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 11-18 of 18 results.

A352507 Number whose representation in the base of Catalan numbers (A014418) is palindromic.

Original entry on oeis.org

0, 1, 3, 6, 8, 15, 22, 43, 48, 53, 59, 64, 69, 133, 152, 171, 177, 196, 215, 430, 444, 458, 477, 491, 505, 524, 538, 552, 564, 578, 592, 611, 625, 639, 658, 672, 686, 1431, 1487, 1543, 1568, 1624, 1680, 1705, 1761, 1817, 1862, 1918, 1974, 1999, 2055, 2111, 2136
Offset: 1

Views

Author

Amiram Eldar, Mar 19 2022

Keywords

Comments

The partial sums of the Catalan numbers with positive index (A014138) are terms, since the representation of A014138(n) is n 1's.

Examples

			The first 10 terms are:
   n  a(n)  A014418(a(n))
  --  ----  -------------
   1     0              0
   2     1              1
   3     3             11
   4     6            101
   5     8            111
   6    15           1001
   7    22           1111
   8    43          10001
   9    48          10101
  10    53          10201
		

Crossrefs

Subsequences: A014138, A141351 \ {2}.

Programs

  • Mathematica
    c[n_] := c[n] = CatalanNumber[n]; q[n_] := Module[{s = {}, m = n, i}, While[m > 0, i = 1; While[c[i] <= m, i++]; i--; m -= c[i]; AppendTo[s, i]]; PalindromeQ @ IntegerDigits[Total[4^(s - 1)], 4]]; Select[Range[0, 2000], q]

A364005 Numbers whose Wythoff representation (A189921, A317208) is palindromic.

Original entry on oeis.org

0, 1, 2, 5, 7, 10, 13, 15, 23, 28, 34, 36, 52, 57, 65, 75, 81, 89, 91, 117, 128, 146, 159, 175, 185, 198, 204, 217, 233, 235, 277, 295, 327, 369, 379, 400, 426, 442, 463, 473, 494, 520, 526, 547, 573, 589, 610, 612, 680, 709, 761, 829, 848, 916, 945, 989, 1023
Offset: 1

Views

Author

Amiram Eldar, Jul 01 2023

Keywords

Comments

Includes all the odd-indexed Fibonacci numbers (A001519), since the Wythoff representation of Fibonacci(1) is 1 and the Wythoff representation of Fibonacci(2*n+1), for n >= 1, is n 0's.
A157725(n) = Fibonacci(n) + 2 is a term for n >= 4, since its Wythoff representation is n-4 1's between 2 0's.
A232970 is a subsequence since the Wythoff representation of A232970(n) = (Fibonacci(3*n+1) + 1)/2 is n 0's and n-1 1's interleaved.

Examples

			The first 10 terms are:
   n  a(n)  A317208(a(n))
  --  ----  -------------
   1     0              0
   2     1              1
   3     2              2
   4     5             22
   5     7            212
   6    10           2112
   7    13            222
   8    15          21112
   9    23         211112
  10    28          21212
		

Crossrefs

Programs

  • Mathematica
    z[n_] := Floor[(n + 1)*GoldenRatio] - n - 1; h[n_] := z[n] - z[n - 1]; w[n_] := Module[{m = n, zm = 0, hm, s = {}}, While[zm != 1, hm = h[m]; AppendTo[s, hm]; If[hm == 1, zm = z[m], zm = z[z[m]]]; m = zm]; s]; w[0] = {0}; Select[Range[0, 1000], PalindromeQ[w[#]] &]

A364214 Numbers whose canonical representation as a sum of distinct Jacobsthal numbers (A280049) is palindromic.

Original entry on oeis.org

1, 2, 4, 5, 6, 10, 12, 15, 18, 21, 22, 30, 34, 42, 44, 49, 58, 63, 66, 71, 80, 85, 86, 102, 110, 126, 130, 146, 154, 170, 172, 183, 198, 209, 218, 229, 244, 255, 258, 269, 284, 295, 304, 315, 330, 341, 342, 374, 390, 422, 430, 462, 478, 510, 514, 546, 562, 594
Offset: 1

Views

Author

Amiram Eldar, Jul 14 2023

Keywords

Comments

The even-indexed Jacobsthal numbers A001045(2*n) = A002450(n) = (4^n-1)/3, for n >= 1, are terms since their representation is 2*n-1 1's.
A001045(2*n+1) - 1 = A020988(n) = (2/3)*(4^n-1) is a term for n >= 1, since its representation is 2*n 1's.
A001045(n) + 1 = A128209(n) is a term for n >= 0, since its representation for n = 0 is 1 and its representation for n >= 1 is n-1 0's between 2 1's.
A160156(n) is a term for n >= 0 since its representation is n 0's interleaved with n+1 1's.

Examples

			The first 10 terms are:
   n  a(n)  A280049(a(n))
  --  ----  -------------
   1     1              1
   2     2             11
   3     4            101
   4     5            111
   5     6           1001
   6    10           1111
   7    12          10001
   8    15          10101
   9    18          11011
  10    21          11111
		

Crossrefs

Programs

  • Mathematica
    Position[Select[Range[1000], EvenQ[IntegerExponent[#, 2]] &], _?(PalindromeQ[IntegerDigits[#, 2]] &)] // Flatten
  • PARI
    s(n) = if(n < 2, n > 0, n = s(n-1); until(valuation(n, 2)%2 == 0, n++); n); \\ A003159
    is(n) = {my(d = binary(s(n))); d == Vecrev(d);}

A354884 Numbers whose skew binary representation (A169683) is palindromic.

Original entry on oeis.org

0, 1, 2, 4, 8, 11, 16, 26, 32, 39, 50, 57, 64, 86, 98, 120, 128, 143, 166, 181, 194, 209, 232, 247, 256, 302, 326, 372, 386, 432, 456, 502, 512, 543, 590, 621, 646, 677, 724, 755, 770, 801, 848, 879, 904, 935, 982, 1013, 1024, 1118, 1166, 1260, 1286, 1380, 1428
Offset: 1

Views

Author

Amiram Eldar, Jun 10 2022

Keywords

Comments

The sequence of powers of 2 (A000079) is a subsequence since A169683(1) = 1, A169683(2) = 2, and for n > 2 A169683(2^n) = 10..01 with n-1 0's between the two 1's.
A000295 is a subsequence since A169683(A000295(0)) = A169683(A000295(1)) = 0 and for n>1 A169683(A000295(n)) is a repunit with n-1 1's.
A144414 is a subsequence since A169683(A144414(1)) = 1 and for n>1 A169683(A144414(n)) = 1010..01 with n-1 0's interleaved with n 1's.

Examples

			The first 10 terms are:
   n  a(n)  A169683(a(n))
  --  ----  -------------
   1    0               0
   2    1               1
   3    2               2
   4    4              11
   5    8             101
   6   11             111
   7   16            1001
   8   26            1111
   9   32           10001
  10   39           10101
		

Crossrefs

Programs

  • Mathematica
    f[0] = 0; f[n_] := Module[{m = Floor@Log2[n + 1], d = n, pos}, Reap[While[m > 0, pos = 2^m - 1; Sow@Floor[d/pos]; d = Mod[d, pos]; --m;]][[2, 1]] // FromDigits]; Select[Range[0, 15000], PalindromeQ[f[#]] &] (* after N. J. A. Sloane at A169683 *)

A364122 Numbers whose Stolarsky representation (A364121) is palindromic.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 13, 15, 18, 21, 23, 34, 36, 40, 45, 50, 55, 66, 71, 89, 91, 95, 108, 113, 120, 128, 136, 144, 159, 176, 196, 204, 233, 235, 239, 261, 273, 286, 291, 298, 319, 327, 338, 351, 364, 377, 400, 426, 464, 490, 518, 550, 563, 610, 612, 616, 654, 667
Offset: 1

Views

Author

Amiram Eldar, Jul 07 2023

Keywords

Comments

The positive Fibonacci numbers (A000045) are terms since the Stolarsky representation of Fibonacci(1) = Fibonacci(2) is 0 and the Stolarsky representation of Fibonacci(n) is n-2 1's for n >= 3.
Fiboancci(2*n+1) + 2 is a term for n >= 3, since its Stolarsky representation is n-1 0's between two 1's.

Examples

			The first 10 terms are:
   n  a(n)  A364121(a(n))
  --  ----  -------------
   1     1  0
   2     2  1
   3     3  11
   4     5  111
   5     6  101
   6     8  1111
   7    13  11111
   8    15  1001
   9    18  11011
  10    21  111111
		

Crossrefs

Programs

  • Mathematica
    stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
    stolPalQ[n_]:= PalindromeQ[stol[n]]; Select[Range[700], stolPalQ]
  • PARI
    stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
    is(n) = {my(s = stol(n)); s == Vecrev(s);}

A364378 Numbers whose representation in Jacobsthal greedy base (A265747) is palindromic.

Original entry on oeis.org

0, 1, 2, 4, 6, 9, 12, 20, 22, 27, 36, 41, 44, 60, 68, 84, 86, 97, 112, 123, 132, 143, 158, 169, 172, 204, 220, 252, 260, 292, 308, 340, 342, 363, 396, 417, 432, 453, 486, 507, 516, 537, 570, 591, 606, 627, 660, 681, 684, 748, 780, 844, 860, 924, 956, 1020, 1028
Offset: 1

Views

Author

Amiram Eldar, Jul 21 2023

Keywords

Comments

A128209(n) = A001045(n) + 1 is a term for n >= 3, since its representation is two 1's with n-3 0's between them.
A084639(n) is a term for n >= 1 since its representation is n 1's.
A014825(n) is a term for n >= 1 since its representation is n-1 0's interleaved with n 1's.

Examples

			The first 10 terms are:
   n  a(n)  A265747(a(n))
  --  ----  -------------
   1     0              0
   2     1              1
   3     2              2
   4     4             11
   5     6            101
   6     9            111
   7    12           1001
   8    20           1111
   9    22          10001
  10    27          10101
		

Crossrefs

Programs

  • Mathematica
    palJacobQ[n_] := PalindromeQ[A265747[n]]; Select[Range[0, 1000], palJacobQ] (* using A265747[n] *)
  • PARI
    is(n) = {my(dig = digits(A265747(n))); dig == Vecrev(dig);} \\ using A265747(n)

A381580 Numbers whose Chung-Graham representation (A381579) is palindromic.

Original entry on oeis.org

0, 1, 2, 4, 9, 12, 15, 18, 22, 33, 44, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 145, 174, 203, 232, 261, 290, 319, 348, 378, 399, 420, 441, 462, 483, 504, 525, 546, 567, 588, 609, 630, 651, 672, 693, 714, 735, 756, 777, 798, 819, 840, 861, 882, 903, 924, 945, 966, 988
Offset: 1

Views

Author

Amiram Eldar, Feb 28 2025

Keywords

Comments

The numbers of the form Fibonacci(2*k) + 1 (A055588) are all terms since A381579(A055588(0)) = 1, A381579(A055588(1)) = 2, and A381579(A055588(k)) = 10^(k-1)+1 (i.e., two 1's with k-2 0's between them) for k >= 2.

Examples

			The first 10 terms are:
   n  a(n) A381579(a(n))
   ---------------------
   1   0               0
   2   1               1
   3   2               2
   4   4              11
   5   9             101
   6  12             111
   7  15             121
   8  18             202
   9  22            1001
  10  33            1111
		

Crossrefs

Subsequence: A055588.
Similar sequences: A002113, A006995, A094202, A331191.

Programs

  • Mathematica
    f[n_] := f[n] = Fibonacci[2*n]; q[n_] := Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2*10^(k-1); m -= 2*f[k], s += 10^(k-1); m -= f[k]]]; PalindromeQ[s]]; Select[Range[0, 1000], q]
  • PARI
    mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
    isok(n) = {my(s = 0, m = n, k, d); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2*10^(k-1); m -= 2*f(k), s += 10^(k-1); m -= f(k))); d = digits(s); Vecrev(d) == d;}

A331192 Numbers whose Zeckendorf representation (A014417) and dual Zeckendorf representation (A104326) are both palindromic.

Original entry on oeis.org

0, 1, 4, 6, 12, 22, 33, 64, 88, 174, 232, 462, 609, 1216, 1596, 3190, 4180, 8358, 10945, 21888, 28656, 57310, 75024, 150046, 196417, 392832, 514228, 1028454, 1346268, 2692534, 3524577, 7049152, 9227464, 18454926, 24157816, 48315630, 63245985, 126491968, 165580140
Offset: 1

Views

Author

Amiram Eldar, Jan 11 2020

Keywords

Comments

Apparently union of numbers of the form F(2*k - 1) - 1 (k > 0) and numbers of the form 2 * F(2*k - 1) - 4 (k > 1), where F(m) is the m-th Fibonacci number.
The numbers of the form F(2*k - 1) - 1 have the same Zeckendorf and dual Zeckendorf representations. For k > 1 the representation is 1010...01, k-1 1's interleaved with k-2 0's.

Examples

			6 is a term since its Zeckendorf representation, 1001, and its dual Zeckendorf representation, 111, are both palindromic.
		

Crossrefs

Programs

  • Mathematica
    mirror[dig_, s_] := Join[dig, s, Reverse[dig]];
    select[v_, mid_] := Select[v, Length[#] == 0 || Last[#] != mid &];
    fib[dig_] := Plus @@ (dig*Fibonacci[Range[2, Length[dig] + 1]]);
    ndig = 12; pals1 = Rest[IntegerDigits /@ FromDigits /@ Select[Tuples[{0, 1}, ndig], SequenceCount[#, {1, 1}] == 0 &]];
    zeckPals = Union @ Join[{0, 1}, fib /@ Join[mirror[#, {}] & /@ (select[pals1, 1]), mirror[#, {1}] & /@ (select[pals1, 1]), mirror[#, {0}] & /@ pals1]];
    pals2 = Join[{{}}, Rest[Select[IntegerDigits[Range[0, 2^ndig - 1], 2], SequenceCount[#, {0, 0}] == 0 &]]];
    dualZeckPals = Union@Join[{0}, fib /@ Join[mirror[#, {}] & /@ (select[pals2, 0]), mirror[#, {0}] & /@ (select[pals2, 0]), mirror[#, {1}] & /@ pals2]];
    Intersection[zeckPals, dualZeckPals]
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