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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A349477 Numbers k such that the sequence of elements of the continued fraction of the harmonic mean of the divisors of k is palindromic.

Original entry on oeis.org

1, 6, 8, 10, 15, 16, 21, 28, 30, 39, 48, 56, 57, 64, 93, 111, 129, 140, 183, 184, 192, 195, 200, 201, 210, 219, 220, 237, 270, 291, 309, 327, 345, 381, 417, 453, 471, 489, 496, 543, 545, 574, 579, 597, 600, 633, 645, 669, 672, 687, 723, 765, 792, 795, 798, 813
Offset: 1

Views

Author

Amiram Eldar, Nov 19 2021

Keywords

Comments

All the harmonic numbers (A001599) are terms of this sequence.

Examples

			8 is a term since the sequence of elements of the continued fraction of the harmonic mean of the divisors of 8, 32/15 = 2 + 1/(7 + 1/2), is {2, 7, 2}, which is palindromic.

Links

Crossrefs

A001599 and A349476 are subsequences.
Cf. A099377, A099378, A349473.

Programs

  • Mathematica
    q[n_] := PalindromeQ[ContinuedFraction[DivisorSigma[0, n] / DivisorSigma[-1, n]]]; Select[Range[1000], q]

A349502 Numbers k such that the continued fraction of the harmonic mean of the divisors of k contains distinct elements.

Original entry on oeis.org

1, 2, 3, 6, 9, 14, 20, 24, 28, 32, 33, 35, 42, 44, 45, 51, 52, 55, 60, 65, 66, 68, 69, 70, 72, 84, 87, 88, 91, 95, 99, 104, 110, 114, 115, 117, 120, 123, 125, 126, 128, 135, 136, 138, 140, 141, 145, 152, 153, 156, 159, 170, 174, 177, 180, 182, 185, 186, 187, 188
Offset: 1

Views

Author

Amiram Eldar, Nov 20 2021

Keywords

Comments

All the harmonic numbers (A001599) are terms of this sequence.

Examples

			2 is a term since the harmonic mean of the divisors of 2 is 4/3 = 1 + 1/3 and the elements of the continued fraction, {1, 3}, are different.
4 is not a term since the harmonic mean of the divisors of 4 is 12/7 = 1 + 1/(1 + 1/(2 + 1/2)) and the elements of the continued fraction, {1, 1, 2, 2}, are not distinct.

Links

Crossrefs

Cf. A001599, A349473, A349476.

Programs

  • Mathematica
    c[n_] := ContinuedFraction[DivisorSigma[0, n]/DivisorSigma[-1, n]]; q[n_] := Length[(cn = c[n])] == Length[DeleteDuplicates[cn]]; Select[Range[200], q]

A349737 a(n) is the common difference of the increasing arithmetic progression C(m,k), C(m,k+1), C(m,k+2) when C(m,k) = A349736(n).

Original entry on oeis.org

14, 1001, 326876, 463991880, 2789279908316, 69923143311577493, 7237577480931700810180, 3072423560706808979836029648, 5323553660882471719158839565113262, 37516291344074264662783594047461175379710, 124094883176124104767115229835643366860919133861769398480
Offset: 1

Views

Author

Bernard Schott, Nov 28 2021

Keywords

Comments

For further information, see A349736.

Examples

			For n = 1, row 7 of Pascal's triangle is 1, 7, 21, 35, 35, 21, 7, 1; C(7,1) = 7, C(7,2) = 21 and C(7,3) = 35 form an arithmetic progression with common difference = 14, hence a(3) = 14.
For n = 2, row 14 is 1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1; C(14,4) = 1001 , C(14,5) = 2002 and C(14,6) = 3003 form an arithmetic progression with common difference = 1001, hence a(4) = 1001.

Crossrefs

Cf. A007318, A349736.

Programs

  • Maple
    Sequence = seq((2/n)*binomial(n^2+4*n+2,(n^2+3*n-2)/2), n=1..16);
  • Mathematica
    nterms=15; Table[2/n*Binomial[n^2+4n+2, (n^2+3n-2)/2], {n, nterms}]  (* Paolo Xausa, Nov 29 2021 *)

Formula

a(n) = (2/n) * binomial(n^2+4n+2,(n^2+3n-2)/2) = (2/n) * A349476(n) for n >= 1.
a(n) ~ c*2^(n^2+4*n)/n^2, where c = 8*sqrt(2/(Pi*e)). - Stefano Spezia, Nov 29 2021
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