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.

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A213500 Rectangular array T(n,k): (row n) = b**c, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and ** = convolution.

Original entry on oeis.org

1, 4, 2, 10, 7, 3, 20, 16, 10, 4, 35, 30, 22, 13, 5, 56, 50, 40, 28, 16, 6, 84, 77, 65, 50, 34, 19, 7, 120, 112, 98, 80, 60, 40, 22, 8, 165, 156, 140, 119, 95, 70, 46, 25, 9, 220, 210, 192, 168, 140, 110, 80, 52, 28, 10, 286, 275, 255, 228, 196, 161, 125, 90
Offset: 1

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Author

Clark Kimberling, Jun 14 2012

Keywords

Comments

Principal diagonal: A002412.
Antidiagonal sums: A002415.
Row 1: (1,2,3,...)**(1,2,3,...) = A000292.
Row 2: (1,2,3,...)**(2,3,4,...) = A005581.
Row 3: (1,2,3,...)**(3,4,5,...) = A006503.
Row 4: (1,2,3,...)**(4,5,6,...) = A060488.
Row 5: (1,2,3,...)**(5,6,7,...) = A096941.
Row 6: (1,2,3,...)**(6,7,8,...) = A096957.
...
In general, the convolution of two infinite sequences is defined from the convolution of two n-tuples: let X(n) = (x(1),...,x(n)) and Y(n)=(y(1),...,y(n)); then X(n)**Y(n) = x(1)*y(n)+x(2)*y(n-1)+...+x(n)*y(1); this sum is the n-th term in the convolution of infinite sequences:(x(1),...,x(n),...)**(y(1),...,y(n),...), for all n>=1.
...
In the following guide to related arrays and sequences, row n of each array T(n,k) is the convolution b**c of the sequences b(h) and c(h+n-1). The principal diagonal is given by T(n,n) and the n-th antidiagonal sum by S(n). In some cases, T(n,n) or S(n) differs in offset from the listed sequence.
b(h)........ c(h)........ T(n,k) .. T(n,n) .. S(n)
h .......... h .......... A213500 . A002412 . A002415
h .......... h^2 ........ A212891 . A213436 . A024166
h^2 ........ h .......... A213503 . A117066 . A033455
h^2 ........ h^2 ........ A213505 . A213546 . A213547
h .......... h*(h+1)/2 .. A213548 . A213549 . A051836
h*(h+1)/2 .. h .......... A213550 . A002418 . A005585
h*(h+1)/2 .. h*(h+1)/2 .. A213551 . A213552 . A051923
h .......... h^3 ........ A213553 . A213554 . A101089
h^3 ........ h .......... A213555 . A213556 . A213547
h^3 ........ h^3 ........ A213558 . A213559 . A213560
h^2 ........ h*(h+1)/2 .. A213561 . A213562 . A213563
h*(h+1)/2 .. h^2 ........ A213564 . A213565 . A101094
2^(h-1) .... h .......... A213568 . A213569 . A047520
2^(h-1) .... h^2 ........ A213573 . A213574 . A213575
h .......... Fibo(h) .... A213576 . A213577 . A213578
Fibo(h) .... h .......... A213579 . A213580 . A053808
Fibo(h) .... Fibo(h) .... A067418 . A027991 . A067988
Fibo(h+1) .. h .......... A213584 . A213585 . A213586
Fibo(n+1) .. Fibo(h+1) .. A213587 . A213588 . A213589
h^2 ........ Fibo(h) .... A213590 . A213504 . A213557
Fibo(h) .... h^2 ........ A213566 . A213567 . A213570
h .......... -1+2^h ..... A213571 . A213572 . A213581
-1+2^h ..... h .......... A213582 . A213583 . A156928
-1+2^h ..... -1+2^h ..... A213747 . A213748 . A213749
h .......... 2*h-1 ...... A213750 . A007585 . A002417
2*h-1 ...... h .......... A213751 . A051662 . A006325
2*h-1 ...... 2*h-1 ...... A213752 . A100157 . A071238
2*h-1 ...... -1+2^h ..... A213753 . A213754 . A213755
-1+2^h ..... 2*h-1 ...... A213756 . A213757 . A213758
2^(n-1) .... 2*h-1 ...... A213762 . A213763 . A213764
2*h-1 ...... Fibo(h) .... A213765 . A213766 . A213767
Fibo(h) .... 2*h-1 ...... A213768 . A213769 . A213770
Fibo(h+1) .. 2*h-1 ...... A213774 . A213775 . A213776
Fibo(h) .... Fibo(h+1) .. A213777 . A001870 . A152881
h .......... 1+[h/2] .... A213778 . A213779 . A213780
1+[h/2] .... h .......... A213781 . A213782 . A005712
1+[h/2] .... [(h+1)/2] .. A213783 . A213759 . A213760
h .......... 3*h-2 ...... A213761 . A172073 . A002419
3*h-2 ...... h .......... A213771 . A213772 . A132117
3*h-2 ...... 3*h-2 ...... A213773 . A214092 . A213818
h .......... 3*h-1 ...... A213819 . A213820 . A153978
3*h-1 ...... h .......... A213821 . A033431 . A176060
3*h-1 ...... 3*h-1 ...... A213822 . A213823 . A213824
3*h-1 ...... 3*h-2 ...... A213825 . A213826 . A213827
3*h-2 ...... 3*h-1 ...... A213828 . A213829 . A213830
2*h-1 ...... 3*h-2 ...... A213831 . A213832 . A212560
3*h-2 ...... 2*h-1 ...... A213833 . A130748 . A213834
h .......... 4*h-3 ...... A213835 . A172078 . A051797
4*h-3 ...... h .......... A213836 . A213837 . A071238
4*h-3 ...... 2*h-1 ...... A213838 . A213839 . A213840
2*h-1 ...... 4*h-3 ...... A213841 . A213842 . A213843
2*h-1 ...... 4*h-1 ...... A213844 . A213845 . A213846
4*h-1 ...... 2*h-1 ...... A213847 . A213848 . A180324
[(h+1)/2] .. [(h+1)/2] .. A213849 . A049778 . A213850
h .......... C(2*h-2,h-1) A213853
...
Suppose that u = (u(n)) and v = (v(n)) are sequences having generating functions U(x) and V(x), respectively. Then the convolution u**v has generating function U(x)*V(x). Accordingly, if u and v are homogeneous linear recurrence sequences, then every row of the convolution array T satisfies the same homogeneous linear recurrence equation, which can be easily obtained from the denominator of U(x)*V(x). Also, every column of T has the same homogeneous linear recurrence as v.

Examples

			Northwest corner (the array is read by southwest falling antidiagonals):
  1,  4, 10, 20,  35,  56,  84, ...
  2,  7, 16, 30,  50,  77, 112, ...
  3, 10, 22, 40,  65,  98, 140, ...
  4, 13, 28, 50,  80, 119, 168, ...
  5, 16, 34, 60,  95, 140, 196, ...
  6, 19, 40, 70, 110, 161, 224, ...
T(6,1) = (1)**(6) = 6;
T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19;
T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.
		

Crossrefs

Cf. A000027.

Programs

  • Mathematica
    b[n_] := n; c[n_] := n
    t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
    TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
    Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
    r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213500 *)
  • PARI
    t(n,k) = sum(i=0, k - 1, (k - i) * (n + i));
    tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k,n - k + 1),", ");); print(););};
    tabl(12) \\ Indranil Ghosh, Mar 26 2017
    
  • Python
    def t(n, k): return sum((k - i) * (n + i) for i in range(k))
    for n in range(1, 13):
        print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017

Formula

T(n,k) = 4*T(n,k-1) - 6*T(n,k-2) + 4*T(n,k-3) - T(n,k-4).
T(n,k) = 2*T(n-1,k) - T(n-2,k).
G.f. for row n: x*(n - (n - 1)*x)/(1 - x)^4.

A001871 Expansion of 1/(1 - 3*x + x^2)^2.

Original entry on oeis.org

1, 6, 25, 90, 300, 954, 2939, 8850, 26195, 76500, 221016, 632916, 1799125, 5082270, 14279725, 39935214, 111228804, 308681550, 853904015, 2355364650, 6480104231, 17786356776, 48715278000, 133167004200, 363372003625, 989900286774
Offset: 0

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Author

Keywords

Comments

Convolution of A001906(n), n >= 1 (even-indexed Fibonacci numbers) with itself.
A001787 and this sequence arise in counting ordered trees of height at most k where only the rightmost branch at the root actually achieves this height and the count is by the number of edges, with k = 3 for A001787 and k = 4 for this sequence.
Gives the number of 3412-avoiding permutations containing exactly one subsequence of type 321. - Dan Daly (ddaly(AT)du.edu), Apr 24 2008

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Partial sums of A001870 (one half of odd-indexed A001629(n), n >= 2, Fibonacci convolution).

Programs

  • Magma
    I:=[1, 6, 25, 90]; [n le 4 select I[n] else 6*Self(n-1)-11*Self(n-2)+6*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 10 2012
    
  • Maple
    f:= gfun:-rectoproc({a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3)-a(n-4),
    a(0)=1,a(1)=6,a(2)=25,a(3)=90},a(n),remember):
    map(f, [$0..50]); # Robert Israel, May 05 2017
    # alternative
    A001871 := proc(n)
        option remember ;
        if n <= 3 then
            op(n+1,[1,6,25,90]) ;
        else
            6*procname(n-1)-11*procname(n-2)+6*procname(n-3)-procname(n-4) ;
        end if;
    end proc:
    seq(A001871(n),n=0..10) ; # R. J. Mathar, Dec 16 2024
  • Mathematica
    CoefficientList[Series[1/(1-3x+x^2)^2,{x,0,40}],x] (* Vincenzo Librandi, Jun 10 2012 *)
  • PARI
    a(n)=((4*n+2)*fibonacci(2*n)+(7*n+5)*fibonacci(2*n+1))/5
    
  • PARI
    Vec(1/(1-3*x+x^2)^2 + O(x^100)) \\ Altug Alkan, Oct 31 2015

Formula

a(n) = (2*(2*n+1)*F(2*(n+1))+3*(n+1)*F(2*n+1))/5 with F(n) = A000045 (Fibonacci numbers).
a(n) = -a(-4-n) = ((4*n+2)*F(2*n) + (7*n+5)*F(2*n+1))/5 with F(n) = A000045 (Fibonacci numbers).
a(n) = (2*a(n-1) + (n+1)*F(2n+4))/3, where F(n) = A000045 (Fibonacci numbers). - Emeric Deutsch, Oct 08 2002
G.f.: 1/(1 - 3*x + x^2)^2. - Simon Plouffe in his 1992 dissertation
a(n) = (Sum_{k=0..n} S(k, 3)*S(n-k, 3)), where S(n, x) = U(n, x/2) is the n-th Chebyshev polynomial of the 2nd kind, A049310. - Paul Barry, Nov 14 2003
a(n) = Sum_{k=1..n+1} F(2k)*F(2(n-k+2)), where F(k) is the k-th Fibonacci number. - Dan Daly (ddaly(AT)du.edu), Apr 24 2008
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). - Vincenzo Librandi, Mar 14 2011
a(n) = 2*A001870(n) - A238846(n). - Philippe Deléham, Mar 06 2014
a(n) ~ (7 + 3*sqrt(5))*n*cos(n*arccos(3/2))/5. - Stefano Spezia, Mar 29 2022
From Peter Bala, Nov 05 2024: (Start)
a(n) = Sum_{k = 0..n} (n + 2*k + 1)*binomial(n+k, 2*k).
a(n) = (n+1) * hypergeom([-n, n+1, (n+3)/2], [1/2, (n+1)/2], -1/4).
Second-order recurrence: n*a(n) = 3*(n + 1)*a(n-1) - (n + 2)*a(n-2) with a(0) = 1 and a(1) = 6. (End)
E.g.f.: exp(3*x/2)*(5*(5 + 18*x)*cosh(sqrt(5)*x/2) + sqrt(5)*(9 + 40*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025

Extensions

Additional comments from Wolfdieter Lang, Apr 07 2000

A030267 Compose the natural numbers with themselves, A(x) = B(B(x)) where B(x) = x/(1-x)^2 is the generating function for natural numbers.

Original entry on oeis.org

1, 4, 14, 46, 145, 444, 1331, 3926, 11434, 32960, 94211, 267384, 754309, 2116936, 5914310, 16458034, 45638101, 126159156, 347769719, 956238170, 2623278946, 7181512964, 19622668679, 53522804976, 145753273225, 396323283724, 1076167858046, 2918447861686
Offset: 1

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Author

Keywords

Comments

Sum of pyramid weights of all nondecreasing Dyck paths of semilength n. (A pyramid in a Dyck word (path) is a factor of the form U^h D^h, where U=(1,1), D=(1,-1) and h is the height of the pyramid. A pyramid in a Dyck word w is maximal if, as a factor in w, it is not immediately preceded by a u and immediately followed by a d. The pyramid weight of a Dyck path (word) is the sum of the heights of its maximal pyramids.) Example: a(4) = 46. Indeed, there are 14 Dyck paths of semilength 4. One of them, namely UUDUDDUD is not nondecreasing because the valleys are at heights 1 and 0. The other 13, with the maximal pyramids shown between parentheses, are: (UD)(UD)(UD)(UD), (UD)(UD)(UUDD), (UD)(UUDD)(UD), (UD)U(UD)(UD)D, (UD)(UUUDDD), (UUDD)(UD)(UD), (UUDD)(UUDD), (UUUDDD)(UD), U(UD)(UD)(UD)D, U(UD)(UUDD)D, U(UUDD)(UD)D, UU(UD)(UD)DD and (UUUUDDDD). The pyramid weights of these paths are 4, 4, 4, 3, 4, 4, 4, 4, 3, 3, 3, 2, and 4, respectively. Their sum is 46. a(n) = Sum_{k = 1..n} k*A121462(n, k). - Emeric Deutsch, Jul 31 2006
Number of 1s in all compositions of n, where compositions are understood with two different kinds of 1s, say 1 and 1' (n >= 1). Example: a(2) = 4 because the compositions of 2 are 11, 11', 1'1, 1'1', 2, having a total of 2 + 1 + 1 + 0 + 0 = 4 1s. Also number of k's in all compositions of n + k (k = 2, 3, ...). - Emeric Deutsch, Jul 21 2008
From Petros Hadjicostas, Jun 24 2019: (Start)
If c = (c(m): m >= 1) is the input sequence and b_k = (b_k(n): n >= 1) is the output sequence under the AIK[k] = INVERT[k] transform (see Bower's web link below), then the bivariate g.f. of the list of sequences (b_k: k >= 1) = ((b_k(n): n >= 1): k >= 1) is Sum_{n, k >= 1} b_k(n)*x^n*y^k = y*C(x)/(1 - y*C(x)), where C(x) = Sum_{m >= 1} c(m)*x^m is the g.f. of the input sequence.
Here, b_k(n) is the number of all (linear) compositions of n with k parts where a part of size m is colored with one of c(m) colors. Thus, Sum_{k = 1..n} k*b_k(n) is the total number of parts in all compositions of n.
If we differentiate the bivariate g.f. function above, i.e., Sum_{n, k >= 1} b_k(n)*x^n*y^k, with respect to y and set y = 1, we get the g.f. of the sequence (Sum_{k = 1..n} k*b_k(n): n >= 1). It is C(x)/(1 - C(x))^2.
When c(m) = m for all m >= 1, we have m-color compositions of n that were first studied by Agarwal (2000). The cyclic version of these m-color compositions were studied by Gibson (2017) and Gibson et al. (2018).
When c(m) = m for each m >= 1, we have C(x) = x/(1 - x)^2, and so C(x)/(1 - C(x))^2 = x * (1 - x)^2/(1 - 3*x + x^2)^2, which is the g.f. of the current sequence.
Hence, a(n) is the total number of parts in all m-color compositions of n (in the sense of Agarwal (2000)).
(End)
Series reversal gives A153294 starting from index 1, with alternating signs: 1, -4, 18, -86, 427, -2180, ... - Vladimir Reshetnikov, Aug 03 2019

Examples

			From _Petros Hadjicostas_, Jun 24 2019: (Start)
Recall that with m-color compositions, a part of size m may be colored with one of m colors.
We have a(1) = 1 because we only have one colored composition, namely 1_1, that has only 1 part.
We have a(2) = 4 because we have the following colored compositions of n = 2: 2_1, 2_2, 1_1 + 1_1; hence, a(2) = 1 + 1 + 2 = 4.
We have a(3) = 14 because we have the following colored compositions of n = 3: 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 2_1 + 1_1, 2_2 + 1_1, 1_1 + 1_1 + 1_1; hence, a(3) = 1 + 1 + 1 + 2 + 2 + 2 + 2 + 3 = 14.
We have a(14) = 46 because we have the following colored compositions of n = 4:
(i) 4_1, 4_2, 4_3, 4_4; with a total of 4 parts.
(ii) 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 3_1 + 1_1, 3_2 + 1_1, 3_3 + 1_1, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_1, 2_2 + 2_2; with a total of 2 x 10 = 20 parts.
(iii) 1_1 + 1_1 + 2_1, 1_1 + 1_1 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 2_1 + 1_1 + 1_1, 2_2 + 1_1 + 1_1; with a total of 3 x 6 = 18 parts.
(iv) 1_1 + 1_1 + 1_1 + 1_1; with a total of 4 parts.
Hence, a(4) = 4 + 20 + 18 + 4 = 46.
(End)
		

References

  • R. P. Grimaldi, Compositions and the alternate Fibonacci numbers, Congressus Numerantium, 186, 2007, 81-96.

Crossrefs

Partial sums of A038731. First differences of A001870.
Cf. A001629 (right-shifted inverse Binomial Transform), A023610 (inverse Binomial Transform of left-shifted sequence), A030279, A045623, A088305, A121462, A153294, A279282, A307415, A308723.

Programs

  • Maple
    with(combinat): L[0]:=2: L[1]:=1: for n from 2 to 60 do L[n]:=L[n-1] +L[n-2] end do: seq(2*fibonacci(2*n)*1/5+(1/5)*n*L[2*n],n=1..30); # Emeric Deutsch, Jul 21 2008
  • Mathematica
    Table[Sum[k Binomial[n+k-1,2k-1],{k,n}],{n,30}] (* or *) LinearRecurrence[ {6,-11,6,-1},{1,4,14,46},30] (* Harvey P. Dale, Aug 01 2011 *)
  • PARI
    a(n)=(2*n*fibonacci(2*n+1)+(2-n)*fibonacci(2*n))/5

Formula

a(n) = -a(-n) = (2n * F(2n+1) + (2 - n) * F(2n))/5 with F(n) = A000045(n) (Fibonacci numbers).
G.f.: x * (1 - x)^2/(1 - 3*x + x^2)^2.
a(n) = Sum_{k = 1..n} k*C(n + k - 1, 2*k - 1).
a(n) = (2/5)*F(2*n) + (1/5)*n*L(2*n), where F(k) are the Fibonacci numbers (F(0)=0, F(1)=1) and L(k) are the Lucas numbers (L(0) = 2, L(1) = 1). - Emeric Deutsch, Jul 21 2008
a(0) = 1, a(1) = 4, a(2) = 14, a(3) = 46, a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). - Harvey P. Dale, Aug 01 2011
a(n) = ((3 - sqrt(5))^n*(5*n - 2*sqrt(5)) + (3 + sqrt(5))^n*(5*n + 2*sqrt(5)))/ (25*2^n). - Peter Luschny, Mar 07 2022
E.g.f.: exp(3*x/2)*(15*x*cosh(sqrt(5)*x/2) + sqrt(5)*(4 + 5*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025

Extensions

Name clarified using a comment of the author by Peter Luschny, Aug 03 2019

A060921 Bisection of Fibonacci triangle A037027: odd-indexed members of column sequences of A037027 (not counting leading zeros).

Original entry on oeis.org

1, 3, 2, 8, 10, 3, 21, 38, 22, 4, 55, 130, 111, 40, 5, 144, 420, 474, 256, 65, 6, 377, 1308, 1836, 1324, 511, 98, 7, 987, 3970, 6666, 6020, 3130, 924, 140, 8, 2584, 11822, 23109, 25088, 16435, 6588, 1554, 192, 9
Offset: 0

Views

Author

Wolfdieter Lang, Apr 20 2001

Keywords

Comments

Row sums give A002450. Column sequences (without leading zeros) give for m=0..5: A001906, 2*A001870, A061182, 4*A061183, A061184, 2*A061185.
Companion triangle (odd-indexed members) A060920.

Examples

			{1}; {3,2}; {8,10,3}; {21,38,22,4}; ...; pFo(2,x) = 2*(1-x).
		

Formula

a(n, m) = A037027(2*n+1-m, m).
a(n, m) = (2*(n-m+1)*A060920(n, m-1)+2*(2*n+1)*a(n-1, m-1))/(5*m), n >= m>0; a(n, 0) := S(n, 3)=A001906(n+1) with Chebyshev's S(n, x) polynomials A049310; else 0.
G.f. for column m >= 0: x^m*pFo(m+1, x)/(1-3*x+x^2)^(m+1), where pFo(n, x) := Sum_{m=0..n-1} A061177(n-1, m)*x^m (row polynomials of signed triangle A061177).
G.f.: 1/(1 - (3+2*y)*x + (1+y)^2*x^2). - Vladeta Jovovic, Oct 11 2003

A054444 Even-indexed terms of A001629(n), n >= 2, (Fibonacci convolution).

Original entry on oeis.org

1, 5, 20, 71, 235, 744, 2285, 6865, 20284, 59155, 170711, 488400, 1387225, 3916061, 10996580, 30737759, 85573315, 237387960, 656451269, 1810142185, 4978643596, 13661617195, 37409025455, 102238082976, 278920277425, 759695287349
Offset: 0

Views

Author

Wolfdieter Lang, Apr 07 2000

Keywords

Comments

8*a(n) is the number of Boolean (equivalently, lattice, modular lattice, distributive lattice) intervals of the form [s,w] in the Bruhat order on S_n, where s is a simple reflection. - Bridget Tenner, Jan 16 2020

Crossrefs

Programs

  • PARI
    a(n) = ((2*n+1)*fibonacci(2*(n+1))+4*(n+1)*fibonacci(2*n+1))/5; \\ Jinyuan Wang, Jul 28 2019

Formula

a(n) = ((2*n+1)*F(2*(n+1)) + 4*(n+1)*F(2*n+1))/5, with F(n) = A000045(n) (Fibonacci numbers).
a(n) = A060920(n+1, 1).
G.f.: (1 - x + x^2)/(1 - 3*x + x^2)^2.
a(n) = Sum_{k=1..n+1} k*binomial(2*n-2*k+2, k). - Emeric Deutsch, Jun 11 2003
a(n) ~ n*(3 + sqrt(5))^(1+n)*2^(-n)/5. - Stefano Spezia, Mar 29 2022
E.g.f.: exp(3*x/2)*(5*(5 + 14*x)*cosh(sqrt(5)*x/2) + sqrt(5)*(7 + 30*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025

A024458 a(n) = s(1)*s(n) + s(2)*s(n-1) + ... + s(k)*s(n+1-k), where k = floor((n+1)/2), s = (Fibonacci numbers).

Original entry on oeis.org

1, 1, 3, 5, 12, 19, 40, 65, 130, 210, 404, 654, 1227, 1985, 3653, 5911, 10720, 17345, 31090, 50305, 89316, 144516, 254568, 411900, 720757, 1166209, 2029095, 3283145, 5684340, 9197455, 15855964, 25655489, 44061862, 71293590, 122032508
Offset: 1

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Author

Keywords

Comments

From Wolfdieter Lang, Jan 02 2012: (Start)
chat(n):=a(n+1), n>=0, is the half-convolution of the sequence A000045(n+1), n>=0, with itself. For the definition of half-convolution see a comment on A201204, where also the rule to find the o.g.f. is given. Here the o.g.f. is obtained from (U(x)^2 + U2(x^2))/2 with U(x)=1/(1-x-x^2), the o.g.f. of A000045(n+1), n>=0, and U2(x):=(1-x)/((1+x)*(1-3*x+x^2)) the o.g.f. of A007598(n+1), n>=0. This coincides with the o.g.f. given below in the formula section after x has been divided.
For the bisection of this half-convolution see A027991(n+1) and A001870(n), n>=0.
(End)

Crossrefs

Programs

  • Magma
    [(&+[Fibonacci(j+1)*Fibonacci(n-j): j in [0..Floor((n-1)/2)]]): n in [1..50]]; // G. C. Greubel, Apr 06 2022
    
  • Mathematica
    Table[((13-5(-1)^n +10n)Fibonacci[n] + (1-(-1)^n +2n)LucasL[n] +8Sin[Pi*n/2])/40, {n, 30}] (* Vladimir Reshetnikov, Oct 03 2016 *)
    LinearRecurrence[{1,3,-2,0,-2,-3,1,1},{1,1,3,5,12,19,40,65},40] (* Harvey P. Dale, Mar 02 2023 *)
  • SageMath
    def A024458(n): return sum(fibonacci(j+1)*fibonacci(n-j) for j in (0..((n-1)//2)) )
    [A024458(n) for n in (1..50)] # G. C. Greubel, Apr 06 2022

Formula

G.f.: x*(1-x^2+x^3)/((1+x^2)*(1+x-x^2)*(1-x-x^2)^2).
a(n) = ((13 - 5*(-1)^n + 10*n)*A000045(n) + (1 - (-1)^n + 2*n)*A000032(n) + 8*sin(Pi*n/2))/40. - Vladimir Reshetnikov, Oct 03 2016
From G. C. Greubel, Apr 06 2022: (Start)
a(2*n) = (1/5)*(n*Lucas(2*n+1) + Fibonacci(2*n)), n >= 1.
a(2*n+1) = (1/5)*((-1)^n + (n+1)*Lucas(2*n+2) + Fibonacci(2*n+1)), n >= 0.
a(n) = Sum_{j=0..floor((n-1)/2)} fibonacci(j+1)*Fibonacci(n-j). (End)

Extensions

More terms from James Sellers, May 03 2000

A094565 Triangle read by rows: binary products of Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 13, 15, 16, 21, 34, 39, 40, 42, 55, 89, 102, 104, 105, 110, 144, 233, 267, 272, 273, 275, 288, 377, 610, 699, 712, 714, 715, 720, 754, 987, 1597, 1830, 1864, 1869, 1870, 1872, 1885, 1974, 2584, 4181, 4791, 4880, 4893, 4895, 4896, 4901, 4935, 5168, 6765
Offset: 1

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Author

Clark Kimberling, May 12 2004

Keywords

Comments

Row n consists of n numbers, first F(2n-1) and last F(2n).
Central numbers: (1,6,40,273,...) = A081016.
Row sums: A001870.
Alternating row sums: 1,1,7,7,48,48,329,329; the sequence b=(1,7,48,329,...) is A004187, given by b(n)=F(4n+2)-b(n-1) for n>=2, with b(1)=1.
In each row, the difference between neighboring terms is a Fibonacci number.

Examples

			Triangle begins:
   1;
   2,   3;
   5,   6    8;
  13,  15,  16,  21;
  34,  39,  40,  42,  55;
  89, 102, 104, 105, 110, 144; ...
		

Crossrefs

Programs

  • GAP
    Flat(List([1..12], n-> List([1..n], k-> Fibonacci(2*k)*Fibonacci(2*n-2*k+1) ))); # G. C. Greubel, Jul 15 2019
  • Magma
    [Fibonacci(2*k)*Fibonacci(2*n-2*k+1): k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 15 2019
    
  • Mathematica
    Table[Fibonacci[2*k]*Fibonacci[2*n-2*k+1], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Jul 15 2019 *)
  • PARI
    row(n) = vector(n, k, fibonacci(2*k)*fibonacci(2*n-2*k+1));
    tabl(nn) = for(n=1, nn, print(row(n))); \\ Michel Marcus, May 03 2016
    
  • Sage
    [[fibonacci(2*k)*fibonacci(2*n-2*k+1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 15 2019
    

Formula

Row n: F(2)F(2n-1), F(4)F(2n-3), ..., F(2n)F(1).

A129722 Number of 0's in even position in all Fibonacci binary words of length n. A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

0, 0, 1, 1, 5, 6, 19, 25, 65, 90, 210, 300, 654, 954, 1985, 2939, 5911, 8850, 17345, 26195, 50305, 76500, 144516, 221016, 411900, 632916, 1166209, 1799125, 3283145, 5082270, 9197455, 14279725, 25655489, 39935214, 71293590, 111228804, 197452746, 308681550
Offset: 0

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Author

Emeric Deutsch, May 13 2007

Keywords

Examples

			a(4)=5 because in 1110', 1111, 1101, 10'10', 10'11, 0110', 0111 and 0101 one has altogether five 0's in even position (marked by ').
		

Crossrefs

Programs

  • Maple
    G:=z^2/(1-z-z^2)^2/(1+z-z^2): Gser:=series(G,z=0,45): seq(coeff(Gser,z,n),n=0..42);
  • Mathematica
    CoefficientList[Series[x^2/((1 + x - x^2)*(1 - x - x^2)^2), {x,0,50}], x] (* G. C. Greubel, Mar 09 2017 *)
    LinearRecurrence[{1,4,-3,-4,1,1},{0,0,1,1,5,6},40] (* Harvey P. Dale, Apr 02 2018 *)
  • PARI
    x='x+O('x^50); concat([0,0], Vec(x^2/((1 + x - x^2)*(1 - x - x^2)^2))) \\ G. C. Greubel, Mar 09 2017

Formula

G.f.: z^2/( (1+z-z^2)*(1-z-z^2)^2 ).
a(2*n+1) = a(2*n) + a(2*n-1) (n>=1).
a(2*n+1) = A001871(n-1) (n>=1).
a(2*n) = A129720(2*n) = A001870(n-1).
a(n) = Sum_{ k=0..floor(n/2)} k*A129721(n,k).
a(n) = F(n)*(n+1)/5 + F(n+1)*(2*n - 5 + 5*(-1)^n)/20 where F = A000045. - Greg Dresden, Jan 01 2021

A238846 Expansion of (1-2*x)/(1-3*x+x^2)^2.

Original entry on oeis.org

1, 4, 13, 40, 120, 354, 1031, 2972, 8495, 24110, 68016, 190884, 533293, 1484020, 4115185, 11375764, 31358376, 86223942, 236540915, 647556620, 1769374931, 4826148314, 13142564448, 35736448200, 97037995225, 263156279524, 712795854421, 1928547574912, 5212430732760
Offset: 0

Views

Author

Philippe Deléham, Mar 05 2014

Keywords

Comments

Convolution of 1, 1, 2, 5, 13, ... (A001519(n)) with 1, 3, 8, 21, 55, ... (A001906(n+1)).

Examples

			a(0) = 1*1 = 1;
a(1) = 1*3 + 1*1 = 4;
a(2) = 1*8 + 1*3 + 2*1 = 13;
a(3) = 1*21 + 1*8 + 2*3 + 5*1 = 40;
a(4) = 1*55 + 1*21 + 2*8 + 5*3 + 13*1 = 120; etc. (from first recurrence formula).
a(0) = 3*0 - 0 + 1 = 1;
a(1) = 3*1 - 0 + 1 = 4;
a(2) = 3*4 - 1 + 2 = 13;
a(3) = 3*13 - 4 + 5 = 40;
a(4) = 3*40 - 13 + 13 = 120; etc (from second recurrence formula).
G.f. = 1 + 4*x + 13*x^2 + 40*x^3 + 120*x^4 + 354*x^5 + 1031*x^6 + ... - _Michael Somos_, Nov 23 2021
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{6, -11, 6, -1}, {1, 4, 13, 40}, 30] (* Bruno Berselli, Mar 06 2014 *)
    a[ n_] := If[n < 0, SeriesCoefficient[ x^3*(2 - x)/(1 - 3*x + x^2)^2, {x, 0, -n}], SeriesCoefficient[ (1 - 2*x)/(1 - 3*x + x^2)^2, {x, 0, n}]]; (* Michael Somos, Nov 23 2021 *)
  • PARI
    {a(n) = if(n<0, polcoeff( x^3*(2-x)/(1-3*x+x^2)^2 + x*O(x^-n), -n), polcoeff( (1-2*x)/(1-3*x+x^2)^2 + x*O(x^n), n))}; /* Michael Somos, Nov 23 2021 */

Formula

a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4) for n>3, a(0)=1, a(1)=4, a(2)=13, a(3)=40.
a(n) = 3*a(n-1) - a(n-2) + A001519(n) for n>1, a(0)=1, a(1)=4.
a(n) = A238731(n+1, 1).
a(n) = -A124037(n+1, 1).
a(n) = (-1)^n*A126126(n+1, 1).
a(n) = ( (3 + sqrt(5))^(1+n)*(8 - (1 - sqrt(5))*(13 + 5*n)) + (3 - sqrt(5))^(1+n)*(8 - (1 + sqrt(5))*(13 + 5*n)) ) / (25*2^(2+n)). - Bruno Berselli, Mar 06 2014
From Philippe Deléham, Mar 06 2014: (Start)
a(n) = 2*A001870(n) - A001871(n).
a(n) = A197649(n+1) - 3*A001871(n-1).
a(n) = A001871(n) - 2*A001871(n-1). (End)
0 = 2 + a(n)*(a(n+1) - a(n+3)) + a(n+1)*(-6*a(n+1) + 12*a(n+2)) + a(n+2)*(-6*a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Nov 23 2021
E.g.f.: exp(3*x/2)*(5*(5 + 4*x)*cosh(sqrt(5)*x/2) + sqrt(5)*(17 + 10*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025

A213777 Rectangular array: (row n) = b**c, where b(h) = F(h), c(h) = F(h+1), F=A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

1, 3, 2, 7, 5, 3, 15, 12, 8, 5, 30, 25, 19, 13, 8, 58, 50, 40, 31, 21, 13, 109, 96, 80, 65, 50, 34, 21, 201, 180, 154, 130, 105, 81, 55, 34, 365, 331, 289, 250, 210, 170, 131, 89, 55, 655, 600, 532, 469, 404, 340, 275, 212, 144, 89, 1164, 1075, 965, 863
Offset: 1

Views

Author

Clark Kimberling, Jun 21 2012

Keywords

Comments

Principal diagonal: A001870
Antidiagonal sums: A152881
row 1, (1,1,2,3,5,8,...)**(1,2,3,5,8,13,...): A023610(k-1)
row 2, (1,1,2,3,5,8,...)**(2,3,5,8,13,21,...): A067331(k-1)
row 3, (1,1,2,3,5,8,...)**(3,5,8,13,21,34,...)
For a guide to related arrays, see A213500.

Examples

			Northwest corner (the array is read by falling antidiagonals):
1....3....7....15....30....58
2....5....12...25....50....96
3....8....19...40....80....154
5....13...31...65....130...250
8....21...50...105...210...404
13...34...81...170...340...654
		

Crossrefs

Cf. A213500.

Programs

  • Mathematica
    b[n_] := Fibonacci[n]; c[n_] := Fibonacci[n + 1];
    t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
    TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
    Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
    r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213777 *)
    Table[t[n, n], {n, 1, 40}] (* A001870 *)
    s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
    Table[s[n], {n, 1, 50}] (* A152881 *)

Formula

T(n,k) = 2*T(n,k-1) + T(n,k-2) - 2*T(n,k-3) - T(n,k-4).
G.f. for row n: f(x)/g(x), where f(x) = F(n-1) + F(n-2)*x and g(x) = (1 - x - x^2)^2.
T(n,k) = (k*Lucas(n+k+1) + Lucas(n)*Fibonacci(k))/5. - Ehren Metcalfe, Jul 10 2019
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