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 1-5 of 5 results.

A082769 a(n) = smallest palindromic prime that begins with A082768(n), or 0 if no such number exists.

Original entry on oeis.org

11, 3, 7, 919, 101, 11, 12421, 131, 14341, 151, 16061, 17471, 181, 191, 30103, 313, 32323, 33533, 34543, 353, 36263, 373, 383, 39293, 70207, 71317, 727, 73037, 74047, 757, 76367, 77377, 787, 797, 90709, 919, 929, 93139, 94049, 95959, 96269, 97379, 98389, 9902099, 1003001, 101, 1022201, 10301, 1043401, 10501, 10601, 1074701, 1082801, 1092901
Offset: 1

Views

Author

Amarnath Murthy, Apr 18 2003

Keywords

Comments

Conjecture: no entry is zero.

Crossrefs

Programs

  • Maple
    ispali := proc(n,b)
        local dgs,i ;
        dgs := convert(n,base,b) ;
        for i from 1 to nops(dgs)/2 do
            if op(i,dgs) <> op(-i,dgs) then
                return false;
            end if;
        end do:
        true;
    end proc:
    L082768 := [seq(A082768(n),n=1..200)] ; # use code in A082768
    L082769 := [seq(0,n=1..200)] ;
    for pi from 2 do
        p :=ithprime(pi) ;
        if ispali(p,10) then
            pdgs := convert(p,base,10) ;
            for sh from 0 do
                restp := add(op(i,pdgs)*10^(i-1),i=1..nops(pdgs)) ;
                for i from 1 to nops(L082768) do
                    if op(i,L082768) = restp then
                        if op(i,L082769) = 0 then
                            L082769 := subsop(i=p,L082769) ;
                            print(L082769) ;
                        end if;
                    end if;
                end do:
                # chop digits from palindromic prime starting at least signif
                pdgs := subsop(1=NULL,pdgs) ;
                if nops(pdgs) = 0 then
                    break ;
                end if;
            end do:
        end if;
    end do: # R. J. Mathar, Aug 27 2025

Extensions

More terms from David Wasserman, Jul 28 2005
12 more terms from R. J. Mathar, Aug 27 2025

A082770 a(n) = smallest palindromic prime that begins with A082768(n) and contains more than twice the number of digits in A082768(n), or 0 if no such number exists.

Original entry on oeis.org

101, 313, 727, 919, 10301, 11311, 12421, 13331, 14341, 15451, 16061, 17471, 18181, 19391, 30103, 31013, 32323, 33533, 34543, 35053, 36263, 37273, 38083, 39293, 70207, 71317, 72227, 73037, 74047, 75557, 76367, 77377, 78487, 79397, 90709
Offset: 1

Views

Author

Amarnath Murthy, Apr 18 2003

Keywords

Comments

Conjecture: no entry is zero. In most cases the number of digits required is 2k+1 where k is the number of digits in A082768(n). What is the first entry that requires more (than 2k+1) digits?
Answer to question: A082768(37) = 92, a(37) = 9200029. - David Wasserman, Jul 28 2005

Crossrefs

Extensions

More terms from David Wasserman, Jul 28 2005

A045572 Numbers that are odd but not divisible by 5.

Original entry on oeis.org

1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99, 101, 103, 107, 109, 111, 113, 117, 119, 121, 123, 127, 129, 131, 133, 137, 139, 141, 143, 147, 149, 151, 153
Offset: 1

Views

Author

Jeff Burch, Dec 11 1999

Keywords

Comments

Contains the repunits R_n, (A000042 or A002275): For any m in the sequence (divisible by neither 2 nor 5), Euler's theorem (i.e., m | 10^m - 1 = 9*R_n) guarantees that R_n is always some multiple of m (see A099679) and thus forms a subsequence. - Lekraj Beedassy, Oct 26 2004
Inverse formula: n = 4*floor(a(n)/10) + floor((a(n) mod 10)/3) + 1. - Carl R. White, Feb 06 2008
Numbers ending with 1, 3, 7 or 9. - Lekraj Beedassy, Apr 04 2009
Complement of A065502. - Reinhard Zumkeller, Nov 15 2009
Union of evenish and oddish numbers, cf. A045797, A045798. - Reinhard Zumkeller, Dec 10 2011
Numbers k such that k^(4*j) mod 10 = 1, for any j. - Gary Detlefs, Jan 03 2012
Numbers coprime to 10. - Charles R Greathouse IV, Sep 05 2013
This is also the sequence of numbers such that all their divisors are the sum of the proper divisors of some number (see A001065 (sum of proper divisors) and A078923 (possible values of sigma(n)-n)). This is due to the fact that in the set of untouchable numbers (A005114) there are only 2 prime numbers (2 and 5) and all other terms are even composite. - Michel Marcus, Jun 14 2014
Numbers n for which A001589(n) is divisible by 5. - Bruno Berselli, Jun 18 2014
For a(n) > 1, positive integers x such that the decimal representation of 1/x is purely periodic after the decimal point (1/x is a repeating decimal with no non-repeating portion). - Doug Bell, Aug 05 2015
The asymptotic density of this sequence is 2/5. - Amiram Eldar, Oct 18 2020

Examples

			a(18) = 10*floor(17/4) + 2*floor( (4*(17 mod 4) + 1)/3 ) + 1
      = 10*4 + 2*floor( (4*(1)+1)/3 ) + 1
      = 40 + 2*floor(5/3) + 1
      = 40 + 2*1 + 1
      = 43.
G.f. = x + 3*x^2 + 7*x^3 + 9*x^4 + 11*x^5 + 13*x^6 + 17*x^7 + 19*x^8 + ...
		

Crossrefs

Relative complement of A017329 in A005408.
Cf. A000035, A000042, A001065, A001589, A002275, A005114, A045797, A045798, A065502, A078923, A079998, A082768 (numbers that begin with 1, 3, 7 or 9), A085820, A099679.

Programs

Formula

a(n) = 10*floor((n-1)/4) + 2*floor( (4*((n-1) mod 4) + 1)/3 ) + 1; a(n) = a(n-1) + 2 + 2*floor(((x+6) mod 10)/9). - Carl R. White, Feb 06 2008
a(n) = 2*n + 2*floor((n-3)/4) + 1. - Kenneth Hammond (weregoose(AT)gmail.com), Mar 07 2008
a(n) = -1 + 2*n + 2*floor((n+1)/4). - Kenneth Hammond (weregoose(AT)gmail.com), Mar 25 2008
From R. J. Mathar, Sep 22 2009: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5).
G.f.: x*(1 + 2*x + 4*x^2 + 2*x^3 + x^4)/((1+x) * (x^2+1) * (x-1)^2). (End)
A000035(a(n))*(1 - A079998(a(n))) = 1. - Reinhard Zumkeller, Nov 15 2009
a(n) = (10*n + 2*(-1)^(n*(n+1)/2) - (-1)^n - 5)/4. - Bruno Berselli, Nov 06 2011
G.f.: x * (1 + 2*x + 4*x^2 + 2*x^3 + x^4) / ((1 - x) * (1 - x^4)). - Michael Somos, Jun 15 2014
a(1 - n) = -a(n) for all n in Z. - Michael Somos, Jun 15 2014
0 = (a(n) - 2*a(n+1) + a(n+2)) * (a(n) - 4*a(n+2) + 3*a(n+3)) for all n in Z. - Michael Somos, Jun 15 2014
From Mikk Heidemaa, Nov 22 2017: (Start)
a(n) = (1/2)*(5*n + ((3*n + 2) mod 4) - 4);
a(n) = (1/4)*((-1)^(n + 1) + 10*n + 2*cos((n*Pi)/2) - 2*sin((n*Pi)/2) - 5);
a(n) = (1/4)*((-1)^(1 + n) + (1 - i)*exp(-(1/2)*i*n*Pi) + (1 + i)*exp(i*n*Pi/2) + 10*n - 5) (for n > 0), where i is the imaginary unit. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(10-2*sqrt(5))*Pi/10. - Amiram Eldar, Dec 12 2021
E.g.f.: (2 + cos(x) + (5*x - 3)*cosh(x) - sin(x) + (5*x - 2)*sinh(x))/2. - Stefano Spezia, Dec 07 2022

A089743 Numbers that are the leading digits of a palindromic prime.

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118
Offset: 1

Views

Author

Amarnath Murthy, Nov 19 2003

Keywords

Comments

Conjecture: every number beginning with 1, 3, 7 or 9 begins infinitely many palindromic primes.

Crossrefs

Extensions

Edited and extended by David Wasserman, Jul 28 2005

A054658 Primes beginning 1, 3, 7, 9 whose reversals are nonprimes.

Original entry on oeis.org

19, 103, 109, 127, 137, 139, 163, 173, 193, 197, 307, 317, 331, 349, 367, 379, 397, 719, 773, 911, 947, 977, 997, 1013, 1019, 1039, 1049, 1051, 1063, 1087, 1093, 1117, 1123, 1129, 1163, 1171, 1187, 1277, 1289, 1291, 1297, 1303, 1307, 1319, 1327, 1361
Offset: 1

Views

Author

Enoch Haga, Apr 18 2000

Keywords

Comments

Or, primes whose reversals are composites ending in 1,3,7,9. - Lekraj Beedassy, Aug 02 2008
A subsequence of A143260. - Lekraj Beedassy, Aug 02 2008

Examples

			a(1)=19 because its reverse is a nonprime, 91.
		

Crossrefs

Programs

  • Mathematica
    pbQ[p_]:=MemberQ[{1,3,7,9},IntegerDigits[p][[1]]]&&CompositeQ[IntegerReverse[p]]; Select[Prime[Range[300]],pbQ] (* Harvey P. Dale, Dec 02 2024 *)

Extensions

Edited by N. J. A. Sloane, Aug 29 2008 at the suggestion of R. J. Mathar
Showing 1-5 of 5 results.