A161705 -id:A161705 - cp's OEIS frontend

A130154 Triangle read by rows: T(n, k) = 1 + 2(n-k)(k-1) (1 <= k <= n).

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 9, 7, 1, 1, 9, 13, 13, 9, 1, 1, 11, 17, 19, 17, 11, 1, 1, 13, 21, 25, 25, 21, 13, 1, 1, 15, 25, 31, 33, 31, 25, 15, 1, 1, 17, 29, 37, 41, 41, 37, 29, 17, 1, 1, 19, 33, 43, 49, 51, 49, 43, 33, 19, 1, 1, 21, 37, 49, 57, 61, 61, 57, 49, 37, 21, 1

Offset: 1

Emeric Deutsch, May 22 2007

Column k, except for the initial k-1 0's, is an arithmetic progression with first term 1 and common difference 2(k-1). Row sums yield A116731. First column of the inverse matrix is A129779.
Studied by Paul Curtz circa 1993.
From Rogério Serôdio, Dec 19 2017: (Start)
T(n, k) gives the number of distinct sums of 2(k-1) elements in {1,1,2,2,...,n-1,n-1}. For example, T(6, 2) = the number of distinct sums of 2 elements in {1,1,2,2,3,3,4,4,5,5}, and because each sum from the smallest 1 + 1 = 2 to the largest 5 + 5 = 10 appears, T(6, 2) = 10 - 1 = 9. [In general: 2*(Sum_{j=1..(k-1)} n-j) - (2*(Sum_{j=1..k-1} j) - 1) = 2*n*(k-1) - 4*(k-1)*k/2 + 1 = 2*(k-1)*(n-k) + 1 = T(n, k). - Wolfdieter Lang, Dec 20 2017]
T(n, k) is the number of lattice points with abscissa x = 2*(k-1) and even ordinate in the closed region bounded by the parabola y = x*(2*(n-1) - x) and the x axis. [That is, (1/2)*y(2*(k-1)) + 1 = T(n, k). - Wolfdieter Lang, Dec 20 2017]
Pascal's triangle (A007318, but with apex in the middle) is formed using the rule South = West + East; the rascal triangle A077028 uses the rule South = (West*East + 1)/North; the present triangle uses a similar rule: South = (West*East + 2)/North. See the formula section for this recurrence. (End)

			The triangle T(n, k) starts:
  n\k  1  2  3  4  5  6  7  8  9 10 ...
  1:   1
  2:   1  1
  3:   1  3  1
  4:   1  5  5  1
  5:   1  7  9  7  1
  6:   1  9 13 13  9  1
  7:   1 11 17 19 17 11  1
  8:   1 13 21 25 25 21 13  1
  9:   1 15 25 31 33 31 25 15  1
 10:   1 17 29 37 41 41 37 29 17  1
 ... reformatted. - _Wolfdieter Lang_, Dec 19 2017

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A116731, A129779. A007318, A077028.

Column sequences (no leading zeros): A000012, A016813, A016921, A017077, A017281, A017533, A131877, A158057, A161705, A215145.

Flat(List([1..12], n-> List([1..n], k-> 1 + 2*(n-k)*(k-1) ))); # G. C. Greubel, Nov 25 2019

[1 + 2*(n-k)*(k-1): k in [1..n], n in [1..12]]; // G. C. Greubel, Nov 25 2019

T:=proc(n,k) if k<=n then 2*(n-k)*(k-1)+1 else 0 fi end: for n from 1 to 14 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form

Flatten[Table[1+2(n-k)(k-1),{n,0,20},{k,n}]] (* Harvey P. Dale, Jul 13 2013 *)

T(n, k) = 1 + 2*(n-k)*(k-1) \\ Iain Fox, Dec 19 2017

first(n) = my(res = vector(binomial(n+1,2)), i = 1); for(r=1, n, for(k=1, r, res[i] = 1 + 2*(r-k)*(k-1); i++)); res \\ Iain Fox, Dec 19 2017

[[1 + 2*(n-k)*(k-1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Nov 25 2019

T(n, k) = 1 + 2*(n-k)*(k-1) (1 <= k <= n).
G.f.: G(t,z) = t*z*(3*t*z^2 - z - t*z + 1)/((1-t*z)*(1-z))^2.
Equals = 2 * A077028 - A000012 as infinite lower triangular matrices. - Gary W. Adamson, Oct 23 2007
T(n, 1) = 1 and  T(n, n) = 1 for n >= 1; T(n, k) = (T(n-1, k-1)*T(n-1, k) + 2)/T(n-2, k-1), for n > 2 and 1 < k < n. See a comment above. - Rogério Serôdio, Dec 19 2017
G.f. column k (with leading zeros): (x^k/(1-x)^2)*(1 + (2*k-3)*x), k >= 1. See the g.f. of the triangle G(t,z) above: (d/dt)^k G(t,x)/k!|{t=0}. - _Wolfdieter Lang, Dec 20 2017

Edited by Wolfdieter Lang, Dec 19 2017

A006137 a(n) = 1 + n/2 + 9*n^2/2.

Original entry on oeis.org

1, 6, 20, 43, 75, 116, 166, 225, 293, 370, 456, 551, 655, 768, 890, 1021, 1161, 1310, 1468, 1635, 1811, 1996, 2190, 2393, 2605, 2826, 3056, 3295, 3543, 3800, 4066, 4341, 4625, 4918, 5220, 5531, 5851, 6180, 6518, 6865, 7221, 7586, 7960, 8343, 8735, 9136, 9546

Offset: 0

N. J. A. Sloane

72*a(n) - 71 = (18*n+1)^2 = A161705(n)^2 is a perfect square. - Klaus Purath, Jan 14 2022

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Nickolas Arustamyan, Christopher Cox, Erik Lundberg, Sean Perry, and Zvi Rosen, On the Number of Equilibria Balancing Newtonian Point Masses with a Central Force, arXiv:2106.11416 [math-ph], 2021.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002378, A003215, A161705, A276819.

Table[1+n/2+9 n^2/2,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{1,6,20},40] (* Harvey P. Dale, Oct 05 2012 *)

a(n)=1+n/2+9*n^2/2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = a(n-1) + 9*n - 4 (with a(0)=1). - Vincenzo Librandi, Nov 18 2010
From Klaus Purath, Jan 14 2022: (Start)
a(n) = A276819(n) + n.
A003215(a(n)) - A003215(a(n)-3) = A002378(9*n). (End)
From Stefano Spezia, Dec 25 2022: (Start)
O.g.f.: (1 + 3*x+ 5*x^2)/(1 - x)^3.
E.g.f.: exp(x)*(2 + 10*x + 9*x^2)/2. (End)

A082286 a(n) = 18*n + 10.

Original entry on oeis.org

10, 28, 46, 64, 82, 100, 118, 136, 154, 172, 190, 208, 226, 244, 262, 280, 298, 316, 334, 352, 370, 388, 406, 424, 442, 460, 478, 496, 514, 532, 550, 568, 586, 604, 622, 640, 658, 676, 694, 712, 730, 748, 766, 784, 802, 820, 838, 856, 874, 892, 910, 928, 946

Offset: 0

Cino Hilliard, May 10 2003

Solutions to (11^x + 13^x) mod 19 = 17.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Tanya Khovanova, Recursive Sequences
Leo Tavares, Illustration: Triangles
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A006370, A008600, A016945, A161705.

Cf. A000217, A017221, A022267, A060544.

[18*n + 10: n in [0..60]]; // Vincenzo Librandi, May 05 2011

Range[10, 1000, 18] (* Vladimir Joseph Stephan Orlovsky, Jun 01 2011 *)

a(n)=18*n+10 \\ Charles R Greathouse IV, Jul 10 2016

a(n) = A006370(A016945(n)). - Reinhard Zumkeller, Apr 17 2008
a(n) = 2*A017221(n). - Michel Marcus, Feb 15 2014
a(n) = A060544(n+2) - 9*A000217(n-1). - Leo Tavares, Oct 15 2022
From Elmo R. Oliveira, Apr 08 2024: (Start)
G.f.: 2*(5+4*x)/(1-x)^2.
E.g.f.: 2*exp(x)*(5 + 9*x).
a(n) = 2*a(n-1) - a(n-2) for n >= 2.
a(n) = 2*(A022267(n+1) - A022267(n)). (End)

More terms from Reinhard Zumkeller, Apr 17 2008

A154515 a(n) = 648n^2 + 72n + 1.

Original entry on oeis.org

721, 2737, 6049, 10657, 16561, 23761, 32257, 42049, 53137, 65521, 79201, 94177, 110449, 128017, 146881, 167041, 188497, 211249, 235297, 260641, 287281, 315217, 344449, 374977, 406801, 439921, 474337, 510049, 547057, 585361, 624961, 665857, 708049, 751537

Offset: 1

Vincenzo Librandi, Jan 11 2009

The identity (648*n^2 + 72*n + 1)^2 - (9*n^2 + n)*(216*n + 12)^2 = 1 can be written as a(n)^2 - A154517(n)*A154519(n)^2 = 1. This is the case s=3 of the identity (8*n^2*s^4 + 8*n*s^2 + 1)^2 - (n^2*s^2 + n)*(8*n*s^3 + 4*s)^2 = 1. - Vincenzo Librandi, Jan 30 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A154517, A154519.

I:=[721, 2737, 6049]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // Vincenzo Librandi, Jan 30 2012

LinearRecurrence[{3, -3, 1}, {721, 2737, 6049}, 50] (* Vincenzo Librandi, Jan 30 2012 *)

a(n)=648*n^2+72*n+1 \\ Charles R Greathouse IV, Dec 27 2011

From Colin Barker, Jan 25 2012: (Start)
G.f.: x*(721 + 574*x + x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=721, a(2)=2737, a(3)=6049. (End)
a(n) = 2*A161705(n)^2 - 1. - Bruno Berselli, Jan 31 2012

A154519 a(n) = 216*n + 12.

Original entry on oeis.org

228, 444, 660, 876, 1092, 1308, 1524, 1740, 1956, 2172, 2388, 2604, 2820, 3036, 3252, 3468, 3684, 3900, 4116, 4332, 4548, 4764, 4980, 5196, 5412, 5628, 5844, 6060, 6276, 6492, 6708, 6924, 7140, 7356, 7572, 7788, 8004, 8220, 8436, 8652

Offset: 1

Vincenzo Librandi, Jan 11 2009

The identity (648*n^2 + 72*n + 1)^2 - (9*n^2 + n)*(216*n + 12)^2 = 1 can be written as A154515(n)^2 - A154517(n)*a(n)^2 = 1 (see also the second comment at A154515).

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A154515, A154517, A161705.

I:=[228, 444]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Jan 30 2012

LinearRecurrence[{2, -1}, {228, 444}, 50] (* Vincenzo Librandi, Jan 30 2012 *)

a(n)=216*n+12 \\ Charles R Greathouse IV, Dec 27 2011

G.f.: x*(228 - 12*x)/(x-1)^2. - Vincenzo Librandi, Jan 30 2012 [corrected by Georg Fischer, May 12 2019]
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Jan 30 2012
a(n) = 12*A161705(n). - Michel Marcus, Aug 19 2018

A215137 a(n) = 17*n + 1.

Original entry on oeis.org

1, 18, 35, 52, 69, 86, 103, 120, 137, 154, 171, 188, 205, 222, 239, 256, 273, 290, 307, 324, 341, 358, 375, 392, 409, 426, 443, 460, 477, 494, 511, 528, 545, 562, 579, 596, 613, 630, 647, 664, 681, 698, 715, 732, 749, 766, 783, 800, 817, 834, 851, 868, 885, 902, 919, 936, 953, 970

Offset: 0

Jeremy Gardiner, Aug 04 2012

Jeremy Gardiner, Table of n, a(n) for n = 0..999
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A129484, A158057, A161705.

I:=[1,18]; [n le 2 select I[n] else 2*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Apr 19 2018

Range[1, 100, 17]
LinearRecurrence[{2,-1}, {1,18}, 50] (* G. C. Greubel, Apr 19 2018 *)

for(n=0, 50, print1(17*n + 1, ", ")) \\ G. C. Greubel, Apr 19 2018

From G. C. Greubel, Apr 19 2018: (Start)
a(n) = 2*a(n-1) - a(n-2).
G.f.: (1+16*x)/(1-x)^2.
E.g.f.: (17*x + 1)*exp(x). (End)

A215144 a(n) = 19*n + 1.

Original entry on oeis.org

1, 20, 39, 58, 77, 96, 115, 134, 153, 172, 191, 210, 229, 248, 267, 286, 305, 324, 343, 362, 381, 400, 419, 438, 457, 476, 495, 514, 533, 552, 571, 590, 609, 628, 647, 666, 685, 704, 723, 742, 761, 780, 799, 818, 837, 856, 875, 894, 913, 932, 951, 970, 989

Offset: 0

Jeremy Gardiner, Aug 04 2012

Jeremy Gardiner, Table of n, a(n) for n = 0..999
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A141868, A161705.

First differences of A051873.

I:=[1,20]; [n le 2 select I[n] else 2*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Apr 19 2018

Range[1, 1000, 19]
19*Range[0,60]+1 (* Harvey P. Dale, Nov 14 2014 *)
LinearRecurrence[{2,-1}, {1,20}, 50] (* G. C. Greubel, Apr 19 2018 *)

for(n=0, 50, print1(19*n + 1, ", ")) \\ G. C. Greubel, Apr 19 2018

From G. C. Greubel, Apr 19 2018: (Start)
a(n) = 2*a(n-1) - a(n-2).
G.f.: (1+18*x)/(1-x)^2.
E.g.f.: (1+19*x)*exp(x). (End)

A382809 a(n) = (6n + 1)(12n + 1)(18*n + 1).

Original entry on oeis.org

1, 1729, 12025, 38665, 89425, 172081, 294409, 464185, 689185, 977185, 1335961, 1773289, 2296945, 2914705, 3634345, 4463641, 5410369, 6482305, 7687225, 9032905, 10527121, 12177649, 13992265, 15978745, 18144865, 20498401, 23047129, 25798825, 28761265, 31942225, 35349481

Offset: 0

Stefano Spezia, Apr 05 2025

a(n) is a Carmichael number if all the three factors (6*n + 1), (12*n + 1), and (18*n + 1) are prime (see Chernick and Ribenboim).

Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 101.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 146.

Jack Chernick, On Fermat's simple theorem, Bulletin of the American Mathematical Society, Vol. 45, No. 4 (1939), pp. 269-274.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002997, A033502, A046025.

Cf. A002476, A002997, A016921, A017533, A068228, A161705.

LinearRecurrence[{4,-6,4,-1},{1,1729,12025,38665},31]

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3.
G.f.: (1 + 1725*x + 5115*x^2 + 935*x^3)/(1 - x)^4.
E.g.f.: exp(x)*(1 + 1728*x + 4284*x^2 + 1296*x^3).
a(n) = A016921(n) * A017533(n) * A161705(n).
a(n) == 1 (mod 72).

A251731 Least k such that k^3 + q is divisible by 3^n where q is the n-th number congruent to 1 or -1 (mod 18).

Original entry on oeis.org

2, 1, 2, 16, 32, 145, 62, 1363, 3458, 19492, 58928, 89308, 70028, 1594318, 1890551, 189871, 31401806, 47918575, 190704887, 163454602, 502048577, 9481323661, 11627845304, 34656488290, 115450061084, 286130228125, 2303721331049, 1569269836240, 22013516320412

Offset: 1

Michel Lagneau, Dec 07 2014

nonn

It is known that k always exists if q is congruent to +-1 mod 18.
The numbers congruent to 1 or -1 (mod 18) are 1, 17, 19, 35, 37, ... = {A161705} UNION {A239129}.
For n >= 2, k^3 == (9 - 18*n - 7*(-1)^n)/2 (mod 3^n) if and only if k - a(n) is divisible by 3^(n-1). - Jinyuan Wang, Feb 13 2020

			a(1) = 2 because the first number of the form +-1 (mod 18) is 1, and 2^3 + 1 = 9 = 3*3^1;
a(2) = 1 because the second number of the form +-1 (mod 18) is 17, and 1^3 + 17 = 18 = 2*3^2;
a(3) = 2 because the third number of the form +-1 (mod 18) is 19, and 2^3 + 19 = 27 = 3^3;
a(4)= 16 because the fourth number of the form +-1 (mod 18) is 35, and 16^3 + 35 = 4131 = 51*3^4.

Robert Israel, Table of n, a(n) for n = 1..2095

Cf. A161705, A239129.

f:= proc(n) local q,R,k;
  if n::odd then q:= 9*n-8 else q:= 9*n-1 fi;
  min(map(subs,[msolve(k^3+q,3^n)],k))
end proc:
map(f, [$1..30]); # Robert Israel, Dec 23 2018

lst1={1};Do[lst1=Union[lst1,Union[{18*n+1},{18*n-1}]],{n,1,10}];lst={};Do[k=1;While[Mod[k^3+lst1[[n]],3^n]!=0,k++];Print[n," ",k],{n,1,10}];lst

a(n) = {if (n % 2, q = 9*(n-1)+1, q = 9*n-1); m = 3^n; k = 1; while ((k^3+q) % m, k++); k;} \\ Michel Marcus, Jan 07 2015

cp's OEIS Frontend

A130154 Triangle read by rows: T(n, k) = 1 + 2*(n-k)*(k-1) (1 <= k <= n).

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A006137 a(n) = 1 + n/2 + 9*n^2/2.

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A082286 a(n) = 18*n + 10.

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A154515 a(n) = 648*n^2 + 72*n + 1.

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A154519 a(n) = 216*n + 12.

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A215137 a(n) = 17*n + 1.

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A215144 a(n) = 19*n + 1.

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A130154 Triangle read by rows: T(n, k) = 1 + 2(n-k)(k-1) (1 <= k <= n).

A154515 a(n) = 648n^2 + 72n + 1.

A382809 a(n) = (6n + 1)(12n + 1)(18*n + 1).