A015518 -id:A015518 - cp's OEIS frontend

A360035 Expansion of e.g.f. xexp(x)cosh(x)*sinh(x).

0, 0, 2, 6, 28, 100, 366, 1274, 4376, 14760, 49210, 162382, 531444, 1727180, 5580134, 17936130, 57395632, 182948560, 581130738, 1840247318, 5811307340, 18305618100, 57531942622, 180441092746, 564859072968, 1765184603000, 5507375961386, 17157594341214, 53379182394916

Offset: 0

Enrique Navarrete, Jan 22 2023

a(n) is the number of ordered set partitions of an n-set into 3 sets such that the first set has an even number of elements, the second set has an odd number of elements, and an element is selected from the third (see example).

			For n = 3, the 6 cases are (where the element selected from the third set is in parenthesis):
{}, {1}, {(2), 3}
{}, {1}, {2, (3)}
{}, {2}, {(1), 3}
{}, {2}, {1, (3)}
{}, {3}, {(1), 2}
{}, {3}, {1, (2)}.

Index entries for linear recurrences with constant coefficients, signature (4,2,-12,-9).

A015518 is the case of no element selected in the 3rd set.

Cf. A360023, A360036.

a(n) = n*A015518(n-1) for n > 0.
a(n) = n*(3^(n-1) - (-1)^(n-1))/4.
G.f.: 2*x^2*(1 - x)/((1 + x)^2*(1 - 3*x)^2). - Stefano Spezia, Jan 23 2023

A366502 Let q = A246655(n) for n >= 2, then a(n) = (q - Kronecker(-4,q))/4 - 1.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 7, 8, 9, 10, 11, 11, 12, 14, 14, 15, 16, 17, 17, 19, 19, 20, 21, 23, 24, 25, 26, 26, 27, 29, 30, 31, 31, 32, 33, 34, 36, 37, 38, 40, 41, 41, 42, 44, 44, 47, 47, 48, 49, 52

Offset: 2

Jianing Song, Oct 12 2023

If q is not a power of 2, then a(n) is the number of pairs of consecutive nonzero squares in the finite field F_q. In other words, a(n) is the number of solutions to x^((q-1)/2) = (x+1)^((q-1)/2) = 1 in F_q. This can be proved by generalizing the argument of Jack D'Aurizio to the Math Stack Exchange question "Existence of Consecutive Quadratic residues" to the case of F_q.

			a(5) = 1 because there is one pair of consecutive nonzero squares in the finite field F_q with q = A246655(5) = 7, namely {1, 2}.
a(7) = 1 because there is one pair of consecutive nonzero squares in the finite field F_q with q = A246655(7) = 9, namely {1, 2} (note that 2 = -1 = i^2 in F_9 = F_3(i)).

Jianing Song, Table of n, a(n) for n = 2..10000
Mathematics Stack Exchange, Existence of Consecutive Quadratic residues.

Cf. A246655, A101455 ({kronecker(-4,n)}), A024698.

A015518(n)-1 and A003463(n)-1 are respectively the number of consecutive nonzero squares in F_{3^n} and in F_{5^n}.

lim_A366502(N) = for(n=3, N, if(isprimepower(n), print1((n - kronecker(-4,n))/4 - 1, ", ")))

from sympy import primepi, integer_nthroot, kronecker_symbol
def A366502(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return int(n+x-sum(primepi(integer_nthroot(x,k)[0]) for k in range(1,x.bit_length())))
    return ((m:=bisection(f,n,n))-kronecker_symbol(-4,m)>>2)-1 # Chai Wah Wu, Jan 19 2025

A093379 Expansion of x(1-2x-2x^2)/((1+x)(1-2x)(1-3x)).

Original entry on oeis.org

0, 1, 2, 5, 12, 31, 82, 225, 632, 1811, 5262, 15445, 45652, 135591, 404042, 1206665, 3609072, 10805371, 32372422, 97029885, 290914892, 872395151, 2616486402, 7848061105, 23541387112, 70618568931, 211844521982, 635511196325

Offset: 0

Paul Barry, Apr 28 2004

Binomial transform is A085280, with leading zero.

Index entries for linear recurrences with constant coefficients, signature (4,-1,-6).

Cf. A087432.

CoefficientList[Series[x (1-2x-2x^2)/((1+x)(1-2x)(1-3x)),{x,0,30}],x] (* or *) LinearRecurrence[{4,-1,-6},{0,1,2,5},40] (* Harvey P. Dale, May 10 2019 *)

a(n)=3^n/12-(-1)^n/12+2^n/3-0^n/3;
a(n)=4a(n-1)-a(n-2)-6a(n-3).
a(n)=A015518(n)/3+A000079(n)/3-0^n/3.

A109191 Triangle read by rows: T(n,k) is number of Grand Motzkin paths of length n having k hills (i.e., ud's starting at level 0). (A Grand Motzkin path is a path in the half-plane x>=0, starting at (0,0), ending at (n,0) and consisting of steps u=(1,1), d=(1,-1) and h=(1,0).).

Original entry on oeis.org

1, 1, 2, 1, 5, 2, 13, 5, 1, 34, 14, 3, 91, 40, 9, 1, 247, 114, 28, 4, 678, 327, 87, 14, 1, 1877, 942, 267, 48, 5, 5233, 2723, 815, 161, 20, 1, 14674, 7892, 2478, 528, 75, 6, 41349, 22924, 7512, 1706, 270, 27, 1, 117001, 66712, 22718, 5452, 941, 110, 7, 332260

Offset: 0

Emeric Deutsch, Jun 21 2005

Row n contains 1 + floor(n/2) terms.
Row sums yield the central trinomial coefficients (A002426).
T(n,0) = A109192(n).
Sum_{k=0..floor(n/2)} k*T(n,k) = A015518(n-1).

			T(3,1)=2 because we have hud and udh, where u=(1,1),d=(1,-1), h=(1,0).
Triangle begins:
   1;
   1;
   2,  1;
   5,  2;
  13,  5,  1;
  34, 14,  3;
  91, 40,  9,  1;

Cf. A001006, A002426, A015518, A109192.

M:=(1-z-sqrt(1-2*z-3*z^2))/2/z^2: G:=1/(1-z+z^2-t*z^2-2*z^2*M): Gser:=simplify(series(G,z=0,16)): P[0]:=1: for n from 1 to 14 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 14 do seq(coeff(t*P[n],t^k),k=1..1+floor(n/2)) od;

G.f.: 1/(1 - z + z^2 - tz^2 - 2z^2*M), where M = 1 + zM + z^2*M^2 = (1 - z - sqrt(1 - 2z - 3z^2))/(2z^2) is the g.f. of the Motzkin numbers (A001006).

A133443 a(n) = Sum_{k=0..n} C(n,floor(k/2))(-1)^k3^(n-k).

Original entry on oeis.org

1, 2, 8, 24, 84, 272, 920, 3040, 10180, 33840, 112968, 376224, 1254696, 4181088, 13939248, 46459584, 154873860, 516229040, 1720795880, 5735921440, 19119861304, 63732624672, 212442552528, 708140901184, 2360471473384, 7868234639072, 26227455730640

Offset: 0

Philippe Deléham, Nov 26 2007, Dec 07 2007

nonn

Hankel transform is 4^n. Second binomial transform is A076035.

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

Cf. A000108, A015518, A053121, A076035, A127362.

Table[Sum[Binomial[n,Floor[k/2]]*(-1)^k*3^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)

a(n) = Sum_{k=0..n} A053121(n,k)*A015518(k+1) = (-1)^n*A127362(n). G.f.: (1/sqrt(1-4*x^2))*(1-x*c(x^2))/(1-3*x*c(x^2)), where c(x) is the g.f. of Catalan numbers A000108.
Recurrence: 3*n*a(n) = 2*(5*n-3)*a(n-1) + 4*(3*n-1)*a(n-2) - 40*(n-2)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2*10^n/3^(n+1). - Vaclav Kotesovec, Oct 20 2012

More terms from Vincenzo Librandi, May 25 2013

A164984 Odd (Jacobsthal) triangle.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 5, 9, 5, 1, 11, 23, 19, 7, 1, 21, 57, 61, 33, 9, 1, 43, 135, 179, 127, 51, 11, 1, 85, 313, 493, 433, 229, 73, 13, 1, 171, 711, 1299, 1359, 891, 375, 99, 15, 1, 341, 1593, 3309, 4017, 3141, 1641, 573, 129, 17, 1

Offset: 1

Mark Dols, Sep 03 2009, Sep 06 2009

Alternate diagonal sums give A008619.

Diagonals sums give A097076. - Philippe Deléham, Oct 13 2013

			1
1,1
3,3,1
5,9,5,1
11,23,19,7,1
21,57,61,33,9,1
Pascal-like triangle based on a right-triangular sum (with the top multiplied by 2): For n=13 a(13)=2*a(3)+a(5)+a(8)+a(9)= 2+3+9+5=19.

Cf. A164981, A001045, A005408, A015518 (row sums), A008619

Cf. A008288, A007318.

Excel formula: C6=2*C4+C5+B5+B4 with C5=a(1)=1 and C6=a(2)

T(n,k) = T(n-1,k) + T(n-1,k-1) + 2*T(n-2,k) + T(n-2,k-1). - Philippe Deléham, Oct 13 2013

A228815 Symmetric triangle, read by rows, related to Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 2, 5, 5, 2, 3, 10, 14, 10, 3, 5, 20, 36, 36, 20, 5, 8, 38, 83, 106, 83, 38, 8, 13, 71, 182, 281, 281, 182, 71, 13, 21, 130, 382, 690, 834, 690, 382, 130, 21, 34, 235, 778, 1606, 2268, 2268, 1606, 778, 235, 34, 55, 420, 1546, 3586, 5780, 6750

Offset: 0

Philippe Deléham, Oct 30 2013

Triangles satisfying the same recurrence: A091533, A091562, A185081, A205575, A209137, A209138.

			Triangle begins :
0
1, 1
1, 2, 1
2, 5, 5, 2
3, 10, 14, 10, 3
5, 20, 36, 36, 20, 5
8, 38, 83, 106, 83, 38, 8
13, 71, 182, 281, 281, 182, 71, 13
21, 130, 382, 690, 834, 690, 382, 130, 21
34, 235, 778, 1606, 2268, 2268, 1606, 778, 235, 34
55, 420, 1546, 3586, 5780, 6750, 5780, 3586, 1546, 420, 55

Cf. A000045 (1st column), A001629 (2nd column), A008998, A152011, A261055 (3rd column).

G.f.: x*(1+y)/(1-x-x*y-x^2-x^2*y-x^2*y^2).
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = 0, T(1,0) = T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n.
Sum_{k = 0..n} T(n,k)*x^k = A000045(n), 2*A015518(n), 3*A015524(n), 4*A200069(n) for x = 0, 1, 2, 3 respectively.
Sum_{k = 0..floor(n/2)} T(n-k,k) = A008998(n+1).

A247584 a(n) = 5a(n-1) - 10a(n-2) + 10a(n-3) - 5a(n-4) + 3*a(n-5) with a(0) = a(1) = a(2) = a(3) = a(4) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 13, 43, 113, 253, 509, 969, 1849, 3719, 8009, 18027, 40897, 91257, 198697, 423777, 894081, 1886011, 4007301, 8594411, 18560081, 40181493, 86872293, 187197193, 402060793, 861827743, 1846685729, 3960390059, 8504658049, 18283290609, 39325827729

Offset: 0

Alexander Samokrutov, Sep 20 2014

a(n)/a(n-1) tends to 2.1486... = 1 + 2^(1/5), the real root of the polynomial x^5 - 5*x^4 + 10*x^3 - 10*x^2 + 5*x - 3.
If x^5 = 2 and n >= 0, then there are unique integers a, b, c, d, g  such that (1 + x)^n = a + b*x + c*x^2 + d*x^3 + g*x^4. The coefficient a is a(n) (from A052102). - Alexander Samokrutov, Jul 11 2015
If x=a(n), y=a(n+1), z=a(n+2), s=a(n+3), t=a(n+4) then x, y, z, s, t satisfies Diophantine equation (see link). - Alexander Samokrutov, Jul 11 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..48 from Alexander Samokrutov)
Alexander Samokrutov, Diophantine equation
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,3).

The following sequences belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A052101, A052102, A052103, A063727, A083098, A083099, A083100, A084057, A093406, A247344.

Cf. A005531.

[n le 5 select 1 else 5*Self(n-1) -10*Self(n-2) +10*Self(n-3) -5*Self(n-4) +3*Self(n-5): n in [1..40]]; // Vincenzo Librandi, Jul 11 2015

m:=50; S:=series( (1-x)^4/(1 -5*x +10*x^2 -10*x^3 +5*x^4 -3*x^5), x, m+1):
seq(coeff(S, x, j), j=0..m); # G. C. Greubel, Apr 15 2021

LinearRecurrence[{5,-10,10,-5,3}, {1,1,1,1,1}, 50] (* Vincenzo Librandi, Jul 11 2015 *)

makelist(sum(2^k*binomial(n,5*k), k, 0, floor(n/5)), n, 0, 50); /* Alexander Samokrutov, Jul 11 2015 */

Vec((1-x)^4/(1-5*x+10*x^2-10*x^3+5*x^4-3*x^5) + O(x^100)) \\ Colin Barker, Sep 22 2014

[sum(2^j*binomial(n, 5*j) for j in (0..n//5)) for n in (0..50)] # G. C. Greubel, Apr 15 2021

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + 3*a(n-5).
a(n) = Sum_{k=0...floor(n/5)} (2^k*binomial(n,5*k)). - Alexander Samokrutov, Jul 11 2015
G.f.: (1-x)^4/(1 -5*x +10*x^2 -10*x^3 +5*x^4 -3*x^5). - Colin Barker, Sep 22 2014

A055036 Min[x] composite zero site for sigma(x+6^n) - sigma(x) - 6^n.

Original entry on oeis.org

104, 125, 195, 415, 2743, 2935, 3535, 19735, 22645, 108703, 977353, 1921033, 2523433, 2425175, 4227575, 85969345, 32606935, 224917033, 1362833713, 716210677, 1557843865, 6226853857, 20369543065

Offset: 1

Labos Elemer, Jun 01 2000

			n = 6: d = 6^6 = 46656, a(n) = a(6) = 2935 because sigma(2935) + 46656 = 1 + 5 + 587 + 2935 + 46656 = sigma(2935 + 46656) = sigma(49591) = 1 + 101 + 491 + 49591 = 50184.

Subsequence of A054904.

Cf. A054902-A054905, A054799, A054982-A054985, A054987, A055009, A015913-A015518.

L = {}; Do[i = 1; While[ ! ((Plus @@ Divisors[i + 6^j] == 6^j + Plus @@ Divisors[i]) && ! PrimeQ[i]), i++ ]; L = Append[L, i], {j, 1, 11}]; L (from Vit Planocka)

a(n) = Min(x) solution for A000203(x+A000400(n)) = A000203(x) + A000400(n) Diophantine equation.

One more term from Vit Planocka (planocka(AT)mistral.cz), Sep 23 2003

a(12)-a(23) from Donovan Johnson, Nov 30 2008

A099091 Riordan array (1,2+3x).

Original entry on oeis.org

1, 0, 2, 0, 3, 4, 0, 0, 12, 8, 0, 0, 9, 36, 16, 0, 0, 0, 54, 96, 32, 0, 0, 0, 27, 216, 240, 64, 0, 0, 0, 0, 216, 720, 576, 128, 0, 0, 0, 0, 81, 1080, 2160, 1344, 256, 0, 0, 0, 0, 0, 810, 4320, 6048, 3072, 512, 0, 0, 0, 0, 0, 243, 4860, 15120, 16128, 6912, 1024, 0, 0, 0, 0, 0, 0

Offset: 0

Paul Barry, Sep 25 2004

Row sums are A015518(n+1). Diagonal sums are A002447. The Riordan array (1,s+tx) defines T(n,k)=binomial(k,n-k)s^k(t/s)^(n-k). The row sums satisfy a(n)=s*a(n-1)+t*a(n-2) and the diagonal sums satisfy a(n)=s*a(n-2)+t*a(n-3).

			Rows begin
1;
0,2;
0,3,4;
0,0,12,8;
0,0,9,36,16;

Number triangle T(n, k)=binomial(k, n-k)2^k*(3/2)^(n-k) Columns have g.f. (2x+3x^2)^k.

cp's OEIS Frontend

A360035 Expansion of e.g.f. x*exp(x)*cosh(x)*sinh(x).

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A366502 Let q = A246655(n) for n >= 2, then a(n) = (q - Kronecker(-4,q))/4 - 1.

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A093379 Expansion of x(1-2x-2x^2)/((1+x)(1-2x)(1-3x)).

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A109191 Triangle read by rows: T(n,k) is number of Grand Motzkin paths of length n having k hills (i.e., ud's starting at level 0). (A Grand Motzkin path is a path in the half-plane x>=0, starting at (0,0), ending at (n,0) and consisting of steps u=(1,1), d=(1,-1) and h=(1,0).).

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A133443 a(n) = Sum_{k=0..n} C(n,floor(k/2))*(-1)^k*3^(n-k).

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A164984 Odd (Jacobsthal) triangle.

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A228815 Symmetric triangle, read by rows, related to Fibonacci numbers.

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A247584 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + 3*a(n-5) with a(0) = a(1) = a(2) = a(3) = a(4) = 1.

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A360035 Expansion of e.g.f. xexp(x)cosh(x)*sinh(x).

A133443 a(n) = Sum_{k=0..n} C(n,floor(k/2))(-1)^k3^(n-k).

A247584 a(n) = 5a(n-1) - 10a(n-2) + 10a(n-3) - 5a(n-4) + 3*a(n-5) with a(0) = a(1) = a(2) = a(3) = a(4) = 1.