cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A038510 Composite numbers with smallest prime factor >= 7.

Original entry on oeis.org

49, 77, 91, 119, 121, 133, 143, 161, 169, 187, 203, 209, 217, 221, 247, 253, 259, 287, 289, 299, 301, 319, 323, 329, 341, 343, 361, 371, 377, 391, 403, 407, 413, 427, 437, 451, 469, 473, 481, 493, 497, 511, 517, 527, 529, 533, 539, 551, 553, 559, 581, 583
Offset: 1

Views

Author

Keywords

Comments

Let A = set of numbers of form 6n + 1, B = numbers of form 6n - 1. Eliminating numbers of form 25 + 30s from A and those of form 35 + 30s from B we obtain sets A* and B*. Removing all terms of the sequence from the union of A* and B*, only prime numbers remain. - Hisanobu Shinya (ilikemathematics(AT)hotmail.com), Jul 14 2002
Divide n by a*b*c where a = 2^(A001511(n)-1), b = 3^(A051064(n)-1) and c = 5^(A055457(n) -1). Then the resulting sequence includes only primes and a(n). - Alford Arnold, Sep 08 2003
Composite numbers not divisible by 2, 3 or 5. - Lekraj Beedassy, Jun 30 2004
Composite numbers k such that k^4 mod 30 = 1. - Gary Detlefs, Dec 09 2012
Composite numbers congruent to 1, 7, 11, 13, -13, -11, -7, -1 (mod 30). Since asymptotically, 100% of integers are composite, we have a(n)/n ~ 30/phi(30) = 30/8 = 3.75. - Daniel Forgues, Mar 16 2013
"John [Conway] recommends the more refined partition [of the positive numbers]: 1, prime, trivially composite, or nontrivially composite. Here, a composite integer is trivially composite if it is divisible by 2, 3, or 5." See link to (van der Poorten, Thomsen, and Wiebe; 2006) pp. 73-74. - Daniel Forgues, Jan 30 2015, Feb 04 2015
For the eight congruences coprime to 30, we can use one byte to encode the "primality/non-primality (unit or composite)" for each [30*n, 30*(n+1)[, n >= 0, closed-open interval, either as little endian binary sequence {01111111, 11111011, 11110111, 01111110, ...}, or as big endian binary sequence {11111110, 11011111, 11101111, 01111110, ...}, which we may then express in base 10. - Daniel Forgues, Feb 05 2015

References

  • J. H. Silverman, A Friendly Introduction to Number Theory, 2nd Edn. "Appendix A: Factorization of Small Composite Integers", Prentice Hall NY 2001.

Crossrefs

Intersection of A002808 and A007775.

Programs

  • Maple
    for n from 1 to 583 do if n^4 mod 30 = 1 and not isprime(n) then print(n)fi od; # Gary Detlefs, Dec 09 2012
  • Mathematica
    Select[Range[1000], ! PrimeQ[#] && FactorInteger[#][[1, 1]] >= 7 &] (* T. D. Noe, Mar 16 2013 *)
  • PARI
    is(n)=gcd(n,30)==1 && !isprime(n) \\ Charles R Greathouse IV, Dec 09 2012

Formula

a(n) ~ 3.75n. - Charles R Greathouse IV, Dec 09 2012

Extensions

Corrected by Ralf Stephan, Apr 04 2003

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

Views

Author

Hieronymus Fischer, Sep 30 2007

Keywords

Comments

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.

Crossrefs

Programs

  • Magma
    [Binomial(n+4,4) mod n: n in [1..100]]; // Vincenzo Librandi, Apr 27 2014
  • Mathematica
    Table[Mod[Binomial[n+4,n],n],{n,90}] (* Harvey P. Dale, Apr 26 2014 *)

Formula

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 6^2=36.

Crossrefs

Programs

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

Formula

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 7^2=49.

Crossrefs

Formula

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)).

A133887 Binomial(n+7,n) mod 7^2.

Original entry on oeis.org

1, 8, 36, 22, 36, 8, 1, 2, 16, 23, 44, 23, 16, 2, 3, 24, 10, 17, 10, 24, 3, 4, 32, 46, 39, 46, 32, 4, 5, 40, 33, 12, 33, 40, 5, 6, 48, 20, 34, 20, 48, 6, 7, 7, 7, 7, 7, 7, 7, 8, 15, 43, 29, 43, 15, 8, 9, 23, 30, 2, 30, 23, 9, 10, 31, 17, 24, 17, 31, 10, 11, 39, 4, 46, 4, 39, 11, 12, 47
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 7^3=343.

Crossrefs

For the sequence regarding binomial(n+7, n) mod 7 see A133877.

Programs

  • Mathematica
    Table[Mod[Binomial[n+7,n],49],{n,0,80}] (* Harvey P. Dale, Apr 08 2018 *)

Formula

a(n)=binomial(n+7,7) mod 7^2.
G.f. g(x)=sum{0<=k<343, a(k)*x^k}/(1-x^343).

A133888 Binomial(n+8,n) mod 8.

Original entry on oeis.org

1, 1, 5, 5, 7, 7, 3, 3, 6, 6, 6, 6, 2, 2, 2, 2, 7, 7, 3, 3, 1, 1, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 1, 1, 3, 3, 7, 7, 2, 2, 2, 2, 6, 6, 6, 6, 3, 3, 7, 7, 5, 5, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 5, 7, 7, 3, 3, 6, 6, 6, 6, 2, 2, 2, 2, 7, 7, 3, 3, 1, 1, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 1, 1, 3, 3, 7, 7, 2
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 8^2=64.

Crossrefs

Programs

  • Mathematica
    Table[Mod[Binomial[n+8,n],8],{n,0,110}] (* Harvey P. Dale, Aug 08 2011 *)

Formula

a(n)=binomial(n+8,8) mod 8.

A133889 a(n) = binomial(n+9,n) mod 9.

Original entry on oeis.org

1, 1, 1, 4, 4, 4, 1, 1, 1, 2, 2, 2, 8, 8, 8, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 7, 7, 7, 4, 4, 4, 5, 5, 5, 2, 2, 2, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 1, 1, 1, 7, 7, 7, 8, 8, 8, 5, 5, 5, 8, 8, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 4, 4, 4, 1, 1, 1, 2, 2, 2, 8, 8, 8, 2, 2, 2, 3, 3, 3, 3, 3, 3
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 9^2 = 81.

Crossrefs

Programs

  • Mathematica
    Table[Mod[Binomial[n+9,n],9],{n,0,110}] (* Harvey P. Dale, Jan 14 2012 *)

Formula

a(n) = binomial(n+9,9) mod 9.

A133878 n modulo 8 repeated 8 times.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 8^2=64.

Crossrefs

Programs

  • Mathematica
    Flatten[Join[Table[PadRight[{},8,n],{n,7}],Table[PadRight[{},8,n],{n,0,7}]]] (* Harvey P. Dale, Nov 06 2011 *)

Formula

a(n)=(1+floor(n/8)) mod 8.
a(n)=1+floor(n/8)-8*floor((n+8)/64).
a(n)=(((n+8) mod 64)-(n mod 8))/8.
a(n)=((n+8-(n mod 8))/8) mod 8.
G.f. g(x)=(1-x^8)(1+2x^8+3x^16+4x^24+5x^32+6x^40+7x^48)/((1-x)(1-x^64)).
G.f. g(x)=(1-x^8)*sum{0<=k<7, (k+1)*x^(8*k)}/((1-x)(1-x^64)).
G.f. g(x)=(7x^64-8x^56+1)/((1-x)(1-x^8)(1-x^64)).

A133879 n modulo 9 repeated 9 times.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3
Offset: 0

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 9^2=81.

Crossrefs

Programs

  • Mathematica
    Rest[Flatten[Table[Table[Table[n,{9}],{n,0,8}],{3}],1]]//Flatten (* Harvey P. Dale, Apr 15 2018 *)

Formula

a(n)=(1+floor(n/9)) mod 9.
a(n)=1+floor(n/9)-9*floor((n+9)/81).
a(n)=(((n+9) mod 81)-(n mod 9))/9.
a(n)=((n+9-(n mod 9))/9) mod 9.
G.f. g(x)=(1-x^9)(1+2x^9+3x^18+4x^27+5x^36+6x^45+7x^56+8x^63)/((1-x)(1-x^81)).
G.f. g(x)=(1-x^9)*sum{0<=k<8, (k+1)*x^(9*k)}/((1-x)(1-x^81)).
G.f. g(x)=(8x^81-9x^72+1)/((1-x)(1-x^9)(1-x^81)).

A249819 Composite natural numbers n for which there are exactly two distinct 0 < k < n^2 such that 2^k - 1 is divisible by n^2.

Original entry on oeis.org

35, 49, 77, 95, 115, 143, 175, 209, 235, 245, 289, 295, 299, 319, 335, 343, 371, 395, 407, 413, 415, 437, 475, 515, 517, 529, 535, 539, 551, 575, 581, 583, 611, 649, 667, 695, 707, 749, 767, 815, 835, 847, 851, 869, 875, 893, 895, 913, 917, 923, 995, 1007
Offset: 1

Views

Author

Antti Karttunen, Nov 15 2014

Keywords

Comments

Equally: odd composite numbers n for which A246702((n+1)/2) = 2.

Examples

			35 = 5*7 is an odd composite. Only cases where 2^k - 1 (with k in range 1 .. 35^2 - 1 = 1 .. 1224) is a multiple of 35 are k = 420 and k = 840, thus 35 is included in this sequence.
		

Crossrefs

Composite terms in A246717.
Seems also to be a subsequence of A038509.
Cf. A246702.

Programs

  • Maple
    isA249819 := proc(n)
        if isprime(n) or n=1 then
            false;
        else
            ct := 0 ;
            for k from 1 to n^2-1 do
                if modp(2 &^ k-1,n^2) = 0 then
                    ct := ct+1 ;
                end if;
                if ct > 2 then
                    return false;
                end if;
            end do:
            return is(ct=2) ;
        end if;
    end proc:
    for n from 1 to 1100 do
        if isA249819(n) then
            printf("%d,\n",n) ;
        end if;
    end do: # R. J. Mathar, Nov 16 2014
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