Displaying 1-10 of 20 results found.
Numbers k such that 8*k^2 + 8*k - 7 is a square.
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1
1, 7, 43, 253, 1477, 8611, 50191, 292537, 1705033, 9937663, 57920947, 337588021, 1967607181, 11468055067, 66840723223, 389576284273, 2270616982417, 13234125610231, 77134136678971, 449570694463597, 2620290030102613, 15272169486152083, 89012726886809887, 518804191834707241
COMMENTS
a(n) is the n-th almost cobalancing number of second type (see Tekcan and Erdem).
FORMULA
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.
EXAMPLE
a(2) = 7 is a term since 8*7^2 + 8*7 - 7 = 441 = 21^2.
MATHEMATICA
LinearRecurrence[{7, -7, 1}, {1, 7, 43}, 24]
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).
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1
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
COMMENTS
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
FORMULA
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)
EXAMPLE
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.
PROG
(JavaScript)
function a(n) {
var sum = 0;
for (var i = 0; i < n; i++)
sum -= i;
while (sum <= 0)
sum += i++;
return sum;
}
(PARI) 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 (Python)
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
a(0) = 3, a(1) = 3, and a(n) = 6*a(n-1) - a(n-2) - 4 for n >= 2.
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9
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
COMMENTS
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.
FORMULA
G.f.: (3 - 18*x + 11*x^2)/((1 - x)*(1 - 6*x + x^2)). - Stefano Spezia, Jan 22 2022
Number of perfect matchings on a triangular lattice of width 4 and length n.
+0
0
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
FORMULA
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)).
Numbers n such that n!/m! is a triangular number for some m < n-1.
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1
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
COMMENTS
According to Melissen's comment in A097571, m > n-7.
FORMULA
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
MATHEMATICA
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 *)
Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.
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9
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
COMMENTS
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).
FORMULA
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..inf} 1/R(k,x) = sqrt((x+2)/x) for either x > 0 or x <= -2.
EXAMPLE
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
CROSSREFS
Cf. A011900, A084159, A085478, A101265 (row sums), A109613, A112373, A123519, A133872 (alt row sums), A146983, A182432, A204021, A208513, A211956, A211957.
Recurrence a(n)*a(n-2) = a(n-1)*(a(n-1)+3) with a(0) = 1, a(1) = 4.
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7
1, 4, 28, 217, 1705, 13420, 105652, 831793, 6548689, 51557716, 405913036, 3195746569, 25160059513, 198084729532, 1559517776740, 12278057484385, 96664942098337, 761041479302308, 5991666892320124, 47172293659258681, 371386682381749321
COMMENTS
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
FORMULA
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 Chebyshev polynomial of the first kind.
Recurrence equation: a(n) = 8*a(n-1) - a(n-2)-3 with a(0) = 1, 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_{n=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
MATHEMATICA
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 *)
PROG
(Magma) 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 (PARI) Vec((1-5*x+x^2)/((1-x)*(1-8*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015
1, 1, 3, 6, 15, 35, 85, 204, 493, 1189, 2871, 6930, 16731, 40391, 97513, 235416, 568345, 1372105, 3312555, 7997214, 19306983, 46611179, 112529341, 271669860, 655869061, 1583407981, 3822685023, 9228778026, 22280241075, 53789260175
COMMENTS
For n >= 1, a(n) is also the edge cover number and edge cut count of the n-Pell graph. - Eric W. Weisstein, Aug 01 2023 Also the independence number, Lovasz number, and Shannon capacity of the n-Pell graph. - Eric W. Weisstein, Aug 01 2023 Floretion Algebra Multiplication Program, FAMP Code: -2jbasejseq[B*C], B = - .5'i + .5'j - .5i' + .5j' - 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj'; C = + .5'i + .5i' + .5'ii' + .5e
REFERENCES
C. Dement, Floretion Integer Sequences (work in progress).
LINKS
Eric Weisstein's World of Mathematics, Edge Cut. Eric Weisstein's World of Mathematics, Pell Graph.
FORMULA
G.f.: y/(y^2-1) where y=x/(x^2+x-1) if offset=1. - Michael Somos, Sep 09 2006 G.f.: (-1+x+x^2)/((1-x)*(x+1)*(x^2+2*x-1)).
MAPLE
seq(iquo(fibonacci(n, 2), 1)-iquo(fibonacci(n, 2), 2), n=1..30); # Zerinvary Lajos, Apr 20 2008 with(combinat):seq(ceil(fibonacci(n, 2)/2), n=1..30); # Zerinvary Lajos, Jan 12 2009
MATHEMATICA
Ceiling[Fibonacci[Range[20], 2]/2]
Table[(1 + (-1)^n + 2 Fibonacci[n + 1, 2])/4, {n, 0, 20}] // Expand
CoefficientList[Series[-(-1 + x + x^2)/(1 - 2 x - 2 x^2 + 2 x^3 + x^4), {x, 0, 20}], x]
LinearRecurrence[{2, 2, -2, -1}, {1, 1, 3, 6}, 20]
PROG
(PARI) {a(n)=local(y); if(n<0, 0, n++; y=x/(x^2+x-1)+x*O(x^n); polcoeff( y/(y^2-1), n))} /* Michael Somos, Sep 09 2006 */
1, 105, 20213, 3998709, 791704585, 156753394977, 31036379835581, 6145046450172525, 1216688160731724433, 240898110778299543129, 47696609245941810082565, 9443687732585695622131557
FORMULA
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
MATHEMATICA
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 *)
PROG
(PARI) 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 (Magma) m:=30; R<x>:=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
Numbers that are both centered triangular numbers ( A005448) and centered hexagonal numbers ( A003215).
+0
1
1, 19, 631, 21421, 727669, 24719311, 839728891, 28526062969, 969046412041, 32919051946411, 1118278719765919, 37988557420094821, 1290492673563457981, 43838762343737476519, 1489227427013510743651, 50589893756115627807601, 1718567160280917834714769
COMMENTS
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.
FORMULA
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
MATHEMATICA
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 *)
PROG
(PARI) Vec(-x*(x^2-16*x+1)/((x-1)*(x^2-34*x+1)) + O(x^100)) \\ Colin Barker, Jan 02 2015
CROSSREFS
Cf. A003215 (Centered hexagonal numbers), A005448 (Centered triangular numbers).
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