)=(a(n-3)+a(n-2))/a(n-1) for n>3.
G.f.: g(x)=x(1+2x-x^2-x^3)/(1-x^2)^2.
G.f.: (Q(0)-1-x)/x^2, where Q(k)= 1 + (k+1)*x/(1 - x/(x + (k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 23 2013
a(n) = 2*a(n-2)-a(n-4) for n > 4. - Chai Wah Wu, May 26 2016
E.g.f.: exp(x) - 1 + x*sinh(x)/2. - Robert Israel, May 27 2016

A133911 Number of prime factors (counted with multiplicity) of the period numbers defined by A133900.

Original entry on oeis.org

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

Offset: 1

Hieronymus Fischer, Oct 20 2007

nonn

			a(6)=5, since A133900(6)=72=2*2*2*3*3.
a(12)=8, since A133900(12)=864=2*2*2*2*2*3*3*3.

Cf. A000040, A001222, A100118, A046363, A133620, A133621, A133623, A133630, A133635.

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

a(n)=A001222(A133900(n)).

A134331 Sum of prime factors (counted with multiplicity) of the period numbers defined by A133900.

Original entry on oeis.org

0, 4, 6, 8, 10, 12, 14, 12, 12, 18, 22, 19, 26, 22, 19, 16, 34, 22, 38, 22, 23, 32, 46, 23, 20, 36, 18, 26, 58, 37, 62, 20, 34, 46, 29, 29, 74, 50, 38, 31, 82, 38, 86, 36, 30, 58, 94, 30, 28, 32, 46, 40, 106, 30, 37, 37, 50, 70, 118, 41, 122, 74, 36, 24, 41, 48, 134, 50, 58, 50

Offset: 1

Hieronymus Fischer, Oct 20 2007

nonn

			a(6)=12, since A133900(6)=72=2*2*2*3*3 and 2+2+2+3+3=12.
a(12)=19, since A133900(12)=864=2*2*2*2*2*3*3*3 and 2+2+2+2+2+3+3+3=19.

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

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

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

Original entry on oeis.org

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.

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

A133905 Least composite number m such that binomial(n+m,m) mod m = 1.

Original entry on oeis.org

4, 9, 25, 10, 26, 9, 9, 9, 6, 4, 4, 34, 34, 85, 289, 4, 4, 57, 87, 8, 8, 25, 25, 25, 134, 4, 4, 15, 15, 111, 111, 4, 4, 8, 8, 10, 10, 121, 121, 82, 86, 4, 4, 49, 49, 49, 49, 4, 4, 265, 68, 10, 10, 8, 8, 6, 9, 4, 4, 194, 194, 469, 249, 4, 4, 44, 44, 146, 146, 16, 16, 6, 6, 4, 4, 162

Offset: 1

Hieronymus Fischer, Oct 20 2007

nonn

			a(1)=4, since binomial(1+4,4) mod 4 = 5 mod 4 = 1 and 4 is the minimal composite number with this property.
a(5)=26 because of binomial(5+26,26)=169911=6535*26+1, but binomial(5+k,k) mod k<>1 for all composite numbers <26.

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

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

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

lcn[n_]:=Module[{m=4},While[PrimeQ[m]||Mod[Binomial[n+m,m],m]!=1,m++];m]; Array[lcn,80] (* Harvey P. Dale, May 13 2022 *)

a(n) = { my(m = 4, ok = 0); until (ok, if (! isprime(m) && (binomial(n+m, m) % m == 1), ok = 1, m++);); return (m);} \\ Michel Marcus, Jul 15 2013

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

cp's OEIS Frontend

A133910 Period numbers of A133900 divided by n^2.

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A133622 a(n) = 1 if n is odd, a(n) = n/2+1 if n is even.

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A133911 Number of prime factors (counted with multiplicity) of the period numbers defined by A133900.

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A134331 Sum of prime factors (counted with multiplicity) of the period numbers defined by A133900.

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

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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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A133905 Least composite number m such that binomial(n+m,m) mod m = 1.

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