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.

Previous Showing 11-15 of 15 results.

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.

Links

Crossrefs

Cf. A048268, A060792, A097856, A097928, A182232, A259374, A097929, A182233, A259375, A259376, A097930, A182234, A259377, A259378, A249156, A097931, A259380, A259381, A259382, A259383, A259384, A099145, A259385, A259386, A259387, A259388, A259389, A259390, A099146, A007632, A007633, A029961, A029962, A029963, A029964, A029804, A029965, A029966, A029967, A029968, A029969, A029970, A029731, A097855, A250408, A250409, A250410, A250411, A099165, A250412.

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.

Links

Crossrefs

Cf. A048268, A060792, A097856, A097928, A182232, A259374, A097929, A182233, A259375, A259376,
A097930, A182234, A259377, A259378, A249156, A097931, A259380, A259381, A259382, A259383,
A259384, A099145, A259385, A259386, A259387, A259388, A259389, A259390, A099146, A007632,
A007633, A029961, A029962, A029963, A029964, A029804, A029965, A029966, A029967, A029968,
A029969, A029970, A029731, A097855, A250408, A250409, A250410, A250411, A099165, A250412.

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.

A259388 Palindromic numbers in bases 5 and 9 written in base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 109, 246, 282, 564, 701, 22386, 32152, 41667, 47653, 48553, 1142597, 1313858, 1412768, 1677684, 12607012902, 19671459008, 20134447808, 24208576998, 24863844904, 26358878059
Offset: 1

Views

Author

Robert G. Wilson v, Jul 16 2015

Keywords

Examples

			246 is in the sequence because 246_10 = 303_9 = 1441_5.

Links

Crossrefs

Cf. A007632, A007633, A029731, A029804, A029961, A029962, A029963, A029964, A029965, A029966, A029967, A029968, A029969, A029970, A048268, A060792, A097855, A097856, A097928, A097929, A097930, A097931, A099145, A099146, A099165, A182232, A182233, A182234, A250408, A250409, A250410, A250411, A250412, A259374, A259375, A259376, A259377, A259378, A249156, A259380, A259381, A259382, A259383, A259384, A259385, A259386, A259387, A259388, A259389, A259390.

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, 5], AppendTo[lst, pp]; Print[pp]]; k++]; lst
b1=5; 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 A029952 and A029955.

A259389 Palindromic numbers in bases 6 and 9 written in base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 80, 154, 191, 209, 910, 3740, 5740, 8281, 16562, 16814, 2295481, 2300665, 2350165, 2439445, 2488945, 2494129, 2515513, 7971580, 48307924, 61281793, 69432517, 123427622, 124091822, 124443290, 55854298990, 184314116750, 185794441250, 187195815770, 327925630018, 7264479038060, 27832011695551
Offset: 1

Views

Author

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

Keywords

Examples

			209 is in the sequence because 209_10 = 252_9 = 545_6.

Links

Crossrefs

Cf. A048268, A060792, A097856, A097928, A182232, A259374, A097929, A182233, A259375, A259376, A097930, A182234, A259377, A259378, A249156, A097931, A259380, A259381, A259382, A259383, A259384, A099145, A259385, A259386, A259387, A259388, A259389, A259390, A099146, A007632, A007633, A029961, A029962, A029963, A029964, A029804, A029965, A029966, A029967, A029968, A029969, A029970, A029731, A097855, A250408, A250409, A250410, A250411, A099165, A250412.

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, 6], AppendTo[lst, pp]; Print[pp]]; k++]; lst
b1=6; b2=9; lst={}; Do[d1=IntegerDigits[n, b1]; d2=IntegerDigits[n, b2]; If[d1==Reverse[d1]&&d2==Reverse[d2], AppendTo[lst, n]], {n, 0, 1000000}]; lst (* Vincenzo Librandi, Jul 17 2015 *)

Formula

Intersection of A029953 and A029955.

A308832 Numbers that are palindromic in bases 3, 9 and 27.

Original entry on oeis.org

0, 1, 2, 4, 8, 10, 20, 364, 728, 730, 1460, 2920, 5840, 7300, 7381, 7462, 14600, 14681, 14762, 265720, 531440, 531442, 532171, 532900, 1062884, 1063613, 1064342, 2125768, 2128684, 2131600, 4251536, 4254452, 4257368, 5314420, 5321710, 5329000, 5373550, 5380840, 5388130, 5432680, 5439970, 5447260
Offset: 1

Views

Author

A.H.M. Smeets, Jun 27 2019

Keywords

Comments

This sequence is infinite because it contains terms of the forms 729^k-1 (k>=0) and 729^k+1 (k>=0).
Let L_27 be the number of digits for a(n) represented in base 27.
If L_27 is odd, the digits are either in the set {0, 1, 2} or in {0, 4, 8} or in {0, 10_10, 20_10}.
If L_27 is even, the digits are either in the set {0, 13_10} or in {0, 26_10} or in {13_10, 26_10}.
Let L_3 be the length of a(n) if represented in base 3, then L_3 in {0,1,2,3} (mod 6).

Links

Crossrefs

Subsequence of A259386.
Cf. A319584 (palindromic in bases 2, 4 and 8).

Programs

  • Mathematica
    palQ[n_,b_] := PalindromeQ[IntegerDigits[n, b]]; aQ[n_] := AllTrue[{3, 9, 27}, palQ[n, #] &]; Select[Range[0, 6*10^6], aQ] (* Amiram Eldar, Jul 04 2019 *)
  • PARI
    nextpal(n, b) = {my(m=n+1, p = 0); while (m > 0, m = m\b; p++;); if (n+1 == b^p, p++); n = n\(b^(p\2))+1; m = n; n = n\(b^(p%2)); while (n > 0, m = m*b + n%b; n = n\b;); m;} \\ after Python
ispal(n, b) = my(d=digits(n, b)); Vecrev(d) == d;
lista(nn) = {my(k=0); while (k <= nn, if (ispal(k, 3) && ispal(k, 9), print1(k, ", ");); k = nextpal(k, 27););} \\ Michel Marcus, Jul 04 2019
  • Python
    def nextpal(n,base): # m is the first palindrome successor of n in base base
    m, pl = n+1, 0
    while m > 0:
        m, pl = m//base, pl+1
    if n+1 == base**pl:
        pl = pl+1
    n = n//(base**(pl//2))+1
    m, n = n, n//(base**(pl%2))
    while n > 0:
        m, n = m*base+n%base, n//base
    return m
def rev(n,b):
    m = 0
    while n > 0:
        n, m = n//b, m*b+n%b
    return m
n, a = 1, 0
while n <= 20000:
    if a == rev(a,9) and a == rev(a,3):
        print(n,a)
        n = n+1
    a = nextpal(a,27)
Previous Showing 11-15 of 15 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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