A107711 -id:A107711 - cp's OEIS frontend

A234042 a(n) = binomial(n+4,4)*gcd(n,5)/5.

1, 1, 3, 7, 14, 126, 42, 66, 99, 143, 1001, 273, 364, 476, 612, 3876, 969, 1197, 1463, 1771, 10626, 2530, 2990, 3510, 4095, 23751, 5481, 6293, 7192, 8184, 46376, 10472, 11781, 13209, 14763, 82251, 18278, 20254, 22386, 24682, 135751, 29799, 32637, 35673, 38916

Offset: 0

Wolfdieter Lang, Feb 24 2014

The sixth column of the triangle A107711.

Indranil Ghosh, Table of n, a(n) for n = 0..10000

Cf. A107711, A208950 (fifth column of A107711), A109009 (gcd(n,5)).

a[n_] := Binomial[n + 4, 4] * GCD[n, 5]/5; Table[a[n], {n, 0, 40}] (* Amiram Eldar, Sep 20 2022 *)

a(n) = binomial(n+4,4)*gcd(n,5)/5 \\ Charles R Greathouse IV, Feb 16 2017

a(n) = A107711(n+5,5) = binomial(n+5,5)*gcd(n,5)/(n+5), with n >= 0.
O.g.f.: ((1+x^20) + x*(1+x^18) + 3*x^2*(1+x^16) + 7*x^3*(1+x^14) + 14*x^4*(1+x^12) + 121*x^5*(1+x^10)+37*x^6*(1+x^8) + 51*x^7*(1+x^6) + 64*x^8*(1+x^4) + 73*x^9*(1+x^2) + 381*x^10)/(1-x^5)^5. From the 5-section using n = 5*k + j, for j = 0, 1, 2, 3, 4.
Sum_{n>=0} 1/a(n) = 20/3 - 16*sqrt(10-22/sqrt(5))*Pi/5. - Amiram Eldar, Sep 20 2022

A208950 a(4n) = n(16n^2-1)/3, a(2n+1) = n(n+1)(2n+1)/6, a(4n+2) = (4n+1)(4n+2)(4*n+3)/6.

Original entry on oeis.org

0, 0, 1, 1, 5, 5, 35, 14, 42, 30, 165, 55, 143, 91, 455, 140, 340, 204, 969, 285, 665, 385, 1771, 506, 1150, 650, 2925, 819, 1827, 1015, 4495, 1240, 2728, 1496, 6545, 1785, 3885, 2109, 9139, 2470, 5330, 2870, 12341, 3311, 7095, 3795, 16215, 4324

Offset: 0

Paul Curtz, Mar 03 2012

a(n+2) is divisible by A060819(floor(n/3)).
a(n) is divisible by A176672(floor(n/3)).
Denominator of a(n)/n is of period 24: 1,1,3,4,1,6,1,4,3,1,1,12,1,2,3,4,1,3,1,4,3,2,1,12 (two successive palindromes).
This is the fifth column of the triangle A107711, hence the formula involving gcd(n+2,4) given below follows. - Wolfdieter Lang, Feb 24 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,4,0,0,0,-6,0,0,0,4,0,0,0,-1).

Cf. A107711 (fifth column).

Cf. A000034, A000292, A002415, A051724, A060819, A061037, A107711, A138190, A145979, A176672, A176895.

[Binomial(n+1,3)*GCD(n+2,4)/4: n in [0..50]]; // G. C. Greubel, Sep 20 2018

CoefficientList[Series[(x^2 + x^3 + 5 x^4 + 5 x^5 + 31 x^6 + 10 x^7 + 22 x^8 + 10 x^9 + 31 x^10 + 5 x^11 + 5 x^12 + x^13 + x^14)/((1 - x)^4 (1 + x)^4 (1 + 4 x^2 + 6 x^4 + 4 x^6 + x^8)), {x, 0, 47}], x] (* Bruno Berselli, Mar 11 2012 *)

A208950(n) := block(
        [a,npr] ,
        if equal(mod(n,4), 0) then (
                a : n/12*(n^2-1)
        ) else if equal(mod(n,2),0) then (
                a : (n-1)*n*(n+1)/6
        ) else (
                npr : (n-1)/2,
                a : npr*(npr+1)*n/6
        ) ,
        return(a)
)$ /* R. J. Mathar, Mar 10 2012 */

vector(50, n, n--; binomial(n+1,3)*gcd(n+2,4)/4) \\ G. C. Greubel, Sep 20 2018

a(n) = 4*a(n-4) - 6*a(n-8) + 4*a(n-12) - a(n-16).
a(n+1) = A002415(n+1)/A145979(n-1).
a(n) = A051724(n-1) * A051724(n) * A051724(n+1).
a(n) = A060819(n-1) * A060819(n) * A060819(n+1) / 3.
a(n) * a(n+4) = A061037(n+1) * A061037(n+2) * A061037(n+3) / 9.
a(n) = A138190(n)/A000034(n) for n > 0.
a(n) = A000292(n-1)/A176895(n+2) for n > 0.
a(n)/a(n+4) = n*(n^2-1)/((n+3)*(n+4)*(n+5)).
a(n)/a(n+12) = (n-1)*n*(n+1)/((n+11)*(n+12)*(n+13)).
G.f.: (x^2 + x^3 + 5*x^4 + 5*x^5 + 31*x^6 + 10*x^7 + 22*x^8 + 10*x^9 + 31*x^10 + 5*x^11 + 5*x^12 + x^13 + x^14) / ((1-x)^4*(1+x)^4*(1 + 4*x^2 + 6*x^4 + 4*x^6 + x^8)). - R. J. Mathar, Mar 10 2012
From Wolfdieter Lang, Feb 24 2014: (Start)
G.f.: (1 + x^12 + x*(1+x^10) + 5*x^2*(1+x^8) + 5*x^3*(1+x^7) + 31*x^4*(1+x^4) + 10*x^5*(1+x^2) + 22*x^6)/(1-x^4)^4. This is the preceding g.f. rewritten.
a(n) = binomial(n+1,3)*gcd(n+2,4)/4, n >= 0. From the g.f., see a comment above on A107711. (End)
a(n) = (n*(n-1)*((n+1)*(4+2*(-1)^n + (1+(-1)^n)*(-1)^((2*n+3+(-1)^n)/4))))/48. - Luce ETIENNE, Jan 01 2015
Sum_{n>=2} 1/a(n) = 12 - 27*log(2)/2. - Amiram Eldar, Aug 12 2022

A027626 Denominator of n(n+5)/((n+2)(n+3)).

Original entry on oeis.org

1, 2, 10, 5, 7, 28, 12, 15, 55, 22, 26, 91, 35, 40, 136, 51, 57, 190, 70, 77, 253, 92, 100, 325, 117, 126, 406, 145, 155, 496, 176, 187, 595, 210, 222, 703, 247, 260, 820, 287, 301, 946, 330, 345, 1081, 376, 392, 1225, 425, 442, 1378, 477, 495, 1540, 532, 551

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Peter Bala, A note on the sequence of numerators of a rational function, 2019.
Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).

Cf. A000292, A027625, A061347, A107711, A234041.

[Denominator(n*(n+5)/((n+2)*(n+3))): n in [0..60]]; // Vincenzo Librandi, Mar 04 2014

CoefficientList[Series[(1+2*x+10*x^2+2*x^3+x^4-2*x^5+x^8)/(1-x^3)^3, {x,0,50}], x] (* Vincenzo Librandi, Mar 04 2014 *)

a(n) = numerator((n+2)*(n+3)/6); \\ Altug Alkan, Apr 18 2018

[numerator(binomial(n+3,2)/3) for n in (0..60)] # G. C. Greubel, Aug 04 2022

a(n) = GCD of n-th and (n+1)st tetrahedral numbers (A000292). - Ross La Haye, Sep 13 2003
G.f.: (1 +2*x +10*x^2 +2*x^3 +x^4 -2*x^5 +x^8)/(1-x^3)^3.
a(n) = A234041(n+1) = A107711(n+4,3) = C(n+3,2)*gcd(n+4,3)/3 for n >= 0. See the o.g.f. of A234041. - Wolfdieter Lang, Feb 26 2014
a(n) = numerator of (n+2)*(n+3)/6. - Altug Alkan, Apr 18 2018
Sum_{n>=0} 1/a(n) = 5 - 4*Pi/(3*sqrt(3)). - Amiram Eldar, Aug 11 2022
a(n) = (n + 2)*(n + 3)*(5 - 2*A061347(n+1))/18. - Stefano Spezia, Oct 16 2023
a(n) is quasi-polynomial in n: a(3*n) = (n+1)*(3*n+2)/2 = A000326(n+1); a(3*n+1) = (n+1)*(3*n+4)/2 = A005449(n+1); a(3*n+2) = (3*n+4)*(3*n+5)/2 = A060544(n+2). - Peter Bala, Nov 20 2024

More terms from Vincenzo Librandi, Mar 04 2014

A234041 a(n) = binomial(n+2,2)*gcd(n,3)/3, n >= 0.

Original entry on oeis.org

1, 1, 2, 10, 5, 7, 28, 12, 15, 55, 22, 26, 91, 35, 40, 136, 51, 57, 190, 70, 77, 253, 92, 100, 325, 117, 126, 406, 145, 155, 496, 176, 187, 595, 210, 222, 703, 247, 260, 820, 287, 301, 946, 330, 345, 1081, 376, 392, 1225, 425, 442, 1378, 477, 495, 1540, 532

Offset: 0

Wolfdieter Lang, Feb 24 2014

Apart from the first term, this is the same as A027626. - Bruno Berselli, Feb 24 2014

This is the sequence of the fourth column of the triangle A107711.

			a(6) = binomial(8,2) = 28 (example for n == 0 (mod 3)),
a(7) = binomial(9,2)/3 = 3*4 = 12 (example for n == 1 (mod 3)),
a(8) = binomial(10,2)/3 = 5*3 = 15 (example for n == 2 (mod 3)).

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

Cf. A027626, A107711, A026741 (third column of A107711), A109007 (gcd(n,3)).

Table[Binomial[n + 2, 2] GCD[n + 3, 3]/3, {n, 0, 60}] (* Bruno Berselli, Feb 24 2014 *)
CoefficientList[Series[(1 + x + 2 x^2 + 7 x^3 + 2 x^4 + x^5 + x^6)/(1 - x^3)^3, {x, 0, 60}], x] (* Vincenzo Librandi, Feb 26 2014 *)

a(n) = numerator((n+1)*(n+2)/6); \\ Altug Alkan, Apr 19 2018

G.f.: (1+x+2*x^2+7*x^3+2*x^4+x^5+x^6)/(1-x^3)^3.
a(n) = A107711(n+3,3) for n >= 0.
a(n) = (2+(-1)^(n+floor((n+1)/3)))*(n+1)*(n+2)/6. - Bruno Berselli, Feb 24 2014
a(n) is the numerator of (n+1)*(n+2)/6. - Altug Alkan, Apr 19 2018
Sum_{n>=0} 1/a(n) = 6 - 4*Pi/(3*sqrt(3)). - Amiram Eldar, Aug 11 2022

A258820 Reversed rows of A178252 presented as diagonals of an irregular triangle.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 5, 2, 1, 1, 3, 10, 1, 1, 7, 5, 5, 1, 1, 4, 7, 5, 1, 1, 9, 28, 35, 3, 1, 1, 5, 12, 14, 7, 1, 1, 11, 15, 21, 14, 7, 1, 1, 6, 55, 30, 126, 28, 1, 1, 13, 22, 165, 42, 21, 4, 1

Offset: 0

Tom Copeland, Jun 18 2015

The diagonals of T are the reversed rows of A178252. E.g., the fifth diagonal of T is (1,2,2,1,1) from the example below, which is the fifth reversed row of A178252.
Factoring out the greatest common divisor (gcd) of the coefficients of the sub-polynomials in the indeterminate q of the polynomials in s presented on p. 12 of the Alexeev et al. link and then evaluating the sub-polynomials at q=1 gives the first nine rows of T given in the example below. E.g., for k=6 (the seventh row), q*s^6 + (6*q + 9*q^2) s^4 + (15*q + 15*q^2) s^2 + 5  = q*s^6 + 3*(2*q + 3*q^2)*s^4 + 15*(q + q^2)*s^2 + 5 generates (1,2+3,1+1,1)=(1,5,2,1).
The row length sequence of this irregular triangle is A008619(n) = 1 + floor(n/2). - Wolfdieter Lang, Aug 25 2015

			The irregular triangle T(n,k) starts
n\k  0 1  2  3 4 5 ...
0:   1
1:   1
2:   1 1
3:   1 1
4:   1 3  1
5:   1 2  1
6:   1 5  2  1
7:   1 3 10  1
8:   1 7  5  5 1
9:   1 4  7  5 1
10:  1 9 28 35 3 1
... reformatted. - _Wolfdieter Lang_, Aug 25 2015

N. Alexeev, J. Andersen, R. Penner, P. Zograf, Enumeration of chord diagrams on many intervals and their non-orientable analogs, arXiv:1307.0967 [math.CO], 2013-2014.

Cf. A050169, A161642, A055151, A088617, A178252, A107711, A165661, A003989, A008619.

max = 15; coes = Table[ PadRight[ CoefficientList[ BernoulliB[n, x], x], max], {n, 0, max-1}]; inv = Inverse[coes] // Numerator; t[n_, k_] := inv[[n, k]]; t[n_, k_] /; k == n+1 = 1; Table[t[n-k+1, k], {n, 2, max+1}, {k, 2, Floor[n/2]+1}] // Flatten (* Jean-François Alcover, Jul 22 2015 *)

T(n,k) = A178252(n-k,n-2k) = A055151(n,k) / A161642(n,k) = A007318(n,2k) * A000108(k) / A161642(n,k) = n! / [(n-2k)! k! (k+1)! A161642(n,k)] = A003989(n-k+1,k+1) * (n-k)! / [ (n-2k)! (k+1)! ], where A003989(j,k) = gcd(j,k).

A165661 Triangle read by rows: denominator of n/binomial(n,m).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 5, 10, 5, 1, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 1, 7, 7, 35, 7, 7, 1, 1, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 1, 1, 9, 12, 21, 126, 21, 12, 9, 1, 1, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 1, 11, 55, 165, 66, 77, 66, 165, 55, 11, 1, 1

Offset: 0

Franklin T. Adams-Watters, Sep 24 2009

First difference from A107711 is at T(10,4) = 21 instead of 42.

Cf. A159335 (numerators), A107711, A007318, A109004.

T(n,m) = binomial(n,m)/gcd(n,binomial(n,m)).

A234040 a(n) = binomial(2(n+1),n) gcd(n,2)/(2*(n+1)).

Original entry on oeis.org

1, 1, 5, 7, 42, 66, 429, 715, 4862, 8398, 58786, 104006, 742900, 1337220, 9694845, 17678835, 129644790, 238819350, 1767263190, 3282060210, 24466267020, 45741281820, 343059613650, 644952073662, 4861946401452, 9183676536076, 69533550916004

Offset: 0

Wolfdieter Lang, Feb 23 2014

This gives the next-to-central entries of the even-indexed rows of the triangle A107711.
For the central entries (of the even-numbered rows) see A001700.
This sequence is composed of the bisection sequences A024492 (even part) and A065097 (odd part).

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000108, A024492, A065097, A107711.

[Binomial(2*(n+1),n)*Gcd(n,2)/(2*(n+1)): n in [0..30]]; // Vincenzo Librandi, Feb 25 2014

Table[Binomial[2 (n + 1), n] GCD[n, 2]/(2 (n + 1)), {n, 0, 40}] (* Vincenzo Librandi, Feb 25 2014 *)

a(n) = binomial(2*(n+1),n)*gcd(n,2)/(2*(n+1)) for n >= 0.
a(n) = A107711(2*(n+1), n) for n >= 0.
G.f.: (3*c(x)- c(-x)-2)/(4*x) =(4*(1-x) - 3*sqrt(1-4*x) - sqrt(1+4*x))/(8*x^2), with c(x) the o.g.f. of the Catalan numbers A000108. See the bisection comment above.

a(26) from Vincenzo Librandi, Feb 25 2014

A239125 Smallest positive integer solution x of (3^3)x - 2^ny = 1 for n >= 0.

Original entry on oeis.org

1, 1, 3, 3, 3, 19, 19, 19, 19, 19, 531, 531, 2579, 6675, 6675, 23059, 55827, 121363, 252435, 252435, 776723, 776723, 776723, 4971027, 4971027, 4971027, 4971027, 4971027, 139188755, 139188755, 676059667, 1749801491, 1749801491, 6044768787, 14634703379

Offset: 0

Wolfdieter Lang, Mar 13 2014

a(n) is the smallest positive integer solution of the linear Diophantine equation 27*a(n) - 2^n*b(n) = 1, n >= 0, with b(n) the period length 18 = phi(27) sequence repeat(26, 13, 20, 10, 5, 16, 8, 4, 2, 1, 14, 7, 17, 22, 11, 19, 23, 25). Here phi(n) = A000010(n) (Euler's totient). These 18 members are a permutation of the smallest nonnegative numbers of the reduced residue system modulo 27.
This is the instance m = 3 of an m-family of sequence pairs [x0(m, n), y0(m, n)], n >= 0, providing a special solution of the linear Diophantine equation 3^m*x - 2^n*y = 1; in fact the one with smallest positive x. The general formula is y0(m, n) = ((3^m+1)/2)^(n+3^(m-1)) (mod 3^m) and x0(m, n) = (1 + 2^n*y0(m, n))/3^m. For m = 0 this is x0(0, n) = 1 + 2^n with y0(0, n) = 1, n >= 0. Obviously, y0(m, n) is a positive integer (y0 = 0 is out). The proof that x0(m, n) is also a positive integer is done by showing that 1 + 2*y0(m, n) == 0 (mod 3^m). Because (3^m+1)/2 == 1/2 (mod 3^m) one shows that ((3^m+1)/2)^(3^(m-1)) + 1 == 0 (mod 3^m). This can be done writing (3^m+1)/2 = 3*q - 1, with q = (3^(m-1) + 1)/2, a natural number for m >= 1. Then the binomial theorem is used. Finally one has to show that binomial(3^(m-1) - 1, n -1)/n is a (positive) integer. Here the triangle A107711 helps (for a nice proof that A107711 is a positive integer triangle see the history with the remark by Peter Bala from Fri Feb 28, after #13).
The general family of positive solutions of 3^m*x - 2^n*y = c (c an integer) is then x(m, n; k) = x0(m, n) + 2^n*tmin(m, n) + 2^n*k and y(m, n; k) = y0(m, n) + 3^m*tmin(m, n) + 3^m*k for k>=0, with tmin(m, n) = ceiling(-c*y0(m, n)/3^m) if c>=0 and tmin(m, n) = ceiling(|c|*x0(m, n)/2^n) if c < 0.
See the Niven-Zuckerman-Montgomery reference, pp. 212-214, for integer solutions of a*x + b*y = c provided gcd(a,b)|c. Note that in their treatment of positive solutions a and b are assumed to be positive, but here we use b < 0.
For this instance m=3 one can prove directly that the a(n) formula given below in terms of b(n) produces (positive) integers. One uses 1/2 (mod 27) = 14 and 14^9 + 1 == 0 (mod 27).

			a(0) = 1 because 27*1 - 1*b(0) = 27 - 26 = 1.
a(1) = 1 because 27*1 - 2*b(1) = 27 - 2*13 = 1.
a(5) = 19, because 27*19 - 32*b(5) = 27*19 - 32*16 = 1.

I. Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wolfdieter Lang, On Collatz' Words, Sequences and Trees, arXiv:1404.2710 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.11.7.
Index entries for linear recurrences with constant coefficients, signature (3,-2,0,0,0,0,0,0,-512,1536,-1024).

Cf. A007583 (comment Feb 15 2014).

LinearRecurrence[{3, -2, 0, 0, 0, 0, 0, 0, -512, 1536, -1024}, {1, 1, 3, 3, 3, 19, 19, 19, 19, 19, 531}, 40] (* Bruno Berselli, Mar 15 2014 *)

a(n) = (1 + 2^n*b(n))/27 with b(n) = 14^(n+9) (mod 27), n >= 0. The sequence b(n) has period length 18, and it is given in a comment above.
a(n) = 3*a(n-1) -2*a(n-2) -512*a(n-9) +1536*a(n-10) -1024*a(n-11) for n>10, with initial values as shown. - Bruno Berselli, Mar 15 2014
G.f.: -(512*x^10-512*x^9+32*x^6-16*x^5+4*x^3-2*x^2+2*x-1) / ((x-1)*(2*x-1)*(2*x+1)*(4*x^2-2*x+1)*(64*x^6-8*x^3+1)). - Colin Barker, Mar 20 2014

cp's OEIS Frontend

A234042 a(n) = binomial(n+4,4)*gcd(n,5)/5.

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A208950 a(4*n) = n*(16*n^2-1)/3, a(2*n+1) = n*(n+1)*(2*n+1)/6, a(4*n+2) = (4*n+1)*(4*n+2)*(4*n+3)/6.

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A027626 Denominator of n*(n+5)/((n+2)*(n+3)).

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A234041 a(n) = binomial(n+2,2)*gcd(n,3)/3, n >= 0.

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A258820 Reversed rows of A178252 presented as diagonals of an irregular triangle.

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A165661 Triangle read by rows: denominator of n/binomial(n,m).

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A234040 a(n) = binomial(2*(n+1),n) * gcd(n,2)/(2*(n+1)).

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A208950 a(4n) = n(16n^2-1)/3, a(2n+1) = n(n+1)(2n+1)/6, a(4n+2) = (4n+1)(4n+2)(4*n+3)/6.

A027626 Denominator of n(n+5)/((n+2)(n+3)).

A234040 a(n) = binomial(2(n+1),n) gcd(n,2)/(2*(n+1)).

A239125 Smallest positive integer solution x of (3^3)x - 2^ny = 1 for n >= 0.