cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

A006095 Gaussian binomial coefficient [n, 2] for q = 2.

Original entry on oeis.org

0, 0, 1, 7, 35, 155, 651, 2667, 10795, 43435, 174251, 698027, 2794155, 11180715, 44731051, 178940587, 715795115, 2863245995, 11453115051, 45812722347, 183251413675, 733006703275, 2932028910251, 11728119835307, 46912487729835
Offset: 0

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Keywords

Comments

Number of 4-block coverings of an n-set where every element of the set is covered by exactly 3 blocks (if offset is 3), so a(n) = (1/4!)*(4^n-6*2^n+8). - Vladeta Jovovic, Feb 20 2001
Number of non-coprime pairs of polynomials (f,g) with binary coefficients where both f and g have degree n+1 and nonzero constant term. - Luca Mariot and Enrico Formenti, Sep 26 2016
Number of triplets found from the integers 1 to 2^n-1 by converting to binary and performing an XOR operation on the corresponding bits of each pair. Defining addition in this carryless way (0+0=1+1=0, 0+1=1+0=1), each triplet (A,B,C) has the property A+B=C, A+C=B and B+C=A. For example, n=3 gives the 7 triplets (1,2,3), (1,4,5), (1,6,7), (2,4,6), (2,5,7), (3,4,7) and (3,5,6). Each integer appears in the set of triplets 2^(n-1)-1 times, for example 3 for n=3. - Ian Duff, Oct 05 2019
Number of 2-dimensional vector subspaces of (Z_2)^n, so also number of Klein subgroups of the group (C_2)^n. - Robert FERREOL, Jul 28 2021

References

  • J. Goldman and G.-C. Rota, The number of subspaces of a vector space, pp. 75-83 of W. T. Tutte, editor, Recent Progress in Combinatorics. Academic Press, NY, 1969.
  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration. Wiley, NY, 1983, p. 99.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • M. Sved, Gaussians and binomials, Ars. Combinatoria, 17A (1984), 325-351.

Crossrefs

First differences: A006516.
Gaussian binomial coefficient [n, k] for q = 2: A000225 (k = 1), this sequence (k = 2), A006096 (k = 3), A006097 (k = 4), A006110 (k = 5), A022189 - A022195 (k = 6 thru 12).

Programs

  • Maple
    a:= n-> add((4^(n-1-j) - 2^(n-1-j))/2, j=0..n-1):
    seq(a(n), n=0..24); # Zerinvary Lajos, Jan 04 2007
    A006095 := -z^2/(z-1)/(2*z-1)/(4*z-1); # Simon Plouffe in his 1992 dissertation. [adapted to offset 0 by Peter Luschny, Jul 20 2021]
    a := n -> (2^n - 2)*(2^n - 1)/6:
    seq(a(n), n = 0..24); # Peter Luschny, Jul 20 2021
  • Mathematica
    Join[{a=0,b=0},Table[c=6*b-8*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011 *)
    CoefficientList[Series[x^2/((1-x)(1-2x)(1-4x)),{x,0,30}],x] (* or *) LinearRecurrence[{7,-14,8},{0,0,1},30] (* Harvey P. Dale, Jul 22 2011 *)
    (* Next, using elementary symmetric functions *)
    f[k_] := 2^(k - 1); t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[2, t[n]]
    Table[a[n], {n, 2, 32}]    (* A203235 *)
    Table[a[n]/2, {n, 2, 32}]  (* A006095 *)
    (* Clark Kimberling, Dec 31 2011 *)
    Table[QBinomial[n, 2, 2], {n, 0, 24}] (* Arkadiusz Wesolowski, Nov 12 2015 *)
  • PARI
    a(n) = (2^n - 1)*(2^(n-1) - 1)/3 \\ Charles R Greathouse IV, Jul 25 2011
    
  • PARI
    concat([0, 0], Vec(x^2/((1-x)*(1-2*x)*(1-4*x)) + O(x^50))) \\ Altug Alkan, Nov 12 2015
  • Sage
    [gaussian_binomial(n,2,2) for n in range(0,25)] # Zerinvary Lajos, May 24 2009
    

Formula

G.f.: x^2/((1-x)(1-2x)(1-4x)).
a(n) = (2^n - 1)*(2^(n-1) - 1)/3 = 4^n/6 - 2^(n-1) + 1/3.
Row sums of triangle A130324. - Gary W. Adamson, May 24 2007
a(n) = Stirling2(n+1,3) + Stirling2(n+1,4). - Zerinvary Lajos, Oct 04 2007; corrected by R. J. Mathar, Mar 19 2011
a(n) = A139250(2^(n-1) - 1), n >= 1. - Omar E. Pol, Mar 03 2011
a(n) = 4*a(n-1) + 2^(n-1) - 1, n >= 2. - Vincenzo Librandi, Mar 19 2011
a(0) = 0, a(1) = 0, a(2) = 1, a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3). - Harvey P. Dale, Jul 22 2011
a(n) = Sum_{k=0..n-2} 2^k*C(2*n-k-2, k), n >= 2. - Johannes W. Meijer, Aug 19 2013
a(n) = Sum_{i=0..n-2, j=i..n-2} 2^{i+j} = 2^0 * (2^0 + 2^1 + ... + 2^(n-2)) + 2^1 * (2^1 + 2^2 + ... + 2^(n-2)) + ... + 2^(n-2) * 2^(n-2), n>1. - J. M. Bergot, May 08 2017
a(n) = a(n-1) + A000217(A000225(n-1)), n > 0. - Ivan N. Ianakiev, Dec 11 2017
E.g.f.: (2*exp(x)-3*exp(2*x)+exp(4*x))/6. - Paul Weisenhorn, Aug 22 2021
From Peter Bala, Jul 01 2025: (Start)
G.f. assuming an offset of 0: exp( Sum_{n >= 1} b(3*n)/b(n)*x^n/n ) = 1 + 7*x + 35*x^2 + ..., where b(n) = A000225(n) = 2^n - 1.
The following are examples of telescoping series:
Sum_{n >= 2} 2^n/a(n) = 6, follows from 1 - (1/6)*Sum_{k = 2..n} 2^k/a(k) = 1/(2^n - 1).
Sum_{n >= 2} 2^n/(a(n)*a(n+2)) = 6/49, follows from 1 - (49/6)*Sum_{k = 2..n} 2^k/(a(k)*a(k+2)) = 1/A006096(n+2);
Sum_{n >= 2} 4^n/(a(n)*a(n+2)) = 26/49, follows from 13 - (49/2)*Sum_{k = 2..n} 4^k/(a(k)*a(k+2)) = A086224(n)/A006096(n+2);
Sum_{n >= 2} 8^n/(a(n)*a(n+2)) = 129/49, follows from 43 - (49/3)*Sum_{k = 2..n} 8^k/(a(k)*a(k+2)) = A171479(n+1)/A006096(n+2). (End)

A014827 a(1)=1, a(n) = 5*a(n-1) + n.

Original entry on oeis.org

1, 7, 38, 194, 975, 4881, 24412, 122068, 610349, 3051755, 15258786, 76293942, 381469723, 1907348629, 9536743160, 47683715816, 238418579097, 1192092895503, 5960464477534, 29802322387690, 149011611938471, 745058059692377, 3725290298461908, 18626451492309564, 93132257461547845
Offset: 1

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Programs

Formula

a(n) = (5^(n+1) - 4*n - 5)/16.
G.f.: x/((1-5*x)*(1-x)^2).
From Paul Barry, Jul 30 2004: (Start)
a(n) = Sum_{k=0..n} (n-k)*5^k = Sum_{k=0..n} k*5^(n-k).
a(n) = Sum_{k=0..n} binomial(n+2,k+2)*4^k [Offset 0]. (End)
From Elmo R. Oliveira, Mar 29 2025: (Start)
E.g.f.: exp(x)*(5*exp(4*x) - 4*x - 5)/16.
a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3) for n > 3. (End)

A016209 Expansion of 1/((1-x)(1-3x)(1-5x)).

Original entry on oeis.org

1, 9, 58, 330, 1771, 9219, 47188, 239220, 1205941, 6059229, 30384718, 152189310, 761743711, 3811110039, 19062724648, 95335146600, 476740303081, 2383895225649, 11920057258978, 59602029687090
Offset: 0

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Comments

For a combinatorial interpretation following from a(n) = A039755(n+2,2) = h^{(3)}A039755.%20-%20_Wolfdieter%20Lang">n, the complete homogeneous symmetric function of degree n in the symbols {1, 3, 5} see A039755. - _Wolfdieter Lang, May 26 2017

Examples

			a(2) = h^{(3)}_2 = 1^2 + 3^2 + 5^2 + 1^1*(3^1 + 5^1) + 3^1*5^1 = 58. - _Wolfdieter Lang_, May 26 2017
		

Crossrefs

Programs

  • Magma
    [(5^(n+2)-2*3^(n+2)+1)/8: n in [0..20]]; // Vincenzo Librandi, Sep 17 2011
  • Maple
    A016209 := proc(n) (5^(n+2)-2*3^(n+2)+1)/8; end proc: # R. J. Mathar, Mar 22 2011
  • Mathematica
    Join[{a=1,b=9},Table[c=8*b-15*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2011 *)
    CoefficientList[Series[1/((1-x)(1-3x)(1-5x)),{x,0,30}],x] (* or *) LinearRecurrence[ {9,-23,15},{1,9,58},30] (* Harvey P. Dale, Feb 20 2020 *)
  • PARI
    a(n)=if(n<0,0,n+=2; (5^n-2*3^n+1)/8)
    

Formula

a(n) = A039755(n+2, 2).
a(n) = (5^(n+2) - 2*3^(n+2)+1)/8 = a(n-1) + A005059(n+1) = 8*a(n-1) - 15*a(n-2) + 1 = (A003463(n+2) - A003462(n+2))/2. - Henry Bottomley, Jun 06 2000
G.f.: 1/((1-x)(1-3*x)(1-5*x)). See the name.
E.g.f.: (25*exp(5*x) - 18*exp(3*x) + exp(x))/8, from the e.g.f. of the third column (k=2) of A039755. - Wolfdieter Lang, May 26 2017

A016218 Expansion of 1/((1-x)*(1-4*x)*(1-5*x)).

Original entry on oeis.org

1, 10, 71, 440, 2541, 14070, 75811, 400900, 2091881, 10808930, 55442751, 282806160, 1436400421, 7271480590, 36715316891, 185008240220, 930767824161, 4676745613050, 23475354034231, 117743274047080, 590182385739101, 2956775990710310, 14807336201610771
Offset: 0

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Programs

Formula

From Vincenzo Librandi, Feb 10 2011: (Start)
a(n) = a(n-1) + 5^(n+1) - 4^(n+1), n >= 1.
a(n) = 9*a(n-1) - 20*a(n-2) + 1, n >= 2. (End)
a(n) = 1/12 - 4^(n+2)/3 + 5^(n+2)/4. - R. J. Mathar, Mar 15 2011

A249997 Expansion of 1/((1-x)*(1+3*x)*(1-4*x)).

Original entry on oeis.org

1, 2, 15, 40, 221, 702, 3355, 11780, 52041, 193402, 817895, 3138720, 12953461, 50618102, 206059635, 813476860, 3286192481, 13047914802, 52482224575, 209057202200, 838843897101, 3347530323502, 13413657088715, 53584020970740, 214547906035321, 857556157684202
Offset: 0

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Author

Alex Ratushnyak, Dec 28 2014

Keywords

Crossrefs

Programs

  • Magma
    [((-1)^n*3^(n+3) +4^(n+3) -7)/84: n in [0..50]]; // G. C. Greubel, Jul 21 2022
    
  • Mathematica
    LinearRecurrence[{2,11,-12}, {1,2,15}, 50] (* G. C. Greubel, Jul 21 2022 *)
  • SageMath
    [((-1)^n*3^(n+3) +4^(n+3) -7)/84 for n in (0..50)] # G. C. Greubel, Jul 21 2022

Formula

G.f.: 1/((1-x) * (1+3*x) * (1-4*x)).
a(n) = (-1)^n*3^(n+2)/28 + 4^(n+2)/21 -1/12. - R. J. Mathar, Jan 09 2015
E.g.f.: (1/84)*(27*exp(-3*x) - 7*exp(x) + 64*exp(4*x)). - G. C. Greubel, Jul 21 2022

A341091 Triangle read by rows: Coefficients for calculation of the sum of all the finite differences from order zero to order k. Sum_{n=0..k} T(n, k)*b(n) = b(0) + b(1) + ... + b(k) + (b(1) - b(0)) + ... + (b(k) - b(k-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... .

Original entry on oeis.org

1, 0, 2, 1, -1, 3, 0, 3, -3, 4, 1, -2, 7, -6, 5, 0, 4, -8, 14, -10, 6, 1, -3, 13, -21, 25, -15, 7, 0, 5, -15, 35, -45, 41, -21, 8, 1, -4, 21, -49, 81, -85, 63, -28, 9, 0, 6, -24, 71, -129, 167, -147, 92, -36, 10, 1, -5, 31, -94, 201, -295, 315, -238, 129, -45, 11
Offset: 0

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Author

Thomas Scheuerle, Feb 13 2022

Keywords

Comments

If we want to calculate the sum of finite differences for a sequence b(n):
b(0)*T(0, n) + ... + b(n)*T(n, n) = b(0) + b(1) + ... + b(n) + (b(1) - b(0)) + ... + (b(n) - b(n-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... This sum includes the sequence b(n) itself. This defines an invertible linear sequence transformation with a deep connection to Bernoulli numbers and other interesting sequences of rational numbers.
From Thomas Scheuerle, Apr 29 2024: (Start)
These are the coefficients of the polynomials defined by the recurrence: P(k, x) = P(k - 1, x) + (x^2 - x)*P(k - 2, x) + 1, with P(-1, x) = 0 and P(0, x) = 1. This can also be expressed as P(k, x) = Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^2 - x)^(m - 1) = Sum_{n=0..k} T(n, k)*x^(k-n). If we would evaluate P(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1+(t-1)*x)*(1 - t*x)).
We may replace (x^2 - x) by (x^(-2) - x^(-1)) to get the coefficients in reverse order: x^k*Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^(-2) - x^(-1))^(m - 1) = Sum_{n=0..k} T(n, k)*x^n = F(k, x). If we would evaluate F(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1 - (t-1)*x)*(1 - t*x)). (End)

Examples

			Triangle begins with T(n, k):
   n=   0,  1,   2,   3,   4,   5,   6,   7,   8
  k=0   1
  k=1   0,  2
  k=2   1, -1,   3
  k=3   0,  3,  -3,   4
  k=4   1, -2,   7,  -6,   5
  k=5   0,  4,  -8,  14, -10,   6
  k=6   1, -3,  13, -21,  25, -15,   7
  k=7   0,  5, -15,  35, -45,  41, -21,   8
  k=8   1, -4,  21, -49,  81, -85,  63, -28,   9
  ...
		

Crossrefs

Cf. A027642, A164555 (Numerators and denominators of Bernoulli numbers).
Cf. A001008, A002805 (Numerators and denominators of harmonic numbers).
Sequences below will be obtained by evaluation of the associated polynomials:

Programs

  • PARI
    A341091(n, k) = sum(m=n, k,(-1)^(m+n)*binomial(m+1, n))
    
  • PARI
    A341091(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*hypergeom([1,k+3],k+3-n,-1)+(-1/2)^n*(2^(n+1)-1)) \\ Thomas Scheuerle, Apr 29 2024

Formula

b(0)*T(0, m) + b(1)*T(1, m) + ... + b(m)*T(m, m)
= Sum_{j=0..m} Sum_{n=0..m-j} Sum_{k=0..n} (-1)^k*binomial(n, k)*b(j+n-k)
= Sum_{n=0..m} b(n)*Sum_{j=n..m}(-1)^(j+n)*binomial(j+1, n).
T(n, k) = Sum_{m=n..k}(-1)^(m+n)*binomial(m+1, n).
T(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*Hypergeometric2F1(1, k+3, k+3-n, -1)+(-1/2)^n*(2^(n+1) - 1)), where Hypergeometric2F1 is the Gaussian hypergeometric function 2F1 as defined in Mathematica. - Thomas Scheuerle, Apr 29 2024
T(k, k) = A000027(k+1) The positive integers.
|T(k-1, k)| = A000217(k) The triangular numbers.
T(k-2, k) = A004006(k).
|T(k-3, k)| = A051744(k).
T(0, k*2) = 1.
T(0, k*2 + 1) = 0.
T(1, k*2 + 1) = k + 2.
T(1, k*2 + 2) = -(k + 1).
T(n, k) with constant n and variable k, a linear recurrence relation with characteristic polynomial (x-1)*(x+1)^(n+1).
Sum_{n=0..k} T(n, k)*B_n = 1. B_n is the n-th Bernoulli number with B_1 = 1/2. B_n = A164555(n)/A027642(n).
Sum_{n=0..k} T(n, k)*(1 - B_n) = k.
Sum_{n=0..k} T(n, k)*(2*n - 3+3*B_n) = k^2.
Sum_{n=0..k} T(n, k)*A032346(n) = A032346(k+1).
From Thomas Scheuerle, Apr 29 2024: (Start)
Sum_{n=0..k} T(n, k)*A000110(n+1) = A000110(k+2) - 1.
Sum_{n=0..k} T(n, k)*(1/(1+n)) = H(1+floor(k/2)), where H(k) is the harmonic number A001008(k)/A002805(k). (End)
Sum_{n=0..k} T(n, k)*c(n) = c(k). C(k) = {-1, 0, 1/2, 1/2, 1/8, -7/20, ...} this sequence of rational numbers can be defined recursively: c(0) = -1, c(m) = (-c(m-1) + Sum_{k=0..m-1} A130595(m+1, k)*c(k))/m.
c(m) is an eigensequence of this transformation, all eigensequences are c(m) multiplied by any factor.
Sum_{n=0..k} T(n, k)*A000045(n) = 2*(A000045(2*floor((k+1)/2) - 1) - 1). A000045 are the Fibonacci numbers.
Sum_{n=0..k} T(n, k)*A000032(n) = A000032(2*floor(k/2)+2) - 2. A000032 are the Lucas numbers.
Sum_{n=0..k} T(n, k)*A001045(n) = A145766(floor((k+1)/2)). A001045 is the Jacobsthal sequence.
This sequence acting as an operator onto a monomial n^w:
Sum_{n=0..k} T(n, k)*n^w = (1/(w+1))*k^(w+1) + Sum_{v=1..w} ((v+B_v)*(w)_v/v!)*k^(w+1-v) - A052875(w) + O_k(w) (w)_v is the falling factorial. If k > w-1 then O_k(w) = 0. If k <= w-1 then O_k(w) is A084416(w, 2+k), the sequence with the exponential generating function: (e^x-1)^(2+k)/(2-e^x).
From Thomas Scheuerle, Apr 29 2024: (Start)
This sequence acting by its inverse operator onto a monomial k^w:
Sum_{n=0..k} T(n, k)*( Sum_{m=0..k} ((-1)^(1+m+k)*binomial(k, m)*(2^(k-m) - 1)*n^m + A344037(m)*B_n) ) = k^w - A372245(w, k+3), note that A372245(w, k+3) = 0 if k+3 > w. B_n is the n-th Bernoulli number with B_1 = 1/2.
How this sequence will act as an operator onto a Dirichlet series may be developed by the formulas below:
Sum_{n=0..k} T(n, k)*2^n = A000295(k+2).
Sum_{n=0..k} T(n, k)*3^n = A000392(k+3).
Sum_{n=0..k} T(n, k)*4^n = A016208(k).
Sum_{n=0..k} T(n, k)*5^n = A016218(k).
Sum_{n=0..k} T(n, k)*6^n = A016228(k).
Sum_{n=0..k} T(n, k)*7^n = A016241(k).
Sum_{n=0..k} T(n, k)*8^n = A016249(k).
Sum_{n=0..k} T(n, k)*9^n = A016256(k).
Sum_{n=0..k} T(n, k)*10^n = A016261(k).
Sum_{n=0..k} T(n, k)*m^n = m^2*m^k/(m-1) - (m-1)^2*(m-1)^k/(m-2) + 1/((m-1)*(m-2)), for m > 2.
Sum_{n=0..k} T(n, k)*( m*B_n + (m-1)*Sum_{t=1..m} t^n )*(1/m^2) = m^k, for m > 0. B_n is the n-th Bernoulli number with B_1 = 1/2.
Sum_{n=0..k} T(n, k) zeta(-n) = Sum_{j=0..k} (-1)^(1+j)/(2+j) = (-1)^(k+1)*LerchPhi(-1, 1, k+3) - 1 + log(2).
Sum_{n=0..k} T(k - n, k)*2^n = A000975(k+1)
Sum_{n=0..k} T(k - n, k)*3^n = A091002(k+2)
Sum_{n=0..k} T(k - n, k)*4^n = A249997(k). (End)

A085277 Expansion of (1+x)^2/((1-2x)(1-3x)).

Original entry on oeis.org

1, 7, 30, 108, 360, 1152, 3600, 11088, 33840, 102672, 310320, 935568, 2815920, 8466192, 25435440, 76380048, 229287600, 688157712, 2065062960, 6196368528, 18591464880, 55779113232, 167346776880, 502059205008, 1506215363760
Offset: 0

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Author

Paul Barry, Jun 25 2003

Keywords

Comments

Inverse binomial transform of A016208. Binomial transform of A085278.

Programs

  • Mathematica
    CoefficientList[Series[(1+x)^2/((1-2x)(1-3x)),{x,0,30}],x] (* or *) Join[{1},LinearRecurrence[{5,-6},{7,30},30]] (* Harvey P. Dale, Sep 30 2012 *)

Formula

a(n)=16*3^n/3+0^n/6-9*2^n/2.
a(0)=1, a(1)=7, a(2)=30, a(n)=5*a(n-1)-6*a(n-2). - Harvey P. Dale, Sep 30 2012

A249998 Expansion of 1/((1+x)*(1+3*x)*(1-4*x)).

Original entry on oeis.org

1, 0, 13, 12, 169, 312, 2341, 6084, 34177, 107184, 517309, 1803516, 8011225, 29653416, 125788117, 481629108, 1991086513, 7770635808, 31663673965, 124911303660, 504875391241, 2003811035160, 8062315730053, 32108048151972, 128855836912609, 514152414736272, 2060422457687581
Offset: 0

Views

Author

Alex Ratushnyak, Dec 28 2014

Keywords

Crossrefs

Formula

a(n) = (-1)^n*3^(n+2)/14 + 4^(n+2)/35 - (-1)^n/10. - R. J. Mathar, Jan 09 2015

A016198 Expansion of g.f. 1/((1-x)*(1-2*x)*(1-5*x)).

Original entry on oeis.org

1, 8, 47, 250, 1281, 6468, 32467, 162590, 813461, 4068328, 20343687, 101722530, 508620841, 2543120588, 12715635707, 63578244070, 317891351421, 1589457019248, 7947285620527, 39736429151210, 198682147853201, 993410743460308, 4967053725690147, 24835268645227950
Offset: 0

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Author

N. J. A. Sloane, Dec 11 1999

Keywords

Crossrefs

Programs

Formula

a(n) = (25*5^n - 16*2^n + 3)/12. - Bruno Berselli, Feb 09 2011
a(n) = [(5^0-2^0) + (5^1-2^1) + ... + (5^n-2^n)]/3. - r22lou(AT)cox.net, Nov 14 2005
a(0)=1, a(n) = 5*a(n-1) + 2^(n+1) - 1. - Vincenzo Librandi, Feb 07 2011
From Elmo R. Oliveira, Mar 26 2025: (Start)
E.g.f.: exp(x)*(25*exp(4*x) - 16*exp(x) + 3)/12.
a(n) = 8*a(n-1) - 17*a(n-2) + 10*a(n-3).
a(n) = A016127(n+1) - A003463(n+2). (End)

Extensions

More terms from Wesley Ivan Hurt, May 05 2014
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