A011900 -id:A011900 - cp's OEIS frontend

A211955 Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.

1, 1, 1, 1, 3, 2, 1, 6, 10, 4, 1, 10, 30, 28, 8, 1, 15, 70, 112, 72, 16, 1, 21, 140, 336, 360, 176, 32, 1, 28, 252, 840, 1320, 1056, 416, 64, 1, 36, 420, 1848, 3960, 4576, 2912, 960, 128, 1, 45, 660, 3696, 10296, 16016, 14560, 7680, 2176, 256

Offset: 0

Peter Bala, Apr 30 2012

Let b(n,x) = Sum_{k = 0..n} binomial(n+k,2*k)*x^k denote the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients (in ascending powers of x) of the related polynomial sequence R(n,x) := (1/2)*b(n,2*x) + 1/2. Several sequences already in the database are of the form (R(n,x))n>=0 for a fixed value of x. These include A101265 (x = 1), A011900 (x = 2), A182432 (x = 3), A054318 (x = 4) as well as signed versions of A133872 (x = -1), A109613(x = -2), A146983 (x = -3) and A084159 (x = -4).

The polynomials R(n,x) factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x). The coefficients of P(n,x) occur in several tables in the database, although without the connection to the Morgan-Voyce polynomials being noted - see A211956 for more details. In terms of T(n,x), the Chebyshev polynomials of the first kind, we have P(2*n,x) = T(2*n,u) and P(2*n+1,x) = 1/u * T(2*n+1,u), where u = sqrt((x+2)/2). Hence R(n,x) = 1/u * T(n,u) * T(n+1,u).

			Triangle begins
.n\k.|..0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = =
..0..|..1
..1..|..1....1
..2..|..1....3....2
..3..|..1....6...10....4
..4..|..1...10...30...28....8
..5..|..1...15...70..112...72...16
..6..|..1...21..140..336..360..176...32

Eric Weisstein's World of Mathematics, Morgan-Voyce polynomials

Cf. A011900, A084159, A085478, A101265 (row sums), A109613, A112373, A123519, A133872 (alt row sums), A146983, A182432, A204021, A208513, A211956, A211957.

T(n,0) = 1; T(n,k) = 2^(k-1)*binomial(n+k,2*k) for k > 0.
O.g.f. for column k (except column 0): 2^(k-1)*x^k/(1-x)^(2*k+1).
O.g.f.: (1-t*(x+2)+t^2)/((1-t)*(1-2*t(x+1)+t^2)) = 1 + (1+x)*t + (1+3*x+2*x^2)*t^2 + ....
Removing the first column from the triangle produces the Riordan array (x/(1-x)^3, 2*x/(1-x)^2).
The row polynomials R(n,x) := 1/2*b(n,2*x) + 1/2 = 1 + x*Sum_{k = 1..n} binomial(n+k,2*k)*(2*x)^(k-1).
Recurrence equation: R(n,x) = 2*(1+x)*R(n-1,x) - R(n-2,x) - x with initial conditions R(0,x) = 1, R(1,x) = 1+x.
Another recurrence is R(n,x)*R(n-2,x) = R(n-1,x)*(R(n-1,x) + x).
With P(n,x) as defined in the Comments section we have (x+2)/x - {Sum_{k = 0..2n} 1/R(k,x)}^2 = 2/(x*P(2*n+1,x)^2); (x+2)/x - {Sum_{k = 0..2n+1} 1/R(k,x)}^2 = (x+2)/(x*P(2*n+2,x)^2); consequently Sum_{k >= 0} 1/R(k,x) = sqrt((x+2)/x) for either x > 0 or x <= -2.
Row sums R(n,1) = A101265(n+1); Alt. row sums R(n,-1) = A133872(n+1);
R(n,2) = A011900(n); R(n,-2) = (-1)^n * A109613(n); R(n,3) = A182432;
R(n,-3) = (-1)^n * A146983(n); R(n,4) = A054318(n+1); R(n,-4) = (-1)^n * A084159(n).

A350921 a(0) = 3, a(1) = 3, and a(n) = 6*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

3, 3, 11, 59, 339, 1971, 11483, 66923, 390051, 2273379, 13250219, 77227931, 450117363, 2623476243, 15290740091, 89120964299, 519435045699, 3027489309891, 17645500813643, 102845515571963, 599427592618131, 3493720040136819, 20362892648202779, 118683635849079851, 691738922446276323

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001653, A011900, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350922, A350923, A350924, A350925, A350926.

G.f.: (3 - 18*x + 11*x^2)/((1 - x)*(1 - 6*x + x^2)). - Stefano Spezia, Jan 22 2022

a(n) = 2*A001653(n) + 1 = 4*A011900(n-1) - 1 for n >= 1. - Hugo Pfoertner, Jan 22 2022

A182432 Recurrence a(n)a(n-2) = a(n-1)(a(n-1) + 3) with a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 28, 217, 1705, 13420, 105652, 831793, 6548689, 51557716, 405913036, 3195746569, 25160059513, 198084729532, 1559517776740, 12278057484385, 96664942098337, 761041479302308, 5991666892320124, 47172293659258681, 371386682381749321

Offset: 0

Peter Bala, Apr 30 2012

The non-linear recurrence equation a(n)*a(n-2) = a(n-1)*(a(n-1) + r) with initial conditions a(0) = 1, a(1) = 1 + r has the solution a(n) = 1/2 + (1/2)*Sum_{k = 0..n} (2*r)^k*binomial(n+k,2*k) = 1/2 + b(n,2*r)/2, where b(n,x) are the Morgan-Voyce polynomials of A085478. The recurrence produces sequences A101265 (r = 1), A011900 (r = 2) and A054318 (r = 4), as well as signed versions of A133872 (r = -1), A109613 (r = -2), A146983 (r = -3) and A084159(r = -4).
Also the indices of centered pentagonal numbers (A005891) which are also centered triangular numbers (A005448). - Colin Barker, Jan 01 2015
Also positive integers y in the solutions to 3*x^2 - 5*y^2 - 3*x + 5*y = 0. - Colin Barker, Jan 01 2015

Seiichi Manyama, Table of n, a(n) for n = 0..1116 (first 201 terms from Vincenzo Librandi)
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Eric Weisstein's World of Mathematics, Morgan-Voyce Polynomials
Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A011900, A054318, A070997, A101265, A109613, A133872, A146983, A211955.

I:=[1, 4, 28]; [n le 3 select I[n] else 9*Self(n-1)-9*Self(n-2)+Self(n-3): n in [1..25]]; // Vincenzo Librandi, May 18 2012

RecurrenceTable[{a[0]==1,a[1]==4,a[n]==(a[-1+n] (3+a[-1+n]))/a [-2+n]}, a[n],{n,30}] (* or *) LinearRecurrence[{9,-9,1},{1,4,28},30] (* Harvey P. Dale, May 14 2012 *)

Vec((1-5*x+x^2)/((1-x)*(1-8*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015

a(n) = 1/2 + (1/2)*Sum_{k = 0..n} 6^k*binomial(n+k,2*k).
a(n) = R(n,3) where R(n,x) denotes the row polynomials of A211955.
a(n) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(5/2) and T(n,x) the n-th Chebyshev polynomial of the first kind.
Recurrence equation: a(n) = 8*a(n-1) - a(n-2) - 3 with a(0) = 1 and a(1) = 4.
O.g.f.: (1 - 5*x + x^2)/((1 - x)*(1 - 8*x + x^2)) = 1 + 4*x + 28*x^2 + ....
Sum_{n >= 0} 1/a(n) = sqrt(5/3); 5 - 3*(Sum_{k = 0..2*n} 1/a(k))^2 = 2/A070997(n)^2.
a(0) = 1, a(1) = 4, a(2) = 28, a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3). - Harvey P. Dale, May 14 2012

A098602 a(n) = A001652(n) * A046090(n).

Original entry on oeis.org

0, 12, 420, 14280, 485112, 16479540, 559819260, 19017375312, 646030941360, 21946034630940, 745519146510612, 25325704946729880, 860328449042305320, 29225841562491651012, 992818284675673829100, 33726595837410418538400, 1145711440187278556476512

Offset: 0

Charlie Marion, Oct 26 2004

From Ron Knott, Nov 25 2013: (Start)
a(n) = 2*r*(r+1) which is also of form s(s+1) where the s is in A053141.
a(n) is an oblong number (A002378) which is twice another oblong number. (End)
2*a(n)+1 and 4*a(n)+1 are both square. - Paul Cleary, Jun 23 2014

			a(1) = 12 = 2(2*3) = 3*4, a(2) = 420 = 2(14*15) = 20*21.

Reinhard Zumkeller, Table of n, a(n) for n = 0..255
Nikola Adžaga, Andrej Dujella, Dijana Kreso, Petra Tadić, On Diophantine m-tuples and D(n)-sets, 2018.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A002378, A053141.

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(12*x/((1-x)*(x^2-34*x+1)))); // G. C. Greubel, Jul 15 2018

2*Table[ Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Ray Chandler, Nov 10 2004, copied incorrect program from A029549, revised Jul 09 2015 *)
RecurrenceTable[{a[n+3] == 35 a[n+2] - 35 a[n+1] + a[n], a[1] == 0, a[2] == 12, a[3] == 420}, a, {n, 1, 10}] (* Ron Knott, Nov 25 2013 *)
LinearRecurrence[{35, -35, 1}, {0, 12, 420}, 25] (* T. D. Noe, Nov 25 2013 *)
Table[(LucasL[4*n+2, 2] - 6)/16, {n,0,30}] (* G. C. Greubel, Jul 15 2018 *)

concat(0, Vec(12*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Mar 02 2016

{a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n};
for(n=0, 30, print1(round((Q(4*n+2) - 6)/16), ", ")) \\ G. C. Greubel, Jul 15 2018

a(n) = 2*A029549(n) = 2*A001109(n)*A001109(n+1).
a(n) = (A001653(n)^2 - 1)/2.
a(n) = A053141(n)^2 + A011900(n)^2 - 1.
For n>0, a(n) = A053141(2n) - 2*A001109(n-1)^2.
For n>0, a(n) = 3*(A001542(n)^2 - A001542(n-1)^2).
For n>0, a(n) = A053141(2n-1) + 2*(A001653(2n-1) - A001109(n-1)^2).
a(n+1) + a(n) = 3*A001542(n+1)^2.
a(n+1) - a(n) = A001542(2*n).
a(n+1)*a(n) = 4*(A001109(n)^4 - A001109(n)^2) = 4*A001110(n)*(A001110(n) - 1).
From Ron Knott, Nov 25 2013: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
G.f.: 12*x / ((1-x)*(x^2-34*x+1)). (End)
a(n) = (-6 + (3-2*sqrt(2))*(17+12*sqrt(2))^(-n)+(3+2*sqrt(2))*(17+12*sqrt(2))^n)/16. -  Colin Barker, Mar 02 2016

More terms from Ray Chandler, Nov 10 2004

Corrected by Bill Lam (bill_lam(AT)myrealbox.com), Feb 27 2006

A111647 a(n) = A001541(n)A001653(n+1)A002315(n).

Original entry on oeis.org

1, 105, 20213, 3998709, 791704585, 156753394977, 31036379835581, 6145046450172525, 1216688160731724433, 240898110778299543129, 47696609245941810082565, 9443687732585695622131557

Offset: 0

Charlie Marion, Aug 24 2005

			a(1) = 105 = 3*5*7.

G. C. Greubel, Table of n, a(n) for n = 0..434
Index entries for linear recurrences with constant coefficients, signature (204,-1190,204,-1).

Cf. A001541, A001653, A002315, A111648, A111649.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)))); // G. C. Greubel, Jul 15 2018

CoefficientList[Series[(1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)), {x, 0, 30}], x] (* G. C. Greubel, Jul 15 2018 *)

x='x+O('x^30); Vec((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1))) \\ G. C. Greubel, Jul 15 2018

2*a(n) = A001109(3*n+1) + A001109(n+1).
a(n) = sqrt(A011900(2*n)*A046090(2*n)*A001109(2*n+1)).
a(n) = A001541(3*n) + 2*A001109(n)*A001541(n-1)*A001541(n).
For n>0, a(n) = A001652(3*n) - A053141(2*n)*A002315(n-1) - A001652(n-1).
G.f.: (1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
2*a(n) = A001109(n+1) + A097731(n) + 6*A097731(n-1). - R. J. Mathar, Jan 31 2024

A107118 Numbers that are both centered triangular numbers (A005448) and centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 19, 631, 21421, 727669, 24719311, 839728891, 28526062969, 969046412041, 32919051946411, 1118278719765919, 37988557420094821, 1290492673563457981, 43838762343737476519, 1489227427013510743651, 50589893756115627807601, 1718567160280917834714769

Offset: 1

Richard Choulet, Sep 18 2007

The centered hexagonal numbers are given by 3*p^2 - 3*p + 1 while the centered triangular numbers are given by (3*r^2 + 3*r + 2)/2. A natural number is both of the above numbers if and only if there exist numbers p and r such that 2*(2p-1)^2 = (2*r+1)^2+1. The Diophantine equation X^2 = 2*Y^2 - 1 has the following solutions: X is given by 1, 7, 41, 239, ..., i.e., A002315, and Y is given by A001653. The first equation gives r with 0, 3, 20, 119, 6906, i.e., A001652, and p with 1, 3, 15, 85, 493, ..., i.e., A011900.

Colin Barker, Table of n, a(n) for n = 1..654
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A003215 (Centered hexagonal numbers), A005448 (Centered triangular numbers).

Cf. A001652, A001653, A002315, A011900.

a[n_] := 17*n - 7 + Sqrt[288*n^2 - 252*n + 45]; NestList[a, 1, 20] (* Stefan Steinerberger, Sep 18 2007 *)
LinearRecurrence[{35,-35,1},{1,19,631},30] (* Harvey P. Dale, Jan 16 2016 *)

Vec(-x*(x^2-16*x+1)/((x-1)*(x^2-34*x+1)) + O(x^100)) \\ Colin Barker, Jan 02 2015

a(n+2) = 34*a(n+1) - a(n) - 14.
a(n+1) = 17*a(n) - 7 + sqrt(288*a(n)^2 - 252*a(n) + 45).
G.f.: h(z)=(z*(1-16*z+z^2))/((1-z)*(1-34*z+z^2)).
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). - Colin Barker, Jan 02 2015
a(n) = (14+(9+6*sqrt(2))*(17+12*sqrt(2))^(-n)+(9-6*sqrt(2))*(17+12*sqrt(2))^n)/32. - Colin Barker, Mar 02 2016

More terms from Stefan Steinerberger, Sep 18 2007

A227026 Numbers k such that k!/m! is a triangular number for some m < k-1.

Original entry on oeis.org

3, 5, 6, 7, 11, 14, 15, 58, 85, 493, 638, 2871, 16731, 97513, 568345, 3312555, 19306983, 112529341, 655869061, 3822685023, 22280241075, 129858761425, 756872327473, 4411375203411, 25711378892991, 149856898154533, 873430010034205, 5090723162050695

Offset: 1

Alex Ratushnyak, Jun 27 2013

A011900 is a subsequence, except A011900(0)=1.

According to Melissen's comment in A097571, m > k-7.

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000217, A011900, A097571.

CoefficientList[Series[(3 - 16 x - 8 x^2 - 3 x^3 - x^4 - 20 x^5 - 13 x^6 + 40 x^7 - 230 x^8 + 289 x^9 - 2276 x^10 + 1771 x^11 + 607 x^12 - 145 x^13) / ((1 - x) (1 - 6 x + x^2)), {x, 0, 30}], x] (* Bruno Berselli, Jun 28 2013 *)

a(n) = 7a(n-1) - 7a(n-2) + a(n-3) for n > 14. - Charles R Greathouse IV, Jun 28 2013

G.f.: x * (3 -16*x -8*x^2 -3*x^3 -x^4 -20*x^5 -13*x^6 +40*x^7 -230*x^8 +289*x^9 -2276*x^10 +1771*x^11 +607*x^12 -145*x^13) / ((1-x)*(1-6*x+x^2)). - Bruno Berselli, Jun 28 2013

A355182 a(n) = t(n) - s(n) where s(n) = n*(n-1)/2 is the sum of the first n nonnegative integers and t(n) is the smallest sum of consecutive integers starting from n such that t(n) > s(n).

Original entry on oeis.org

1, 1, 4, 3, 1, 6, 3, 10, 6, 1, 10, 4, 15, 8, 21, 13, 4, 19, 9, 26, 15, 3, 22, 9, 30, 16, 1, 24, 8, 33, 16, 43, 25, 6, 35, 15, 46, 25, 3, 36, 13, 48, 24, 61, 36, 10, 49, 22, 63, 35, 6, 49, 19, 64, 33, 1, 48, 15, 64, 30, 81, 46, 10, 63, 26, 81, 43, 4, 61, 21, 80, 39, 100, 58, 15, 78, 34, 99

Offset: 1

Andrea La Rosa, Jun 23 2022

nonn

Record high values of a(n)/n approach sqrt(2) and occur at values of n that are terms of A011900; a(A011900(k)) = A046090(k). - Jon E. Schoenfield, Jun 23 2022

It appears that the sequence 1,2,4,5,6,8,... (the largest integer in the t(n) sum) is A288998. - Michel Marcus, Jun 24 2022

			a(6) = -s(6) + t(6):
s(6) is the sum of the first 6 nonnegative integers = 6*5 / 2 = 15.
t(6) is the smallest sum k of consecutive integers starting from n = 6 such that k > s(6) = 15.
The first few sets of consecutive integers starting from 6 are
  {6}, whose elements add up to 6,
  {6, 7}, whose elements add up to 13,
  {6, 7, 8}, whose elements add up to 21,
  {6, 7, 8, 9}, whose elements add up to 30,
  ...
the smallest sum that is > 15 is 21, so t(6) = 21, so a(6) = -15 + 21 = 6.

Cf. A000217, A001477, A093001.

Cf. A288998.

function a(n) {
    var sum = 0;
    for (var i = 0; i < n; i++)
        sum -= i;
    while (sum <= 0)
        sum += i++;
    return sum;
}

a(n) = my(t=0, s=n*(n-1)/2, k=n); until (t > s, t += k; k ++); t-s; \\ Michel Marcus, Jun 24 2022

from math import isqrt
def A355182(n): return ((m:=(isqrt(((k:=n*(n-1))<<3)+1)+1)>>1)*(m+1)>>1)-k # Chai Wah Wu, Jul 14 2022

From Jon E. Schoenfield, Jun 23 2022: (Start)
a(n) = t(n) - s(n) where
   s(n) = n*(n-1)/2,
      j = floor(sqrt(8*n^2 - 8*n + 1)),
      m = ceiling(j/2) - n + 1, and
   t(n) = (m*(m + 2*n - 1))/2. (End)

A358682 Numbers k such that 8k^2 + 8k - 7 is a square.

Original entry on oeis.org

1, 7, 43, 253, 1477, 8611, 50191, 292537, 1705033, 9937663, 57920947, 337588021, 1967607181, 11468055067, 66840723223, 389576284273, 2270616982417, 13234125610231, 77134136678971, 449570694463597, 2620290030102613, 15272169486152083, 89012726886809887, 518804191834707241

Offset: 1

Stefano Spezia, Nov 26 2022

a(n) is the n-th almost cobalancing number of second type (see Tekcan and Erdem).

			a(2) = 7 is a term since 8*7^2 + 8*7 - 7 = 441 = 21^2.

Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A002315, A006452, A077443, A077446, A106328, A106329, A216134.

Mathematica
```
LinearRecurrence[{7,-7,1},{1,7,43},24]
```

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) for n > 3.
a(n) = (3*(3 - 2*sqrt(2))^n*(2 + sqrt(2)) + 3*(2 - sqrt(2))*(3 + 2*sqrt(2))^n - 4)/8.
O.g.f.: x*(1 + x^2)/((1 - x)*(1 - 6*x + x^2)).
E.g.f.: (3*(2 + sqrt(2))*(cosh(3*x - 2*sqrt(2)*x) + sinh(3*x - 2*sqrt(2)*x)) + 3*(2 - sqrt(2))*(cosh(3*x + 2*sqrt(2)*x) + sinh(3*x + 2*sqrt(2)*x)) - 4*(cosh(x) + sinh(x)) - 8)/8.
a(n) = 3*A011900(n) - 2 = 6*A053142(n) + 1. - Hugo Pfoertner, Nov 26 2022

A309117 Number of perfect matchings on a triangular lattice of width 4 and length n.

Original entry on oeis.org

1, 1, 5, 15, 56, 203, 749, 2777, 10293, 38240, 141997, 527593, 1960029, 7282483, 27057400, 100531559, 373522965, 1387822193, 5156442953, 19158736256, 71184183353, 264484479633, 982690786037, 3651182836279, 13565952140920, 50404229548515, 187276671274621

Offset: 0

Sergey Perepechko, Jul 13 2019

S. N. Perepechko, Number of perfect matchings on triangular lattices of fixed width, DIMA'2015 slides.

Cf. A011900, A307349.

G.f.: (1-z)*(1+z)*(1-z-5*z^2-z^3+z^4)/((1+z-3*z^2-3*z^3+z^4)*(1-3*z-3*z^2+z^3+z^4)).

cp's OEIS Frontend

A211955 Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A350921 a(0) = 3, a(1) = 3, and a(n) = 6*a(n-1) - a(n-2) - 4 for n >= 2.

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

A182432 Recurrence a(n)*a(n-2) = a(n-1)*(a(n-1) + 3) with a(0) = 1, a(1) = 4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A098602 a(n) = A001652(n) * A046090(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

Extensions

A111647 a(n) = A001541(n)*A001653(n+1)*A002315(n).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A107118 Numbers that are both centered triangular numbers (A005448) and centered hexagonal numbers (A003215).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A227026 Numbers k such that k!/m! is a triangular number for some m < k-1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A355182 a(n) = t(n) - s(n) where s(n) = n*(n-1)/2 is the sum of the first n nonnegative integers and t(n) is the smallest sum of consecutive integers starting from n such that t(n) > s(n).

A182432 Recurrence a(n)a(n-2) = a(n-1)(a(n-1) + 3) with a(0) = 1, a(1) = 4.

A111647 a(n) = A001541(n)A001653(n+1)A002315(n).

A358682 Numbers k such that 8k^2 + 8k - 7 is a square.