A087799 -id:A087799 - cp's OEIS frontend

A174503 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A087799(n)) ), where A087799(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.

1, 8, 1, 96, 1, 968, 1, 9600, 1, 95048, 1, 940896, 1, 9313928, 1, 92198400, 1, 912670088, 1, 9034502496, 1, 89432354888, 1, 885289046400, 1, 8763458109128, 1, 86749292044896, 1, 858729462339848, 1, 8500545331353600, 1

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A087799(n)) or, more explicitly,
L = 1/10 + 1/(2*98) + 1/(3*970) + 1/(4*9602) + 1/(5*95050) +...
so that L = 0.1054740177896236251618898675297390156061405857647...
then exp(L) = 1.1112372317482311056432125938345153306039099019639...
equals the continued fraction given by this sequence:
exp(L) = [1;8,1,96,1,968,1,9600,1,95048,1,940896,1,...]; i.e.,
exp(L) = 1 + 1/(8 + 1/(1 + 1/(96 + 1/(1 + 1/(968 + 1/(1 +...)))))).
Compare these partial quotients to A087799(n), n=1,2,3,...:
[10,98,970,9602,95050,940898,9313930,92198402,912670090,9034502498,...].

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,11,0,-11,0,1).

Cf. A087799, A174500, A174502, A174504, A174509.

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((5+sqrt(24))^m+(5-sqrt(24))^m))));contfrac(exp(L))[n]}

a(2n-2) = 1, a(2n-1) = A087799(n) - 2, for n>=1 [conjecture].
The above conjectures are correct. See the Bala link for details. - Peter Bala, Jan 08 2013
a(n) = 11*a(n-2)-11*a(n-4)+a(n-6). G.f.: -(x^4+8*x^3-10*x^2+8*x+1) / ((x-1)*(x+1)*(x^4-10*x^2+1)). [Colin Barker, Jan 20 2013]

A204518 Numbers such that floor(a(n)^2 / 6) is a square.

Original entry on oeis.org

0, 1, 2, 3, 5, 10, 27, 49, 98, 267, 485, 970, 2643, 4801, 9602, 26163, 47525, 95050, 258987, 470449, 940898, 2563707, 4656965, 9313930, 25378083, 46099201, 92198402, 251217123, 456335045, 912670090, 2486793147, 4517251249, 9034502498, 24616714347

Offset: 1

M. F. Hasler, Jan 15 2012

Or: Numbers whose square, with its last base-6 digit dropped, is again a square. (For the three initial terms whose square has only one digit in base 6, this is then meant to yield zero.)

Index entries for linear recurrences with constant coefficients, signature (0,0,10,0,0,-1).

Cf. A023110 (base 10), A204502 (base 9), A204514 (base 8), A204516 (base 7), A204520 (base 5), A004275 (base 4), A055793 (base 3), A055792 (base 2).

b=6;for(n=0,2e9,issquare(n^2\b) & print1(n","))

concat(0, Vec(-x^2*(x+1)*(3*x^4+7*x^3-2*x^2-x-1)/(x^6-10*x^3+1) + O(x^100))) \\ Colin Barker, Sep 18 2014

a(n) = sqrt(A055851(n)).
From Colin Barker, Sep 18 2014: (Start)
a(n) = 10*a(n-3) - a(n-6) for n > 7.
G.f.: -x^2*(x+1)*(3*x^4 + 7*x^3 - 2*x^2 - x - 1) / (x^6-10*x^3+1). (End)
a(3n+2) = A001079(n). a(3n) = A087799(n-1). - R. J. Mathar, Feb 05 2020

More terms from Colin Barker, Sep 18 2014

A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

Original entry on oeis.org

2, 2, 2, 2, 3, 2, 2, 4, 7, 2, 2, 5, 14, 18, 2, 2, 6, 23, 52, 47, 2, 2, 7, 34, 110, 194, 123, 2, 2, 8, 47, 198, 527, 724, 322, 2, 2, 9, 62, 322, 1154, 2525, 2702, 843, 2, 2, 10, 79, 488, 2207, 6726, 12098, 10084, 2207, 2, 2, 11, 98, 702, 3842, 15127, 39202, 57965, 37634, 5778, 2

Offset: 0

William W. Collier, Feb 18 2018

Note the similarity in form of the recursive steps in the array definition above and the polynomial definition under FORMULA.

			i\j |0  1   2    3      4       5        6          7           8            9
----+-------------------------------------------------------------------------
   0|2  2   2    2      2       2        2          2           2            2
   1|2  3   7   18     47     123      322        843        2207         5778
   2|2  4  14   52    194     724     2702      10084       37634       140452
   3|2  5  23  110    527    2525    12098      57965      277727      1330670
   4|2  6  34  198   1154    6726    39202     228486     1331714      7761798
   5|2  7  47  322   2207   15127   103682     710647     4870847     33385282
   6|2  8  62  488   3842   30248   238142    1874888    14760962    116212808
   7|2  9  79  702   6239   55449   492802    4379769    38925119    345946302
   8|2 10  98  970   9602   95050   940898    9313930    92198402    912670090
   9|2 11 119 1298  14159  154451  1684802   18378371   200477279   2186871698
  10|2 12 142 1692  20162  240252  2862862   34114092   406506242   4843960812
  11|2 13 167 2158  27887  360373  4656962   60180133   777684767  10049721838
  12|2 14 194 2702  37634  524174  7300802  101687054  1416317954  19726764302
  13|2 15 223 3330  49727  742575 11088898  165590895  2472774527  36926027010
  14|2 16 254 4048  64514 1028176 16386302  261152656  4162056194  66331746448
  15|2 17 287 4862  82367 1395377 23639042  400468337  6784322687 114933017342
  16|2 18 322 5778 103682 1860498 33385282  599074578 10749957122 192900153618
  17|2 19 359 6802 128879 2441899 46267202  876634939 16609796639 314709501202
  18|2 20 398 7940 158402 3160100 63043598 1257711860 25091193602 500566160180
  19|2 21 439 9198 192719 4037901 84603202 1772629341 37140612959 778180242798

William W. Collier, a(i,j) = f(i+2,j)
William W. Collier, Experimental Mathematics on Wisteria Tables, Talk to Poughkeepsie ACM Chapter.
OEIS Wiki, The (1,2) Pascal Triangle.

The array first appeared in A298675.
Rows 1 through 29 of the array appear in these OEIS entries: A005248, A003500, A003501, A003499, A056854, A086903, A056918, A087799, A057076, A087800, A078363, A067902, A078365, A090727, A078367, A087215, A078369, A090728, A090729, A090730, A090731, A090732, A090733, A090247, A090248, A090249, A090251. Also entries occur for rows 45, 121, and 320:  A087265, A065705, A089775. Each of these entries asserts that a(i,j)=f(i+2,j) is true for that row.
A few of the columns appear in the OEIS: A008865 (for column 2), A058794 and A007754 (for column 3), and A230586 (for column 5).
Main diagonal gives A343261.

A:= proc(i, j) option remember; `if`(min(i, j)=0, 2,
      `if`(j=1, i+2, (i+2)*A(i, j-1)-A(i, j-2)))
    end:
seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Mar 05 2019

a[, 0] = a[0, ] = 2; a[i_, 1] := i + 2;
a[i_, j_] := a[i, j] =(i + 2) a[i, j - 1] - a[i, j - 2];
Table[a[i - j, j], {i, 0, 10}, {j, 0, i}] // Flatten (* Jean-François Alcover, Dec 07 2019 *)

Let k be an integer, and let r1 and r2 be the roots of x + 1/x = k. Then f(k,n) = r1^n + r2^n is an integer, for integer n >= 0. Theorem: a(i,j) = f(i+2,j), for i,j >= 0. Proof: See the Collier link.
Define polynomials recursively by:
    p[0](n) = 2, for n >= 0 ( [ and ] demark subscripts).
    p[1](n) = n + 2, for n >= 0.
    p[j](n) = p[j-1](n) * p[1](n) - p[j-2](n), for j > 1, n >= 0. The coefficients of these polynomials occur as the even numbered, upward diagonals in the OEIS Wiki link. Conjecture: a(i,j) = p[j](i), i,j >= 0.

Edited by N. J. A. Sloane, Apr 04 2018

A171640 a(n) = 10*a(n-1)-a(n-2)-4 with a(1)=1 and a(2)=3.

Original entry on oeis.org

1, 3, 25, 243, 2401, 23763, 235225, 2328483, 23049601, 228167523, 2258625625, 22358088723, 221322261601, 2190864527283, 21687323011225, 214682365584963, 2125136332838401, 21036680962799043, 208241673295152025, 2061380051988721203, 20405558846592060001

Offset: 1

Mark Dols, Dec 13 2009

			a(2+1) = [5-sqrt(24)+5+sqrt(24)]^2/4 = 100/4 = 25.

Colin Barker, Table of n, a(n) for n = 1..1000
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A132596, A087799, A001263.

RecurrenceTable[{a[n] == 10 a[n - 1] - a[n - 2] - 4, a[1] == 1, a[2] == 3}, a, {n, 1, 21}] (* Michael De Vlieger, Oct 02 2015 *)
LinearRecurrence[{11,-11,1},{1,3,25},30] (* Harvey P. Dale, May 05 2018 *)

Vec(-x*(3*x^2-8*x+1)/((x-1)*(x^2-10*x+1)) + O(x^30)) \\ Colin Barker, Oct 02 2015

a(n) = A132596(n-1)+1.
2*a(n) + 2*A132596(n-1) = A087799(n-1).
a(n+1) = [(sqrt(3)-sqrt(2))^n +(sqrt(3)+ sqrt(2))^n]^2 / 4.
From Colin Barker, Oct 02 2015: (Start)
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) for n>3.
G.f.: -x*(3*x^2-8*x+1) / ((x-1)*(x^2-10*x+1)).
(End)
6*a(n)*(a(n)-1) = [A122653(n-1)]^2. - Jean-Luc Manguin, Jun 02 2020

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A174503 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A087799(n)) ), where A087799(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.

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A204518 Numbers such that floor(a(n)^2 / 6) is a square.

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A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

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A171640 a(n) = 10*a(n-1)-a(n-2)-4 with a(1)=1 and a(2)=3.

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A171659 Lucas-version of A165293.

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