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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A259390 Palindromic numbers in bases 7 and 9 written in base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 40, 50, 100, 164, 200, 264, 300, 328, 400, 2000, 3550, 8200, 10252, 14510, 14762, 22800, 45600, 164900, 201720, 400200, 532900, 555013, 738100, 2756120, 2913368, 3344352, 3501600, 4084000, 12990350, 22674550, 194062432, 1684866370, 2225211080, 13575144288, 15127811455, 20404027400, 20537111057, 22668403353, 30862471355, 83714515310, 84668107250, 796259955485, 1202029647736, 2088800185930, 20268849562000
Offset: 1

Views

Author

Eric A. Schmidt and Robert G. Wilson v, Jul 17 2015

Keywords

Examples

			264 is in the sequence because 264_10 = 323_9 = 525_7.
		

Crossrefs

Programs

  • Mathematica
    (* first load nthPalindromeBase from A002113 *) palQ[n_Integer, base_Integer] := Block[{}, Reverse[ idn = IntegerDigits[n, base]] == idn]; k = 0; lst = {}; While[k < 21000000, pp = nthPalindromeBase[k, 9]; If[palQ[pp, 7], AppendTo[lst, pp]; Print[pp]]; k++]; lst

Formula

Intersection of A029954 and A029955.

A332120 a(n) = 2*(10^(2n+1)-1)/9 - 2*10^n.

Original entry on oeis.org

0, 202, 22022, 2220222, 222202222, 22222022222, 2222220222222, 222222202222222, 22222222022222222, 2222222220222222222, 222222222202222222222, 22222222222022222222222, 2222222222220222222222222, 222222222222202222222222222, 22222222222222022222222222222, 2222222222222220222222222222222
Offset: 0

Views

Author

M. F. Hasler, Feb 09 2020

Keywords

Crossrefs

Cf. A002275 (repunits R_n = (10^n-1)/9), A002276 (2*R_n), A011557 (10^n).
Cf. A138148 (cyclops numbers with binary digits), A002113 (palindromes).
Cf. A332130 .. A332190 (variants with different repeated digit 3, ..., 9).
Cf. A332121 .. A332129 (variants with different middle digit 1, ..., 9).

Programs

  • Maple
    A332120 := n -> 2*((10^(2*n+1)-1)/9-10^n);
  • Mathematica
    Array[2 ((10^(2 # + 1)-1)/9 - 10^#) &, 15, 0]
  • PARI
    apply( {A332120(n)=(10^(n*2+1)\9-10^n)*2}, [0..15])
    
  • Python
    def A332120(n): return (10**(n*2+1)//9-10**n)*2

Formula

a(n) = 2*A138148(n) = A002276(2n+1) - 2*10^n.
G.f.: 2*x*(101 - 200*x)/((1 - x)(1 - 10*x)(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.
E.g.f.: 2*exp(x)*(10*exp(99*x) - 9*exp(9*x) - 1)/9. - Stefano Spezia, Jul 13 2024

A332190 a(n) = 10^(2n+1) - 1 - 9*10^n.

Original entry on oeis.org

0, 909, 99099, 9990999, 999909999, 99999099999, 9999990999999, 999999909999999, 99999999099999999, 9999999990999999999, 999999999909999999999, 99999999999099999999999, 9999999999990999999999999, 999999999999909999999999999, 99999999999999099999999999999, 9999999999999990999999999999999
Offset: 0

Views

Author

M. F. Hasler, Feb 08 2020

Keywords

Crossrefs

Cf. A002275 (repunits R_n = (10^n-1)/9), A002283 (9*R_n), A011557 (10^n).
Cf. A138148 (cyclops numbers with binary digits only), A002113 (palindromes).
Cf. A332120 .. A332180 (variants with different repeated digit 2, ..., 8).
Cf. A332191 .. A332197, A181965 (variants with different middle digit 1, ..., 8).

Programs

  • Maple
    A332190 := n -> 10^(2*n+1)-1-9*10^n;
  • Mathematica
    Array[10^(2 # + 1)-1-9*10^# &, 15, 0]
    LinearRecurrence[{111,-1110,1000},{0,909,99099},20] (* Harvey P. Dale, May 28 2021 *)
  • PARI
    apply( {A332190(n)=10^(n*2+1)-1-9*10^n}, [0..15])
    
  • Python
    def A332190(n): return 10**(n*2+1)-1-9*10^n

Formula

a(n) = 9*A138148(n) = A002283(2n+1) - A011557(n).
G.f.: 9*x*(101 - 200*x)/((1 - x)(1 - 10*x)(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.

A050250 Number of nonzero palindromes less than 10^n.

Original entry on oeis.org

9, 18, 108, 198, 1098, 1998, 10998, 19998, 109998, 199998, 1099998, 1999998, 10999998, 19999998, 109999998, 199999998, 1099999998, 1999999998, 10999999998, 19999999998, 109999999998, 199999999998, 1099999999998, 1999999999998, 10999999999998
Offset: 1

Views

Author

Eric W. Weisstein, Dec 11 1999

Keywords

Crossrefs

Programs

  • Maple
    A050250List := proc(len);  local s, egf, ser; s:= 11/(2*sqrt(10));
    egf := -2*exp(x) + (1-s)*exp(-sqrt(10)*x) + (1+s)*exp(sqrt(10)*x);
    ser := series(egf, x, len+2): seq(simplify(n!*coeff(ser,x,n)), n = 1..len) end:
    A050250List(25); # Peter Luschny, Jun 11 2022 after Stefano Spezia
  • Mathematica
    LinearRecurrence[{1,10,-10},{9,18,108},30] (* Harvey P. Dale, Jan 29 2012 *)
    CoefficientList[Series[2Cosh[Sqrt[10]x]-2(Cosh[x]+Sinh[x])+11Sinh[Sqrt[10]x]/Sqrt[10],{x,0,25}],x]Table[n!,{n,0,25}] (* Stefano Spezia, Jun 11 2022 *)
  • PARI
    a(n)=10^(n\2)*(13-9*(-1)^n)/2-2 \\ Charles R Greathouse IV, Jun 25 2017
    
  • Python
    def a(n):
      m = 10 ** (n >> 1)
      if n & 1 == 0:
        return (m - 1) << 1
      else:
        return (11 * m) - 2 # DarĂ­o Clavijo, Oct 16 2023

Formula

a(2*k) = 2*10^k - 2, a(2*k + 1) = 11*10^k - 2. - Sascha Kurz, Apr 14 2002
From Jonathan Vos Post, Jun 18 2008: (Start)
a(n) = Sum_{i=1..n} A050683(i).
a(n) = Sum_{i=1..n} 9*10^floor((i-1)/2).
a(n) = 9*Sum_{i=1..n} 10^floor((i-1)/2). (End)
From Bruno Berselli, Feb 15 2011: (Start)
G.f.: 9*x*(1+x)/((1-x)*(1-10*x^2)).
a(n) = (1/2)*10^((2*n + (-1)^n - 1)/4)*(13 - 9*(-1)^n) - 2. (End)
a(1)=9, a(2)=18, a(3)=108; for n>3, a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3). - Harvey P. Dale, Jan 29 2012
a(n) = 10*a(n-2) + 18. - R. J. Mathar, Nov 07 2015
E.g.f.: 2*cosh(sqrt(10)*x) - 2*(cosh(x) + sinh(x)) + 11*sinh(sqrt(10)*x)/sqrt(10). - Stefano Spezia, Jun 11 2022

Extensions

More terms from Patrick De Geest, Dec 15 1999
a(24)-a(25) from Jonathan Vos Post, Jun 18 2008

A118594 Palindromes in base 3 (written in base 3).

Original entry on oeis.org

0, 1, 2, 11, 22, 101, 111, 121, 202, 212, 222, 1001, 1111, 1221, 2002, 2112, 2222, 10001, 10101, 10201, 11011, 11111, 11211, 12021, 12121, 12221, 20002, 20102, 20202, 21012, 21112, 21212, 22022, 22122, 22222, 100001, 101101, 102201, 110011, 111111, 112211, 120021
Offset: 1

Views

Author

Martin Renner, May 08 2006

Keywords

Comments

The number of n-digit terms is given by A225367. - M. F. Hasler, May 05 2013 [Moved here on May 08 2013]
Digit-wise application of A000578 (and also superposition of a(n) with its horizontal OR vertical reflection) yields A006072. - M. F. Hasler, May 08 2013
Equivalently, palindromes k (written in base 10) such that 4*k is a palindrome. - Bruno Berselli, Sep 12 2018

Crossrefs

Programs

  • Mathematica
    (* get NextPalindrome from A029965 *) Select[NestList[NextPalindrome, 0, 1110], Max@IntegerDigits@# < 3 &] (* Robert G. Wilson v, May 09 2006 *)
    Select[FromDigits/@Tuples[{0,1,2},8],IntegerDigits[#]==Reverse[ IntegerDigits[ #]]&] (* Harvey P. Dale, Apr 20 2015 *)
  • PARI
    {for(l=1,5,u=vector((l+1)\2,i,10^(i-1)+(2*i-11&&i==1,2]), print1(v*u",")))} \\ The n-th term could be produced by using (partial sums of) A225367 to skip all shorter terms, and then skipping the adequate number of vectors v until n is reached.  - M. F. Hasler, May 08 2013
    
  • Python
    from itertools import count, islice, product
    def agen(): # generator of terms
        yield from [0, 1, 2]
        for d in count(2):
            for start in "12":
                for rest in product("012", repeat=d//2-1):
                    left = start + "".join(rest)
                    for mid in [[""], ["0", "1", "2"]][d%2]:
                        yield int(left + mid + left[::-1])
    print(list(islice(agen(), 42))) # Michael S. Branicky, Mar 29 2022
    
  • Python
    from sympy import integer_log
    from gmpy2 import digits
    def A118594(n):
        if n == 1: return 0
        y = 3*(x:=3**integer_log(n>>1,3)[0])
        return int((s:=digits(n-x,3))+s[-2::-1] if nChai Wah Wu, Jun 14 2024
  • Sage
    [int(n.str(base=3)) for n in (0..757) if Word(n.digits(3)).is_palindrome()] # Peter Luschny, Sep 13 2018
    

Extensions

More terms from Robert G. Wilson v, May 09 2006
a(40) and beyond from Michael S. Branicky, Mar 29 2022

A134810 Giza numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 121, 232, 343, 454, 565, 676, 787, 898, 12321, 23432, 34543, 45654, 56765, 67876, 78987, 1234321, 2345432, 3456543, 4567654, 5678765, 6789876, 123454321, 234565432, 345676543, 456787654, 567898765, 12345654321, 23456765432
Offset: 1

Views

Author

Omar E. Pol, Nov 25 2007, Nov 26 2007

Keywords

Comments

For n > 9 the structure of digits represents the pyramids of Giza. Also the top of a mountain. The first digit is equal to the last digit. The first digits are in consecutive increasing order. The last digits are in consecutive decreasing order. The largest digit is the central digit. The number of digits is odd. This sequence has 45 terms. The final term is 12345678987654321. Giza numbers are mountain numbers A134941 and palindromes A002113.
There are 10 - k numbers with 2*k - 1 digits. - Omar E. Pol, Aug 04 2011

Examples

			Illustration using the final term of this sequence:
  . . . . . . . . 9 . . . . . . . .
  . . . . . . . 8 . 8 . . . . . . .
  . . . . . . 7 . . . 7 . . . . . .
  . . . . . 6 . . . . . 6 . . . . .
  . . . . 5 . . . . . . . 5 . . . .
  . . . 4 . . . . . . . . . 4 . . .
  . . 3 . . . . . . . . . . . 3 . .
  . 2 . . . . . . . . . . . . . 2 .
  1 . . . . . . . . . . . . . . . 1
		

Crossrefs

Programs

  • Mathematica
    ups = Flatten[Table[Range[i, j - 1], {i, 1, 9}, {j, i + 1, 10}], 1];afull = Sort[  Map[ToExpression@StringJoin@Map[ToString, #[[;; -2]] ~Join~ Reverse[#]] &, ups]];afull (* James C. McMahon, Apr 11 2025 *)
  • Python
    ups = [tuple(range(i, j)) for i in range(1, 10) for j in range(i+1, 11)]
    afull = sorted(int("".join(map(str, u[:-1] + u[::-1]))) for u in ups)
    print(afull) # Michael S. Branicky, Aug 02 2022

Formula

A178333(a(n))*A136522(a(n)) = 1. - Reinhard Zumkeller, May 25 2010

A181965 a(n) = 10^(2n+1) - 10^n - 1.

Original entry on oeis.org

8, 989, 99899, 9998999, 999989999, 99999899999, 9999998999999, 999999989999999, 99999999899999999, 9999999998999999999, 999999999989999999999, 99999999999899999999999, 9999999999998999999999999, 999999999999989999999999999, 99999999999999899999999999999, 9999999999999998999999999999999
Offset: 0

Views

Author

Ivan Panchenko, Apr 04 2012

Keywords

Comments

n 9's followed by an 8 followed by n 9's.
See A183187 = {26, 378, 1246, 1798, 2917, ...} for the indices of primes.

Crossrefs

Cf. (A077794-1)/2 = A183187 (indices of primes).
Cf. A002275 (repunits R_n = (10^n-1)/9), A002283 (9*R_n), A011557 (10^n).
Cf. A138148 (cyclops numbers with binary digits only), A002113 (palindromes).
Cf. A332190 .. A332197 (variants with different middle digit 0, ..., 7).

Programs

  • Maple
    A181965 := n -> 10^(2*n+1)-1-10^n; # M. F. Hasler, Feb 08 2020
  • Mathematica
    Array[10^(2 # + 1) - 1- 10^# &, 15, 0] (*  M. F. Hasler, Feb 08 2020 *)
    Table[With[{c=PadRight[{},n,9]},FromDigits[Join[c,{8},c]]],{n,0,20}] (* Harvey P. Dale, Jun 07 2021 *)
  • PARI
    apply( {A181965(n)=10^(n*2+1)-1-10^n}, [0..15]) \\ M. F. Hasler, Feb 08 2020
    
  • Python
    def A181965(n): return 10**(n*2+1)-1-10^n # M. F. Hasler, Feb 08 2020

Formula

From M. F. Hasler, Feb 08 2020: (Start)
a(n) = 9*A138148(n) + 8*10^n = A002283(2n+1) - A011557(10^n).
G.f.: (8 + 101*x - 1000*x^2)/((1 - x)(1 - 10*x)(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2. (End)

Extensions

Edited and extended to a(0) = 8 by M. F. Hasler, Feb 10 2020

A259381 Palindromic numbers in bases 3 and 8 written in base 10.

Original entry on oeis.org

0, 1, 2, 4, 121, 130, 203, 316, 8578, 9490, 17492, 944035, 1141652, 1276916, 1554173, 58961443, 67470916, 4099065139, 5691134677, 81452592329, 81473867465, 419572845958, 21056462595764, 363376288168081
Offset: 1

Views

Author

Eric A. Schmidt and Robert G. Wilson v, Jul 16 2015

Keywords

Examples

			121 is in the sequence because 121_10 = 171_8 = 11111_3.
		

Crossrefs

Programs

  • Mathematica
    (* first load nthPalindromeBase from A002113 *) palQ[n_Integer, base_Integer] := Block[{}, Reverse[ idn = IntegerDigits[n, base]] == idn]; k = 0; lst = {}; While[k < 21000000, pp = nthPalindromeBase[k, 8]; If[palQ[pp, 3], AppendTo[lst, pp]; Print[pp]]; k++]; lst
    b1=3; b2=8; lst={}; Do[d1=IntegerDigits[n, b1]; d2=IntegerDigits[n, b2]; If[d1==Reverse[d1]&&d2==Reverse[d2], AppendTo[lst, n]], {n, 0, 10000000}]; lst (* Vincenzo Librandi, Jul 17 2015 *)

Formula

Intersection of A014190 and A029803.

A259383 Palindromic numbers in bases 5 and 8 written in base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 18, 36, 186, 438, 2268, 2709, 11898, 18076, 151596, 228222, 563786, 5359842, 32285433, 257161401, 551366532, 621319212, 716064597, 2459962002, 5018349804, 5067084204, 7300948726, 42360367356, 139853034114, 176616961826, 469606524278, 669367713609, 1274936571666, 1284108810066, 5809320306961, 8866678870082, 11073162740322, 14952142559323, 325005646077513
Offset: 1

Views

Author

Eric A. Schmidt and Robert G. Wilson v, Jul 16 2015

Keywords

Examples

			186 is in the sequence because 186_10 = 272_8 = 1221_5.
		

Crossrefs

Programs

  • Mathematica
    (* first load nthPalindromeBase from A002113 *) palQ[n_Integer, base_Integer] := Block[{}, Reverse[ idn = IntegerDigits[n, base]] == idn]; k = 0; lst = {}; While[k < 21000000, pp = nthPalindromeBase[k, 8]; If[palQ[pp, 5], AppendTo[lst, pp]; Print[pp]]; k++]; lst
    b1=5; b2=8; lst={}; Do[d1=IntegerDigits[n, b1]; d2=IntegerDigits[n, b2]; If[d1==Reverse[d1]&&d2==Reverse[d2], AppendTo[lst, n]], {n, 0, 10000000}]; lst (* Vincenzo Librandi, Jul 17 2015 *)

Formula

Intersection of A029952 and A029803.

A259387 Palindromic numbers in bases 4 and 9 written in base 10.

Original entry on oeis.org

0, 1, 2, 3, 5, 10, 255, 273, 373, 546, 2550, 2730, 2910, 16319, 23205, 54215, 1181729, 1898445, 2576758, 3027758, 3080174, 4210945, 9971750, 163490790, 2299011170, 6852736153, 6899910553, 160142137430, 174913133450, 204283593150, 902465909895, 1014966912315, 2292918574418, 9295288254930, 11356994802010, 11372760382810, 38244097345762
Offset: 1

Views

Author

Eric A. Schmidt and Robert G. Wilson v, Jul 16 2015

Keywords

Examples

			273 is in the sequence because 273_10 = 333_9 = 10101_4.
		

Crossrefs

Programs

  • Mathematica
    (* first load nthPalindromeBase from A002113 *) palQ[n_Integer, base_Integer] := Block[{}, Reverse[ idn = IntegerDigits[n, base]] == idn]; k = 0; lst = {}; While[k < 21000000, pp = nthPalindromeBase[k, 9]; If[palQ[pp, 4], AppendTo[lst, pp]; Print[pp]]; k++]; lst
    b1=4; b2=9; lst={}; Do[d1=IntegerDigits[n, b1]; d2=IntegerDigits[n, b2]; If[d1==Reverse[d1]&&d2==Reverse[d2], AppendTo[lst, n]], {n, 0, 10000000}]; lst (* Vincenzo Librandi, Jul 17 2015 *)

Formula

Intersection of A014192 and A029955.
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