A144454 -id:A144454 - cp's OEIS frontend

A145909 First 6-fold decimation of A061039. First bisection of A144454.

0, 8, 8, 16, 80, 40, 56, 224, 32, 40, 440, 176, 208, 728, 280, 320, 1088, 136, 152, 1520, 560, 616, 2024, 736, 800, 2600, 312, 336, 3248, 1160, 1240, 3968, 1408, 1496, 4760, 560, 592, 5624, 1976, 2080, 6560, 2296, 2408, 7568, 880, 920

Offset: 0

Paul Curtz, Oct 24 2008

G. C. Greubel, Table of n, a(n) for n = 0..5000

Cf. A061039, A144454.

A145909 := proc(n) ; A061039(6*n+3) end proc:
seq(A145909(n),n=0..80) ; # R. J. Mathar, Jan 06 2011

Numerator[(1/9)*(1 - 1/(2*Range[0,75]+1)^2)] (* G. C. Greubel, Mar 08 2022 *)

[numerator((1/9)*(1 - 1/(2*n+1)^2)) for n in (0..75)] # G. C. Greubel, Mar 08 2022

a(n) = A061039(6*n+3).

A013656 a(n) = n(9n-2).

Original entry on oeis.org

0, 7, 32, 75, 136, 215, 312, 427, 560, 711, 880, 1067, 1272, 1495, 1736, 1995, 2272, 2567, 2880, 3211, 3560, 3927, 4312, 4715, 5136, 5575, 6032, 6507, 7000, 7511, 8040, 8587, 9152, 9735, 10336, 10955, 11592, 12247, 12920, 13611, 14320, 15047, 15792, 16555

Offset: 0

David W. Wilson

For n>0, numbers such that sqrt(a(n)) has the continued fraction {k;[1,1,1,2k]}, where the part in [] is repeated and k is of the form 3m+2 (A016789). - Bruno Berselli, May 30 2013

For n >= 1, the continued fraction expansion of sqrt(4*a(n)) is [6n-1; {3, 3n-1, 3, 12n-2}]. - Magus K. Chu, Sep 18 2022

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A010701, A017257, A185019.

Cf. A016789, A061039, A144454.

[n*(9*n-2): n in [0..60]]; // G. C. Greubel, Mar 11 2022

Table[n(9n-2),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,32},50] (* Harvey P. Dale, Jul 07 2012 *)

a(n)=n*(9*n-2) \\ Charles R Greathouse IV, Jun 17 2017

a(n+1) = A144454(9*n+7) = A061039(27*n+21). - Paul Curtz, Nov 05 2008
a(n) = a(n-1) + 18*n - 11 with n>0, a(0)=0. - Vincenzo Librandi, Nov 22 2010
a(0)=0, a(1)=7, a(2)=32, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jul 07 2012
From G. C. Greubel, Mar 11 2022: (Start)
G.f.: x*(7 - 11*x)/(1-x)^3.
E.g.f.: x*(7 + 9*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = -(psi(7/9)+gamma)/2 = (A354640-A001620)/2 = 0.22000753... - R. J. Mathar, Apr 22 2024

A147296 a(n) = n(9n+2).

Original entry on oeis.org

0, 11, 40, 87, 152, 235, 336, 455, 592, 747, 920, 1111, 1320, 1547, 1792, 2055, 2336, 2635, 2952, 3287, 3640, 4011, 4400, 4807, 5232, 5675, 6136, 6615, 7112, 7627, 8160, 8711, 9280, 9867, 10472, 11095, 11736, 12395, 13072, 13767, 14480, 15211, 15960

Offset: 0

Paul Curtz, Nov 05 2008

For n >= 1, the continued fraction expansion of sqrt(4*a(n)) is [6n; {1, 1, 1, 3n-1, 1, 1, 1, 12n}]. - Magus K. Chu, Sep 17 2022

Reply to V. Librandi, A147296 (SeqFan list). - _M. F. Hasler_, Mar 01 2009
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Equals first 9-fold decimation of A144454.

Cf. A010701, A017173, A147296, A185019.

Table[n(9n+2),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,11,40},50] (* Harvey P. Dale, Dec 19 2014 *)

A147296(n) = n*(9*n + 2) \\ M. F. Hasler, Mar 01 2009

a(n) = n*(9*n + 2), as conjectured by V. Librandi. - M. F. Hasler, Mar 01 2009
G.f.: x*(11+7*x)/(1-x)^3. - Jaume Oliver Lafont, Aug 30 2009
a(n) = floor((3*n + 1/3)^2). - Reinhard Zumkeller, Apr 14 2010
From Elmo R. Oliveira, Dec 15 2024: (Start)
E.g.f.: exp(x)*x*(11 + 9*x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

More terms from M. F. Hasler, Mar 01 2009

A300295 Denominator of (1/3)n(n + 2)/((1 + 2n)(3 + 2*n)).

Original entry on oeis.org

1, 15, 105, 63, 99, 429, 195, 85, 969, 133, 483, 1725, 675, 783, 2697, 1023, 385, 3885, 481, 1599, 5289, 1935, 2115, 6909, 2499, 901, 8745, 1045, 3363, 10797, 3843, 4095, 13065, 4623, 1633, 15549, 1825, 5775, 18249, 6399, 6723, 21165, 7395, 2581, 24297, 2821, 8835, 27645, 9603, 9999, 31209, 10815, 3745

Offset: 0

Wolfdieter Lang, Mar 15 2018

The corresponding numerators are given in A144454(n+1).
r(n) = A144454(n+1)/a(n) = Sum_{k=0..n-1} 1/(A(k)*A(k+1)*A(k+2)), with A(j) = 1 + 2*j = A005408(n) for n >= 1, and r(0) = 0. This can be written as r(n) = 1/12 - 1/(4*A(n)*A(n+1)) = (1/3)*n*(n + 2)/(A(n)*A(n+1)). See Jolley, p. 40/41, (209), and the general remark on p. 38, (201). The value of the infinite series is therefore 1/12.
For the proof that numerator(r(n)) = A144454(n+1) one checks the formula with (mod 9) and (mod 3) given there. E.g., if n = 1 + 9*k then r(n-1) = k*(2 + 9*k)/((1 + 6*k)*(1 + 18*k)) and numerator(r(n-1)) = k*(2 + 9*k) = ((n-1)^2 - )/9 as claimed, because this ratio for r(n-1) is in lowest terms.

			The series begins: 1/(1*3*5) + 1/(3*5*7) + 1/(5*7*9) + ...
The partial sums are r(n) = A144454(n+1)/a(n), n >= 1, and with r(0) = 0 they begin with 0/1, 1/15, 8/105, 5/63, 8/99, 35/429, 16/195, 7/85, 80/969, 11/133, 40/483, 143/1725, 56/675, 65/783, 224/2697, 85/1023, 32/385,...

L. B. W. Jolley, Summation of Series, Dover Publications, 2nd rev. ed., 1961, pp. 38, 40, 41.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,-3,0,0,0,0,0,0,0,0,1).

Cf. A005408, A144454(n+1) (numerators).

Table[(n(n+2))/(3(1+2n)(3+2n)),{n,0,60}]//Denominator (* Harvey P. Dale, Jun 19 2021 *)

a(n) = denominator((1/3)*n*(n + 2)/((1 + 2*n)*(3 + 2*n))); \\ Michel Marcus, Mar 15 2018

a(n) = denominator(r(n)), with r(n) = (1/3)*n*(n + 2)/((1 + 2*n)*(3 + 2*n)), n >= 0. r(n-1) = (1/3)*(n^2 - 1)/((2*n)^2 -1), n >= 1.
G.f. for r(n) = A144454(n+1)/a(n): G(x) = (1/12)*(1 - hypergeometric([1, 2], [5/2], -x/(1-x)))/(1-x) = ((-3 + 5*x)*sqrt(x)/sqrt(1 - x) + 3*sqrt(1 - x)*(1 - x)*arcsinh(sqrt(x)/sqrt(1 - x)))/(24*x*(1 - x)*sqrt(x)/sqrt(1 - x))
  = ((-3 + 5*x)*sqrt(x/(1-x)) + 3*(1 - x)*sqrt(1 - x)*log((1 + sqrt(x))/sqrt(1 - x)))/(24*x*(1 - x)*sqrt(x/(1 - x))).

A147650 First trisection of A061040.

Original entry on oeis.org

1, 12, 81, 48, 75, 324, 147, 64, 729, 100, 363, 1296, 507, 588, 2025, 768, 289, 2916, 361, 1200, 3969, 1452, 1587, 5184, 1875, 676, 6561, 784, 2523, 8100, 2883, 3072, 9801, 3468, 1225, 11664, 1369, 4332, 13689, 4800, 5043

Offset: 1

Paul Curtz, Nov 09 2008

a(n) gives the denominator of (n-1)*(n+1)/(9*n^2), for n >= 1. The numerator is given by A144454(n). - Wolfdieter Lang, Mar 16 2018

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,-3,0,0,0,0,0,0,0,0,1).

Cf. A061040, A017198 (2nd trisection), A017234 (3d trisection).

Table[Which[MemberQ[{1, 8}, Mod[n, 9]], n^2, Mod[n, 3] != 0, 3 n^2, True, 9 n^2], {n, 41}] (* Michael De Vlieger, Mar 16 2018 *)

a(n) = denominator((n-1)*(n+1)/(9*n^2)); \\ Michel Marcus, Mar 17 2018

For n >= 1: a(n) = n^2 if n == 1 (mod 9) or == 8 (mod 9). For other n: a(n) = 3*n^2 if n == 1 (mod 3) or == 2 (mod 3), and a(n) = 9*n^2 if n == 0 (mod 3). From the denominator comment above. - Wolfdieter Lang, Mar 16 2018

Offset changed from 0 to 1, and extended by Wolfdieter Lang, Mar 16 2018

A147651 First trisection of A028560.

Original entry on oeis.org

0, 27, 72, 135, 216, 315, 432, 567, 720, 891, 1080, 1287, 1512, 1755, 2016, 2295, 2592, 2907, 3240, 3591, 3960, 4347, 4752, 5175, 5616, 6075, 6552, 7047, 7560, 8091, 8640, 9207, 9792, 10395, 11016

Offset: 0

Paul Curtz, Nov 09 2008

Nonnegative k such that k/9 + 1 is a square. - Bruno Berselli, Apr 10 2018

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005563, A028560, A067725, A144396, A144454.

Table[9 n (n + 2), {n, 0, 40}]  (* Harvey P. Dale, Feb 20 2011 *)

a(n)=9*n*(n+2) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 9*n*(n+2).
a(n) = a(n-1) + 9*A144396(n) for n > 0.
From Elmo R. Oliveira, Nov 29 2024: (Start)
G.f.: 9*x*(3 - x)/(1-x)^3.
E.g.f.: 9*x*(3 + x)*exp(x).
a(n) = 9*A005563(n) = 3*A067725(n) = A028560(3*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

More terms from Vincenzo Librandi, Nov 26 2010

A154629 Period 9: repeat [9, 3, 1, 3, 3, 1, 3, 9, 1].

Original entry on oeis.org

9, 3, 1, 3, 3, 1, 3, 9, 1, 9, 3, 1, 3, 3, 1, 3, 9, 1, 9, 3, 1, 3, 3, 1, 3, 9, 1, 9, 3, 1, 3, 3, 1, 3, 9, 1, 9, 3, 1, 3, 3, 1, 3, 9, 1, 9, 3, 1, 3, 3, 1, 3, 9, 1, 9, 3, 1, 3, 3, 1, 3, 9, 1, 9, 3, 1, 3, 3, 1, 3, 9, 1, 9, 3, 1, 3, 3, 1, 3, 9, 1, 9, 3, 1, 3, 3

Offset: 0

Paul Curtz, Jan 13 2009

Greatest common divisor of (n+1)^2-1 and (n+1)^2+8. - Bruno Berselli, Mar 08 2017

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,1).

Cf. A005563, A144454, A147674.

With[{n=5}, PadLeft[{}, 9n, {9,3,1,3,3,1,3,9,1}]] (* Harvey P. Dale, Oct 22 2011 *)

Vec((9 + 3*x + x^2 + 3*x^3 + 3*x^4 + x^5 + 3*x^6 + 9*x^7 + x^8) / ((1 - x)*(1 + x + x^2)*(1 + x^3 + x^6)) + O(x^100)) \\ Colin Barker, Dec 21 2017

a(n) = A147674(n)/9.
a(n) = A005563(n) / A144454(n+1) for n>0.
From Colin Barker, Dec 21 2017: (Start)
G.f.: (9 + 3*x + x^2 + 3*x^3 + 3*x^4 + x^5 + 3*x^6 + 9*x^7 + x^8) / ((1 - x)*(1 + x + x^2)*(1 + x^3 + x^6)).
a(n) = a(n-9) for n>8.
(End)

cp's OEIS Frontend

A145909 First 6-fold decimation of A061039. First bisection of A144454.

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A013656 a(n) = n*(9*n-2).

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Magma

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A147296 a(n) = n*(9*n+2).

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A300295 Denominator of (1/3)*n*(n + 2)/((1 + 2*n)*(3 + 2*n)).

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A147650 First trisection of A061040.

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A147651 First trisection of A028560.

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A154629 Period 9: repeat [9, 3, 1, 3, 3, 1, 3, 9, 1].

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A013656 a(n) = n(9n-2).

A147296 a(n) = n(9n+2).

A300295 Denominator of (1/3)n(n + 2)/((1 + 2n)(3 + 2*n)).