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.

Showing 1-10 of 18 results. Next

A156695 Odd numbers that are not of the form p + 2^a + 2^b, a, b > 0, p prime.

Original entry on oeis.org

1, 3, 5, 6495105, 848629545, 1117175145, 2544265305, 3147056235, 3366991695, 3472109835, 3621922845, 3861518805, 4447794915, 4848148485, 5415281745, 5693877405, 6804302445, 7525056375, 7602256605, 9055691835, 9217432215
Offset: 1

Views

Author

Keywords

Comments

Crocker shows that this sequence is infinite.
All members above 5 found so far (up to 2.5 * 10^11) are divisible by 255 = 3 * 5 * 17, and many are divisible by 257. I conjecture that all members of this sequence greater than 5 are divisible by 255. This implies that all odd numbers (greater than 7) are the sum of a prime and at most three positive powers of two.
Pan shows that, for every c > 1, a(n) << x^c. More specifically, there are constants C,D > 0 such that there are at least Dx/exp(C log x log log log log x/log log log x) members of this sequence up to x. - Charles R Greathouse IV, Apr 11 2016
All terms > 5 are numbers k > 3 such that k - 2^n is a de Polignac number (A006285) for every n > 0 with 2^n < k. Are there numbers K such that |K - 2^n| is a Riesel number (A101036) for every n > 0? If so, ||K - 2^n| - 2^m| is composite for every pair m,n > 0, by the dual Riesel conjecture. - Thomas Ordowski, Jan 06 2024
In keeping with the example's connection to A000215, the lowest ki for ki * Product_{i=0..11} (F(i)) to belong to A156695 are 1, 433007, 25471, 17047, 1291, 7, 101, 807, 83, 347, 9, 179. So for example, 433007*(3*5) is a term. This implies a variant of the first commented conjecture accordingly. - Bill McEachen, Apr 17 2025

Examples

			Prime factorization of terms:
F_0 = 3, F_1 = 5, F_2 = 17, F_3 = 257 are Fermat numbers (cf. A000215)
6495105    = 3   * 5   * 17               * 25471
848629545  = 3   * 5   * 17               * 461      * 7219
1117175145 = 3   * 5   * 17         * 257 * 17047
2544265305 = 3^2 * 5   * 17         * 257 * 12941
3147056235 = 3^2 * 5   * 17         * 257 * 16007
3366991695 = 3   * 5   * 17   * 83  * 257 * 619
3472109835 = 3   * 5   * 17         * 257 * 52981
3621922845 = 3   * 5   * 17^2       * 257 * 3251
3861518805 = 3^3 * 5   * 17         * 257 * 6547
4447794915 = 3^3 * 5   * 17         * 257 * 7541
4848148485 = 3^4 * 5   * 17               * 704161
5415281745 = 3   * 5   * 17               * 21236399
5693877405 = 3^2 * 5   * 17         * 257 * 28961
6804302445 = 3^2 * 5   * 17   * 53  * 257 * 653
7525056375 = 3^2 * 5^3 * 17         * 257 * 1531
7602256605 = 3   * 5   * 17         * 257 * 311      * 373
9055691835 = 3   * 5   * 17         * 257 * 138181
9217432215 = 3^2 * 5   * 17   * 173 * 257 * 271
		

Crossrefs

Programs

  • PARI
    is(n)=if(n%2==0,return(0)); for(a=1,log(n)\log(2), for(b=1,a, if(isprime(n-2^a-2^b),return(0)))); 1 \\ Charles R Greathouse IV, Nov 27 2013
    
  • Python
    from itertools import count, islice
    from sympy import isprime
    def A156695_gen(startvalue=1): # generator of terms >= startvalue
        for n in count(max(startvalue+(startvalue&1^1),1),2):
            l = n.bit_length()-1
            for a in range(l,0,-1):
                c = n-(1<A156695_list = list(islice(A156695_gen(),4)) # Chai Wah Wu, Nov 29 2023

Extensions

Factorizations added by Daniel Forgues, Jan 20 2011

A304081 Number of ways to write n as p + 2^k + (1+(n mod 2))*5^m, where p is an odd prime, and k and m are nonnegative integers with 2^k + (1+(n mod 2))*5^m squarefree.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 2, 1, 3, 1, 2, 2, 2, 1, 3, 3, 3, 2, 4, 2, 3, 2, 5, 2, 4, 2, 3, 3, 3, 2, 4, 3, 5, 1, 7, 4, 4, 3, 7, 2, 4, 3, 8, 4, 7, 4, 6, 3, 7, 3, 6, 4, 5, 3, 5, 4, 5, 2, 7, 3, 5, 4, 8, 4, 5, 3, 5, 5, 8, 6, 6, 6, 9, 3, 9, 7, 6, 6, 8, 5, 6, 4, 6, 8, 7, 6, 8, 7, 4
Offset: 1

Views

Author

Zhi-Wei Sun, May 06 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 7.
This has been verified for n up to 2*10^10.
See also A303821, A303934, A303949, A304031 and A304122 for related information, and A304034 for a similar conjecture.
The author would like to offer 2500 US dollars as the prize to the first proof of the conjecture, and 250 US dollars as the prize to the first explicit counterexample. - Zhi-Wei Sun, May 08 2018

Examples

			a(6) = 1 since 6 = 3 + 2^1 + 5^0 with 3 an odd prime and 2^1 + 5^0 = 3 squarefree.
a(15) = 1 since 15 = 5 + 2^3 + 2*5^0 with 5 an odd prime and 2^3 + 2*5^0 = 2*5 squarefree.
a(35) = 1 since 35 = 29 + 2^2 + 2*5^0 with 29 an odd prime and 2^2 + 2*5^0 = 2*3 squarefree.
a(91) = 1 since 91 = 17 + 2^6 + 2*5^1 with 17 an odd prime and 2^6 + 2*5^1 = 2*37 squarefree.
a(9574899) = 1 since 9574899 = 9050609 + 2^19 + 2*5^0 with 9050609 an odd prime and 2^19 + 2*5^0 = 2*5*13*37*109 squarefree.
a(6447154629) = 2 since 6447154629 = 6447121859 + 2^15 + 2*5^0 with 6447121859 prime and 2^15 + 2*5^0 = 2*5*29*113 squarefree, and 6447154629 = 5958840611 + 2^15 + 2*5^12 with 5958840611 prime and 2^15 + 2*5^12 = 2*17*41*433*809 squarefree.
		

Crossrefs

Programs

  • Mathematica
    PQ[n_]:=n>2&&PrimeQ[n];
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+(1+Mod[n,2])*5^m]&&PQ[n-2^k-(1+Mod[n,2])*5^m],r=r+1],{k,0,Log[2,n]},{m,0,If[2^k==n,-1,Log[5,(n-2^k)/(1+Mod[n,2])]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A304034 Number of ways to write n as p + 2^k + (1+(n mod 2))*3^m with p prime, where k and m are positive integers with 2^k + (1+(n mod 2))*3^m squarefree.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 2, 1, 3, 1, 4, 2, 5, 1, 3, 2, 5, 1, 7, 3, 3, 4, 4, 4, 6, 2, 3, 5, 6, 2, 7, 3, 5, 5, 6, 5, 9, 3, 4, 6, 7, 2, 12, 2, 5, 6, 7, 4, 10, 3, 3, 5, 8, 2, 8, 3, 4, 6, 8, 5, 9, 4, 2, 7, 7, 3, 13, 5, 5, 9, 7, 5, 13, 3, 6, 10, 7, 5, 10, 5, 7, 7, 9, 8, 13
Offset: 1

Views

Author

Zhi-Wei Sun, May 06 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 11.
This has been verified for n up to 10^10.
See also A304081 for a similar conjecture.

Examples

			a(8) = 1 since 8 = 3 + 2^1 + 3^1 with 3 prime and 2^1 + 3^1 = 5 squarefree.
a(13) = 1 since 13 = 3 + 2^2 + 2*3^1 with 3 prime and 2^2 + 2*3^1 = 2*5 squarefree.
a(19) = 1 since 19 = 5 + 2^3 + 2*3^1 with 5 prime and 2^3 + 2*3^1 = 2*7 squarefree.
a(23) = 1 since 23 = 13 + 2^2 + 2*3^1 with 13 prime and 2^2 + 2*3 = 2*5 squarefree.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+(1+Mod[n,2])*3^m]&&PrimeQ[n-2^k-(1+Mod[n,2])*3^m],r=r+1],{k,1,Log[2,n]},{m,1,If[2^k==n,-1,Log[3,(n-2^k)/(1+Mod[n,2])]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A109925 Number of primes of the form n - 2^k.

Original entry on oeis.org

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

Views

Author

Amarnath Murthy, Jul 17 2005

Keywords

Comments

Erdős conjectures that the numbers in A039669 are the only n for which n-2^r is prime for all 2^rT. D. Noe and Robert G. Wilson v, Jul 19 2005
a(A006285(n)) = 0. - Reinhard Zumkeller, May 27 2015

Examples

			a(21) = 4, 21-2 =19, 21-4 = 17, 21-8 = 13, 21-16 = 5, four primes.
127 is the smallest odd number > 1 such that a(n) = 0: A006285(2) = 127. - _Reinhard Zumkeller_, May 27 2015
		

Crossrefs

Programs

  • Haskell
    a109925 n = sum $ map (a010051' . (n -)) $ takeWhile (< n)  a000079_list
    -- Reinhard Zumkeller, May 27 2015
    
  • Magma
    a109925:=function(n); count:=0; e:=1; while e le n do if IsPrime(n-e) then count+:=1; end if; e*:=2; end while; return count; end function; [ a109925(n): n in [1..105] ]; // Klaus Brockhaus, Oct 30 2010
    
  • Maple
    A109925 := proc(n)
        a := 0 ;
        for k from 0 do
            if n-2^k < 2 then
                return a ;
            elif isprime(n-2^k) then
                a := a+1 ;
            end if;
        end do:
    end proc:
    seq(A109925(n),n=1..80) ; # R. J. Mathar, Mar 07 2022
  • Mathematica
    Table[cnt=0; r=1; While[rRobert G. Wilson v, Jul 21 2005 *)
    Table[Count[n - 2^Range[0, Floor[Log2[n]]], ?PrimeQ], {n, 110}] (* _Harvey P. Dale, Oct 21 2024 *)
  • PARI
    a(n)=sum(k=0,log(n)\log(2),isprime(n-2^k)) \\ Charles R Greathouse IV, Feb 19 2013
    
  • Python
    from sympy import isprime
    def A109925(n): return sum(1 for i in range(n.bit_length()) if isprime(n-(1<Chai Wah Wu, Nov 29 2023

Formula

a(A118954(n))=0, a(A118955(n))>0; A118952(n)<=a(n); A078687(n)=a(A000040(n)). - Reinhard Zumkeller, May 07 2006
G.f.: ( Sum_{i>=0} x^(2^i) ) * ( Sum_{j>=1} x^prime(j) ). - Ilya Gutkovskiy, Feb 10 2022

Extensions

Corrected and extended by T. D. Noe and Robert G. Wilson v, Jul 19 2005

A303821 Number of ways to write 2*n as p + 2^x + 5^y, where p is a prime, and x and y are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 3, 3, 4, 4, 5, 3, 6, 5, 5, 6, 6, 4, 7, 6, 7, 7, 10, 4, 9, 10, 6, 10, 8, 5, 8, 6, 7, 7, 9, 5, 8, 11, 6, 10, 11, 6, 11, 8, 6, 8, 11, 4, 9, 9, 7, 6, 11, 6, 7, 11, 7, 10, 11, 5, 11, 9, 6, 7, 6, 6, 5, 12, 7, 10, 15, 8, 15, 10, 11, 13, 11, 7, 9, 8, 9, 12, 14
Offset: 1

Views

Author

Zhi-Wei Sun, May 01 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. Moreover, for any integer n > 4, we can write 2*n as p + 2^x + 5^y, where p is an odd prime, and x and y are positive integers.
This has been verified for n up to 10^10.
See also A303934 and A304081 for further refinements, and A303932 and A304034 for similar conjectures.

Examples

			a(2) = 1 since 2*2 = 2 + 2^0 + 5^0 with 2 prime.
a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 prime.
a(5616) = 2 since 2*5616 = 9059 + 2^11 + 5^3 = 10979 + 2^7 + 5^3 with 9059 and 10979 both prime.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A303934 Number of ways to write 2*n as p + 2^k + 5^m with p prime and 2^k + 5^m squarefree, where k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 3, 3, 2, 2, 3, 3, 4, 3, 5, 4, 4, 3, 4, 5, 7, 4, 7, 4, 8, 7, 6, 7, 6, 5, 5, 5, 7, 5, 8, 5, 5, 8, 6, 9, 9, 6, 8, 6, 6, 7, 8, 4, 7, 8, 7, 3, 10, 6, 7, 8, 7, 7, 9, 5, 8, 7, 6, 5, 5, 6, 3, 11, 7, 9, 12, 8, 12, 10, 11, 11, 9, 7, 9, 7, 8, 8, 11, 7, 11, 8, 9, 15, 11, 8, 9, 8, 9
Offset: 1

Views

Author

Zhi-Wei Sun, May 03 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This has been verified for all n = 2..10^10.
Note that a(n) <= A303821(n).

Examples

			a(2) = 1 since 2*2 = 2 + 2^0 + 5^0 with 2 prime and 2^0 + 5^0 squarefree.
a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 prime and 2^1 + 5^0 squarefree.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+5^m]&&PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A081311 Numbers that can be written as sum of a prime and an 3-smooth number.

Original entry on oeis.org

3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75
Offset: 1

Views

Author

Reinhard Zumkeller, Mar 17 2003

Keywords

Comments

A081308(a(n))>0; complement of A081310.
Up to 10^n this sequence has 8, 95, 916, 8871, 86974, 858055, 8494293, 84319349, 838308086, ... terms. The lower density is of this sequence is greater than 0.59368 (see Pintz), but seems to be less than 1; can this be proved? Charles R Greathouse IV, Sep 01 2015

Crossrefs

A118955 is a subsequence.
Union of A081312 and A081313.

Programs

  • Haskell
    a081310 n = a081310_list !! (n-1)
    a081310_list = filter ((== 0) . a081308) [1..]
    -- Reinhard Zumkeller, Jul 04 2012
    
  • Mathematica
    nmax = 1000;
    S = Select[Range[nmax], Max[FactorInteger[#][[All, 1]]] <= 3 &];
    A081308[n_] := Count[TakeWhile[S, # < n &], s_ /; PrimeQ[n - s]];
    Select[Range[nmax], A081308[#] > 0 &] (* Jean-François Alcover, Oct 13 2021 *)
  • PARI
    is(n)=for(i=0, logint(n,3), my(k=3^i); while(kCharles R Greathouse IV, Sep 01 2015

A118954 Numbers that cannot be written as 2^k + prime.

Original entry on oeis.org

1, 2, 16, 22, 26, 28, 36, 40, 46, 50, 52, 56, 58, 64, 70, 76, 78, 82, 86, 88, 92, 94, 96, 100, 106, 112, 116, 118, 120, 122, 124, 126, 127, 134, 136, 142, 144, 146, 148, 149, 154, 156, 160, 162, 166, 170, 172, 176, 178, 184, 186, 188, 190, 196, 202, 204, 206, 208
Offset: 1

Views

Author

Reinhard Zumkeller, May 07 2006

Keywords

Comments

A109925(a(n)) = 0.

Crossrefs

Complement of A118955. Subsequence of A118956. Supersequence of A006285.

Programs

  • Haskell
    a118954 n = a118954_list !! (n-1)
    a118954_list = filter f [1..] where
       f x = all (== 0) $ map (a010051 . (x -)) $ takeWhile (< x) a000079_list
    -- Reinhard Zumkeller, Jan 03 2014
    
  • Magma
    lst:=[]; for n in [1..208] do k:=-1; repeat k+:=1; a:=n-2^k; until a lt 1 or IsPrime(a); if a lt 1 then Append(~lst, n); end if; end for; lst; // Arkadiusz Wesolowski, Sep 02 2016
  • PARI
    is(n)=my(k=1);while(kCharles R Greathouse IV, Sep 01 2015
    

Formula

n < a(n) < kn for some k < 2 and all large enough n, see Romanoff and either Erdős or van der Corput. - Charles R Greathouse IV, Sep 01 2015

A304031 Number of ways to write 2*n as p + 2^k + 5^m with p prime and 2^k + 5^m a product of at most three distinct primes, where k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 3, 3, 2, 2, 3, 3, 4, 3, 5, 4, 4, 3, 4, 5, 7, 4, 7, 4, 8, 7, 6, 7, 6, 5, 5, 5, 7, 5, 8, 5, 5, 8, 6, 9, 9, 6, 8, 6, 6, 7, 8, 4, 7, 8, 7, 3, 10, 6, 7, 8, 7, 7, 9, 5, 8, 7, 6, 5, 5, 6, 3, 11, 7, 9, 12, 8, 12, 10, 11, 11, 9, 7, 9, 7, 8, 8, 11, 7, 11, 8, 9, 15, 11, 8, 9, 8, 9
Offset: 1

Views

Author

Zhi-Wei Sun, May 04 2018

Keywords

Comments

a(n) > 0 for all 1 < n <= 10^10 with the only exception n = 3114603841, and 2*3114603841 = 6219442049 + 2^3 + 5^10 with 6219442049 prime and 2^3 + 5^10 = 3*17*419*457 squarefree.
Note that a(n) <= A303934(n) <= A303821(n).

Examples

			a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 = 2^1 + 5^0  prime.
a(7) = 2 since 2*7 = 7 + 2^1 + 5^1 with 7 = 2^1 + 5^1 prime, and 2*7 = 11 + 2^1 + 5^0 with 11 and 2^1 + 5^0 both prime.
a(42908) = 2 since 2*42908 = 85751 + 2^6 + 5^0 with 85751 prime and 2^6 + 5^0 = 5*13, and 2*42908 = 69431 + 2^14 + 5^0 with 69431 prime and 2^14 + 5^0 = 5*29*113.
		

Crossrefs

Programs

  • Mathematica
    qq[n_]:=qq[n]=SquareFreeQ[n]&&Length[FactorInteger[n]]<=3;
    tab={};Do[r=0;Do[If[qq[2^k+5^m]&&PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A118957 Numbers of the form 2^k + p, where p is a prime less than 2^k.

Original entry on oeis.org

6, 7, 10, 11, 13, 15, 18, 19, 21, 23, 27, 29, 34, 35, 37, 39, 43, 45, 49, 51, 55, 61, 63, 66, 67, 69, 71, 75, 77, 81, 83, 87, 93, 95, 101, 105, 107, 111, 117, 123, 125, 130, 131, 133, 135, 139, 141, 145, 147, 151, 157, 159, 165, 169, 171, 175, 181, 187, 189, 195, 199
Offset: 1

Views

Author

Reinhard Zumkeller, May 07 2006

Keywords

Crossrefs

Complement of A118956; subsequence of A118955.

Programs

  • Maple
    isA118957 := proc(n)
        local twok,p ;
        twok := 1 ;
        while twok < n-1 do
            p := n-twok ;
            if isprime(p) and p < twok then
                return true;
            end if;
            twok := twok*2 ;
        end do:
        return false;
    end proc:
    for n from 1 to 200 do
        if isA118957(n) then
            printf("%d,",n) ;
        end if;
    end do: # R. J. Mathar, Feb 27 2015
  • Mathematica
    okQ[n_] := Module[{k, p}, For[k = Ceiling[Log[2, n]], k>1, k--, p = n-2^k; If[2 <= p < 2^k && PrimeQ[p], Return[True]]]; False]; Select[Range[200], okQ] (* Jean-François Alcover, Mar 11 2019 *)
  • PARI
    is(n)=isprime(n-2^logint(n,2)) \\ Charles R Greathouse IV, Sep 01 2015; edited Jan 24 2024
    
  • Python
    from sympy import primepi
    def A118957(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum(primepi(min(x-(m:=1<Chai Wah Wu, Feb 23 2025

Formula

A118952(a(n)) = 1.
Showing 1-10 of 18 results. Next