A011848 -id:A011848 - cp's OEIS frontend

A160636 Hankel transform of A114464.

1, 0, -1, -2, -8, 0, 128, 1024, 16384, 0, -4194304, -134217728, -8589934592, 0, 35184372088832, 4503599627370496, 1152921504606846976, 0, -75557863725914323419136, -38685626227668133590597632

Offset: 0

Paul Barry, May 21 2009

Hankel transform of A114464(n+1) is A160637.

G. C. Greubel, Table of n, a(n) for n = 0..115

Cf. A004524, A160637.

R:= RealField(); [Round(2^Floor(Binomial(n,2)/2)*((Sqrt(2)/2 -1/2)*Sin(3*Pi(R)*n/4+Pi(R)/4)+(Sqrt(2)/2+1/2)*Cos(Pi(R)*n/4+Pi(R)/4))): n in [0..50]]; // G. C. Greubel, May 03 2018

Table[Round[2^Floor[Binomial[n, 2]/2]*((Sqrt[2]-1)*Sin[(3*n+1)*Pi/4]/2 + (Sqrt[2]+1)*Cos[(n+1)*Pi/4]/2)], {n, 0, 50}] (* G. C. Greubel, May 03 2018 *)
a[ n_] := -Sign[Mod[n - 1, 4]]*(-1)^Quotient[n - 1, 4]*2^Quotient[n (n - 1), 4]; (* Michael Somos, Mar 14 2020 *)

for(n=0,50, print1(round(2^floor(binomial(n,2)/2)*((sqrt(2)-1)*sin((3*n+1)*Pi/4)/2 +(sqrt(2)+1)*cos((n+1)*Pi/4)/2)), ", ")) \\ G. C. Greubel, May 03 2018

A160636(n)=if(n%4!=1,(-1)^((n+2)\4)<<(binomial(n,2)\2),0) \\ M. F. Hasler, May 09 2018

a(n) = 2^floor(C(n,2)/2)*((sqrt(2)-1)*sin((3*n+1)*Pi/4)/2 +(sqrt(2)+1)*cos((n+1)*Pi/4)/2).
a(4k+1) = 0, a(n) = (-1)^floor((n+2)/4) * 2^A011848(n) if n !== 1 (mod 4), where A011848(n) = floor(C(n,2)/2). - M. F. Hasler, May 09 2018
a(n) = -a(2-n) * 2^A004524(n) for all n in Z. - Michael Somos, Mar 14 2020

Comment with an incorrect formula deleted by M. F. Hasler, May 09 2018

A174102 Triangle read by rows: T(n, m) = floor(binomial(n+1, m)* binomial(n+2, m)/(2*m+2)), 1 <= m <= n.

Original entry on oeis.org

1, 3, 3, 5, 10, 5, 7, 25, 25, 7, 10, 52, 87, 52, 10, 14, 98, 245, 245, 98, 14, 18, 168, 588, 882, 588, 168, 18, 22, 270, 1260, 2646, 2646, 1260, 270, 22, 27, 412, 2475, 6930, 9702, 6930, 2475, 412, 27, 33, 605, 4537, 16335, 30492, 30492, 16335, 4537, 605, 33

Offset: 1

Roger L. Bagula, Mar 07 2010

Row sums are {1, 6, 20, 64, 211, 714, 2430, 8396, 29390, 104004, 371448, 1337216, ...}.

			Triangle begins as:
   1;
   3,   3;
   5,  10,    5;
   7,  25,   25,    7;
  10,  52,   87,   52,   10;
  14,  98,  245,  245,   98,   14;
  18, 168,  588,  882,  588,  168,   18;
  22, 270, 1260, 2646, 2646, 1260,  270,  22;
  27, 412, 2475, 6930, 9702, 6930, 2475, 412, 27;

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

Cf. A166454.

Cf. A011848 (right diagonal).

[[Floor(Binomial(n+1, k)*Binomial(n+2, k)/(2*k+2)): k in [1..n]]: n in [1..12]]; // G. C. Greubel, Apr 13 2019

T[n_, k_] = Floor[Binomial[n+1, k]*Binomial[n+2, k]/(2*(k+1))];
Table[T[n, k], {n,1,12}, {k,1,n}]//Flatten (* modified by G. C. Greubel, Apr 13 2019 *)

{T(n,k) = (binomial(n+1,k)*binomial(n+2,k)/(2*k+2))\1};
for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Apr 13 2019

[[floor(binomial(n+1,k)*binomial(n+2,k)/(2*k+2)) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Apr 13 2019

T(n, m) = floor(binomial(n+1, m-1)*binomial(n+2, m-1)/(2*m)).

Partially edited by Jon E. Schoenfield, Dec 02 2013

Edited by G. C. Greubel, Apr 13 2019

A215099 a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is prime.

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 8, 10, 11, 13, 18, 24, 25, 29, 34, 38, 39, 41, 44, 48, 53, 55, 56, 58, 71, 73, 78, 84, 85, 89, 94, 102, 103, 109, 120, 124, 131, 133, 138, 144, 145, 149, 162, 164, 169, 173, 178, 180, 181, 187, 192, 196, 197, 201

Offset: 0

Alex Ratushnyak, Aug 03 2012

nonn

For n>0 and (n mod 4)<2, a(n) is odd.
Same definition, but k+a(n-2) is a
Fibonacci number: A006498 except first two terms,
Lucas number: A000045 except first two terms,
Pell number: A089928(n-1),
Jacobsthal number: A215095,
factorial: A215096,
square: A194274,
cube: A215097,
triangular number: A011848(n+2),
oblong number: A215098.
Example of a related sequence definition: a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is a cube.

Iain Fox, Table of n, a(n) for n = 0..10000

Cf. A062042: a(1) = 2, a(n) = least k>a(n-1) such that k+a(n-1) is a prime.

Cf. A073627, A073628.

first(n) = my(res = vector(n, i, i-1), k); for(x=3, n, k=res[x-1]+1; while(!isprime(k+res[x-2]), k++); res[x]=k); res \\ Iain Fox, Apr 22 2019 (corrected by Iain Fox, Apr 25 2019)

from sympy import prime
prpr = 0
prev = 1
for n in range(77):
    print(prpr, end=', ')
    b = c = 0
    while c<=prev:
        c = prime(b+1) - prpr
        b+=1
    prpr = prev
    prev = c

A231559 a(n) = floor( A000326(n)/2 ).

Original entry on oeis.org

0, 0, 2, 6, 11, 17, 25, 35, 46, 58, 72, 88, 105, 123, 143, 165, 188, 212, 238, 266, 295, 325, 357, 391, 426, 462, 500, 540, 581, 623, 667, 713, 760, 808, 858, 910, 963, 1017, 1073, 1131, 1190, 1250, 1312, 1376, 1441, 1507, 1575, 1645, 1716, 1788, 1862, 1938

Offset: 0

Bruno Berselli, Nov 11 2013

First trisection of A011865.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. pentagonal numbers: A000326.
Cf. A011848 for the triangular numbers: floor(A000217/2); A007590 for the squares: floor(A000290/2); A156859 for the hexagonal numbers: floor(A000384/2).
First differences: A047262.

Magma
```
[Floor(n*(3*n-1)/4): n in [0..60]];
```

Table[Floor[n (3 n - 1)/4], {n, 0, 60}]
CoefficientList[Series[x^2(2+x^2)/((1+x^2)(1-x)^3),{x,0,70}],x] (* or *) LinearRecurrence[{3,-4,4,-3,1},{0,0,2,6,11},70] (* Harvey P. Dale, Jan 28 2022 *)

G.f.: x^2*(2 + x^2)/((1 + x^2)*(1 - x)^3).

a(n) = ( n*(3*n-1) + i^(n*(n+1)) - 1 )/4, where i=sqrt(-1).

A373584 a(n) is equal to the number of shaded cells in a regular hexagon with side n drawn on a hexagonal grid.

Original entry on oeis.org

1, 7, 13, 19, 31, 49, 67, 85, 109, 139, 169, 199, 235, 277, 319, 361, 409, 463, 517, 571, 631, 697, 763, 829, 901, 979, 1057, 1135, 1219, 1309, 1399, 1489, 1585, 1687, 1789, 1891, 1999, 2113, 2227, 2341, 2461, 2587, 2713, 2839, 2971, 3109, 3247, 3385, 3529

Offset: 1

Nicolay Avilov, Jun 10 2024

On a hexagonal grid, cells are colored as follows: one cell and all those located along three straight lines passing through the center of the original cell and forming six 60° angles between each other are painted. In each of these corners, cells are painted over so that a V-shaped arrangement of cells repeats ad infinitum. The number of shaded cells in regular hexagons centered on the starting cell determines the sequence a(n).

			a(3) = 19 - 6*1 = 13;
a(4) = 37 - 6*3 = 19.
                                                   o . o . o
                                 o . . o          . o . . o .
                   o . o        . o . o .        o . o . o . o
         o o      . o o .      . . o o . .      . . . o o . . .
   o    o o o    o o o o o    o o o o o o o    o o o o o o o o o
         o o      . o o .      . . o o . .      . . . o o . . .
                   o . o        . o . o .        o . o . o . o
                                 o . . o          . o . . o .
                                                   o . o . o
   1      7         13             19                 31

Paolo Xausa, Table of n, a(n) for n = 1..10000
Nicolay Avilov, Members of the sequence a(1) - a(7).
Nicolay Avilov, Problem 2663. Snowflakes (in Russian).
Nicolay Avilov, Illustration a(13) and a(16)
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. A003215, A011848, A054925.

Table[6*Ceiling[n*(n - 1)/4] + 1, {n, 100}] (* Paolo Xausa, Jul 01 2024 *)

a(n+4) = a(n) + 12*n + 18.
a(n) = 6*ceiling(n*(n - 1)/4) + 1.
a(n) = A003215(n) - 6*A011848(n+1).
a(n) = 6*A054925(n) + 1.
G.f.: (1 + 4*x - 4*x^2 + 4*x^3 + x^4)/((1 - x)^3*(1 + x^2)). - Stefano Spezia, Jun 11 2024
E.g.f.: (exp(x)*(5 + 6*x + 3*x^2) - 3*cos(x) + 3*sin(x))/2. - Stefano Spezia, Aug 31 2025

A024174 a(n) is floor((4th elementary symmetric function of 1,2,..,n)/(3rd elementary symmetric function of 1,2,...,n)).

Original entry on oeis.org

0, 0, 1, 2, 3, 4, 6, 8, 10, 13, 16, 19, 22, 25, 29, 33, 37, 42, 47, 52, 57, 62, 68, 74, 80, 87, 94, 101, 108, 115, 123, 131, 139, 148, 157, 166, 175, 184, 194, 204, 214, 225, 236, 247, 258, 269, 281, 293, 305, 318, 331, 344, 357, 370, 384, 398, 412, 427, 442

Offset: 3

Clark Kimberling

nonn

			G.f. = x^5 + 2*x^6 + 3*x^7 + 4*x^8 + 6*x^9 + 8*x^10 + 10*x^11 + 13*x^12 + ...

Ivan Neretin, Table of n, a(n) for n = 3..10000

Cf. A000915, A001303, A011848, A000217.

Table[Floor[(n - 3) (15 n^3 + 15 n^2 - 10 n - 8)/(120 n (n + 1))], {n, 3, 45}] (* Ivan Neretin, Nov 25 2016 *)
Insert[Table[Floor[1/8 (-2 - 3 n + n^2)], {n, 4, 45}], 0, 1] (* Ralf Steiner, Oct 27 2021 *)

{a(n) = if( n<4, 0, (n-3) * (15*n^3 + 15*n^2 - 10*n - 8) \ (120 * n * (n+1)))}; /* Michael Somos, Nov 25 2016 */

Empirical g.f.: x^5*(x^7-2*x^6+2*x^5-2*x^4+x^3-x^2+x-1) / ((x-1)^3*(x^2+1)*(x^4+1)). - Colin Barker, Aug 16 2014
a(n) = floor( A000915(n-3)/A001303(n-2) ). - R. J. Mathar, Sep 23 2016
a(n) = floor((n - 3)*(15n^3 + 15n^2 - 10n - 8)/(120*n*(n + 1))). - Ivan Neretin, Nov 25 2016
a(n) = floor((A000217(n-2)/2 - 1)/2) = floor((n^2 - 3*n - 2)/8), n >= 4. - Ralf Steiner, Oct 25 2021

Offset set to 3 by R. J. Mathar, Sep 23 2016

A104563 A floretion-generated sequence relating to centered square numbers.

Original entry on oeis.org

0, 1, 3, 5, 8, 13, 19, 25, 32, 41, 51, 61, 72, 85, 99, 113, 128, 145, 163, 181, 200, 221, 243, 265, 288, 313, 339, 365, 392, 421, 451, 481, 512, 545, 579, 613, 648, 685, 723, 761, 800, 841, 883, 925, 968, 1013, 1059, 1105, 1152, 1201, 1251

Offset: 0

Creighton Dement, Mar 15 2005

Floretion Algebra Multiplication Program, FAMP Code: a(n) = 1vesrokseq[A*B] with A = - .5'i - .5i' + .5'ii' + .5e, B = + .5'ii' - .5'jj' + .5'kk' + .5e. RokType: Y[sqa.Findk()] = Y[sqa.Findk()] + Math.signum(Y[sqa.Findk()])*p (internal program code). Note: many slight variations of the "RokType" already exist, such that it has become difficult to assign them all names.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. A001844, A004525, A104564, A098180, A059100, A010000, A047599, A007877.

LinearRecurrence[{3, -4, 4, -3, 1}, {0, 1, 3, 5, 8}, 60] (* Amiram Eldar, Dec 14 2024 *)

concat(0, Vec(x*(1 + x)*(1 - x + x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ Colin Barker, Apr 29 2019

G.f.: x*(1 + x^3)/((1 + x^2)*(1 - x)^3).
FAMP result: 2*a(n) + 2*A004525(n+1) = A104564(n) + a(n+1).
Superseeker results:
a(2*n+1) = A001844(n) = 2*n*(n+1) + 1 (Centered square numbers);
a(n+1) - a(n) = A098180(n) (Odd numbers with two times the odd numbers repeated in order between them);
a(n) + a(n+2) = A059100(n+1) = A010000(n+1);
a(n+2) - a(n) = A047599(n+1) (Numbers that are congruent to {0, 3, 4, 5} mod 8);
a(n+2) - 2*a(n+1) + a(n) = A007877(n+3) (Period 4 sequence with initial period (0, 1, 2, 1));
Coefficients of g.f.*(1-x)/(1+x) = convolution of this with A280560 gives A004525;
Coefficients of g.f./(1+x) = convolution of this with A033999 gives A054925.
a(n) = (1/2)*(n^2 + 1 - cos(n*Pi/2)). - Ralf Stephan, May 20 2007
From Colin Barker, Apr 29 2019: (Start)
a(n) = (2 - (-i)^n - i^n + 2*n^2) / 4 where i=sqrt(-1).
a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n>4. (End)
a(n) = A011848(n-1)+A011848(n+2). - R. J. Mathar, Sep 11 2019
Sum_{n>=1} 1/a(n) = Pi^2/48 + (Pi/2) * tanh(Pi/2) + (Pi/(4*sqrt(2)) * tanh(Pi/(2*sqrt(2)))). - Amiram Eldar, Dec 14 2024

Stephan's formula corrected by Bruno Berselli, Apr 29 2019

A215095 a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is a Jacobsthal number.

Original entry on oeis.org

0, 1, 3, 4, 8, 17, 35, 68, 136, 273, 547, 1092, 2184, 4369, 8739, 17476, 34952, 69905, 139811, 279620, 559240, 1118481, 2236963, 4473924, 8947848, 17895697, 35791395, 71582788, 143165576, 286331153, 572662307, 1145324612, 2290649224, 4581298449, 9162596899

Offset: 0

Alex Ratushnyak, Aug 03 2012

nonn

Same definition, but k+a(n-2) is a
Fibonacci number: A006498 except first two terms,
Lucas number: A000045 except first two terms,
Pell number: A089928(n-1),
factorial: A215096,
square: A194274,
cube: A215097,
triangular number: A011848(n+2),
oblong number: A215098,
prime number: A215099.
Example of a related sequence definition: a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is a cube.

Cf. A001045, A006498, A089928, A000045.

prpr = 0
prev = 1
jac = [0]*10000
for n in range(10000):
    jac[n] = prpr
    curr = prpr*2 + prev
    prpr = prev
    prev = curr
prpr, prev = 0, 1
for n in range(1, 44):
    print(prpr, end=', ')
    b = c = 0
    while c<=prev:
        c = jac[b] - prpr
        b+=1
    prpr = prev
    prev = c

Conjecture: G.f. (x+2*x^2)/(1-x-x^2-x^3-2*x^4). - David Scambler, Aug 06 2012

A289708 Number of matchings in the n-triangular honeycomb queen graph.

Original entry on oeis.org

1, 4, 51, 2468, 516950, 514413280, 2620954569792

Offset: 1

Eric W. Weisstein, Jul 14 2017

Eric Weisstein's World of Mathematics, Independent Edge Set
Eric Weisstein's World of Mathematics, Matching

Cf. A011848 (matching number), A289878, A289884, A289709.

Name corrected by Andrew Howroyd, Jul 17 2017

cp's OEIS Frontend

A160636 Hankel transform of A114464.

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A174102 Triangle read by rows: T(n, m) = floor(binomial(n+1, m)* binomial(n+2, m)/(2*m+2)), 1 <= m <= n.

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A215099 a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is prime.

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A231559 a(n) = floor( A000326(n)/2 ).

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A373584 a(n) is equal to the number of shaded cells in a regular hexagon with side n drawn on a hexagonal grid.

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A024174 a(n) is floor((4th elementary symmetric function of 1,2,..,n)/(3rd elementary symmetric function of 1,2,...,n)).

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A104563 A floretion-generated sequence relating to centered square numbers.

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